a, \(\left(\sqrt{x-1}-2\right)^2+\)\(\left(\sqrt{x-1}-3\right)^2\)

xog xét 2 TH

b, bình phương 

2

GTLN : 2 dấu = xra \(2\le x\le4\)

Hà Thị Thế pạn làm ra lun giúp mjk dx k ạ

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)

4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)

\(\Leftrightarrow\left(x+3\right)\cdot\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\2\left(x^2-4x+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=3\end{matrix}\right.\)

a, ĐK: \(x\ge11\)

\(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)

\(\Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{x^2-x+11}=16\)

\(\Leftrightarrow2x+2\sqrt{x^2-x+11}=16\)

\(\Leftrightarrow x+\sqrt{x^2-x+11}=8\)

Ta thấy \(x+\sqrt{x^2-x+11}>11>\text{​​}8\)

\(\Rightarrow\) phương trình vô nghiệm.

\(a,\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\left(x\ge11\right)\\ \Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{\left(x+\sqrt{x-11}\right)\left(x-\sqrt{x-11}\right)}=16\\ \Leftrightarrow2x+2\sqrt{x^2-x+11}=16\\ \Leftrightarrow x+\sqrt{x^2-x+11}=8\\ \Leftrightarrow\sqrt{x^2-x+11}=8-x\\ \Leftrightarrow x^2-x+11=x^2-16x+64\\ \Leftrightarrow15x=53\\ \Leftrightarrow x=\dfrac{53}{15}\left(ktm\right)\)

\(b,\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\\ \Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\\ \Leftrightarrow\left|\sqrt{2x-5}-1\right|=1-\sqrt{2x-5}\\ \Leftrightarrow\sqrt{2x-5}-1\le0\\ \Leftrightarrow\sqrt{2x-5}\le1\\ \Leftrightarrow2x-5\le1\Leftrightarrow x\le\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{2}\)

a) Áp dụng bđt AM-GM có:

\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)

\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)

Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)

Vậy...

b)Đk:\(x\ge2\)

Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)

\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)

Do \(x\ge2\Rightarrow x-1>0\)

Chia cả hai vế của pt cho x-1 ta được:

\(\left(x-1\right)\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy S={2}

c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)

Thay x=3 vào pt thấy thỏa mãn

Vậy S={3}

a) Quên mất, ko áp dụng đc AM-GM, xin lỗi

Pt \(\Leftrightarrow\sqrt[3]{9-x}-2=2-\sqrt[3]{7+x}\)

\(\Leftrightarrow\dfrac{9-x-8}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{8-\left(7-x\right)}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\)

\(\Leftrightarrow\dfrac{1-x}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{1-x}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{1}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4=4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}\left(1\right)\end{matrix}\right.\)

Từ (1) \(\Leftrightarrow\sqrt[3]{\left(9-x\right)^2}-\sqrt[3]{\left(7+x\right)^2}+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)

\(\Leftrightarrow\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)\left(\sqrt[3]{9-x}+\sqrt[3]{7+x}\right)+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)

\(\Leftrightarrow\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right).4+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)

\(\Leftrightarrow\sqrt[3]{9-x}-\sqrt[3]{7+x}=0\)

\(\Leftrightarrow\sqrt[3]{9-x}=\sqrt[3]{7+x}\)\(\Leftrightarrow9-x=7+x\)

\(\Leftrightarrow x=1\)

Vậy S={1}

 ĐKXĐ:....

\(\sqrt{4-\sqrt{1-x}}=\sqrt{2-x}\)

\(\Rightarrow4-\sqrt{1-x}=2-x\)

\(\Rightarrow\sqrt{1-x}=2+x\)

\(\Rightarrow1-x=4+4x+x^2\)

\(\Rightarrow1-x-4-4-x^2=0\)

\(\Rightarrow x^2+x+7=0\)

Đến đây dễ rồi làm nốt nha bạn !

 ĐKXĐ:....

\sqrt{4-\sqrt{1-x}}=\sqrt{2-x}4−1−x​​=2−x​

\Rightarrow4-\sqrt{1-x}=2-x⇒4−1−x​=2−x

\Rightarrow\sqrt{1-x}=2+x⇒1−x​=2+x

\Rightarrow1-x=4+4x+x^2⇒1−x=4+4x+x2

\Rightarrow1-x-4-4-x^2=0⇒1−x−4−4−x2=0

\Rightarrow x^2+x+7=0⇒x2+x+7=0

Đến đây dễ rồi làm nốt nha bạn !

a) đk: \(1\le x\le5\)

 \(\sqrt[4]{5-x}+\sqrt[4]{x-1}=\sqrt{2}\)

<=> \(\left(\sqrt[4]{5-x}+\sqrt[4]{x-1}\right)^4=\sqrt{2}^4\)

<=> \(5-x+x-1+4\sqrt[4]{5-x}^3.\sqrt[4]{x-1}+6\sqrt[4]{5-x}^2.\sqrt[4]{x-1}^2+4\sqrt[4]{5-x}.\sqrt[4]{x-1}^3=4\)

<=> \(\sqrt[4]{\left(5-x\right)\left(x-1\right)}.\left(2\sqrt[4]{5-x}^2+3\sqrt[4]{5-x}.\sqrt[4]{x-1}+2\sqrt[4]{x-1}^2\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt[4]{\left(5-x\right)\left(x-1\right)}=0\left(2\right)\\2\sqrt[4]{5-x}^2+3\sqrt[4]{\left(5-x\right)\left(x-1\right)}+2\sqrt[4]{x-1}^2=0\left(1\right)\end{matrix}\right.\)

Giải (2) <=> \(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\left(tm\right)\)

Giải (1) : Đặt \(\sqrt[4]{5-x}=a;\sqrt[4]{x-1}=b\)(đk : a, b \(\ge\)0)

Khi đó, ta có: \(2a^2+3ab+2b^2=0\)

<=> 2(a² + 3/2ab + 9/16b²) + \(\dfrac{7}{8}b^2=0\)

<=> \(2\left(a+\dfrac{3}{4}b\right)^2+\dfrac{7}{8}b^2=0\)

<=> \(\left\{{}\begin{matrix}a+\dfrac{3}{4}b=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\sqrt[4]{x-1}=0\\\sqrt[4]{5-x}=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)(vô lí)

 sao cách này rắc rối quá vậy , có cách nào đơn giản hơn không?  mà pt này rõ ràng có nghiệm chứ có phải vô nghiệm đâu 

\(\sqrt{x-4\sqrt{x-1}+3}+\sqrt{x-6\sqrt{x-1}+8}=1\\ < =>\sqrt{x-1-2\sqrt{x-1}.2+4}+\sqrt{x-1-2\sqrt{x-1}.3+9}=1\\ < =>\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)ĐK: x>=1

\(< =>|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1\\ < =>\left(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\right)^2=1\\ < =>\sqrt{x-1}-2+2\left|\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}-3\right)\right|+\sqrt{x-1}-3=1\\ < =>2\sqrt{x-1}-5+2\left|x+5-5\sqrt{x-1}\right|=1\\ < =>2\left|x+5-5\sqrt{x-1}\right|=6-2\sqrt{x-1}\\ < =>\left|x+5-5\sqrt{x-1}\right|=3-\sqrt{x-1}\)

\(< =>\left[{}\begin{matrix}x+5-5\sqrt{x-1}=3-\sqrt{x-1}\left(1\right)\\x+5-5\sqrt{x-1}=\sqrt{x-1}-3\left(2\right)\end{matrix}\right.\)

Giải (1): \(x+5-5\sqrt{x-1}=3-\sqrt{x-1}\\ < =>x+2-4\sqrt{x-1}=0\\ < =>x-1-2\sqrt{x-1}.2+4=1\\ < =>\left(\sqrt{x-1}-2\right)^2=1\\ < =>\left[{}\begin{matrix}\sqrt{x-1}-2=1\\\sqrt{x-1}-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=8\\x=0\left(loại\right)\end{matrix}\right.\)

Giải (2) cũng ra x=8

a. \(\sqrt[3]{1-2x}+3=0\left(ĐK:x\le\dfrac{1}{2}\right)\)

<=> \(\sqrt[3]{1-2x}=-3\)

<=> \(1-2x=\left(-3\right)^3\)

<=> \(1-2x=-27\)

<=> \(-2x=-28\)

<=> \(x=14\left(TM\right)\)

1) \(\sqrt{x^2+1}=\sqrt{5}\)

\(\Leftrightarrow x^2+1=5\)

\(\Leftrightarrow x^2=5-1\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2=2^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\)) 

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=3+1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=\dfrac{4}{2}\)

\(\Leftrightarrow x=2\left(tm\right)\)

3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))

\(\Leftrightarrow43-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1=43-x\)

\(\Leftrightarrow x^2-x-42=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))

\(\Leftrightarrow\sqrt{4x-3}=x-2\)

\(\Leftrightarrow4x-3=\left(x-2\right)^2\)

\(\Leftrightarrow x^2-4x+4=4x-3\)

\(\Leftrightarrow x^2-8x+7=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))

\(\Leftrightarrow\sqrt{x}+3=2\sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1^2\)

\(\Leftrightarrow x=1\left(tm\right)\)

1)

\(\sqrt{x^2+1}=\sqrt{5}\\ \Leftrightarrow x^2+1=5\\ \Leftrightarrow x^2=5-1=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy PT có nghiệm `x=2` hoặc `x=-2`

2)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)

Vậy PT có nghiệm `x=2`

3)

\(ĐKXĐ:x\le43\)

PT trở thành:

\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm `x=-6` hoặc `x=7`

4)

ĐKXĐ: \(x\ge\dfrac{3}{4}\)

PT trở thành:

\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)

5) 

ĐKXĐ: \(x\ge0\)

PT trở thành:

\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)

Khi đó:

(1)\(\Leftrightarrow3t^2+8t+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)

Vậy PT vô nghiệm.