\(\sqrt{x+1}-4x^2=\sqrt{3x}-1\left(x\ge0\right)\left(1\right)\)

\(\Leftrightarrow-4x^2+1+\sqrt{x+1}-\dfrac{\sqrt{6}}{2}=\sqrt{3x}-\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow-\left(2x-1\right)\left(2x+1\right)+\dfrac{x+1-\dfrac{3}{2}}{\sqrt{x+1}+\dfrac{\sqrt{6}}{2}}=\dfrac{3x-\dfrac{3}{2}}{\sqrt{3x}+\dfrac{\sqrt{6}}{2}}\)

\(\Leftrightarrow-4\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)+\dfrac{x-\dfrac{1}{2}}{\sqrt{x+1}+\dfrac{\sqrt{6}}{2}}-\dfrac{3\left(x-\dfrac{1}{2}\right)}{\sqrt{3x}+\dfrac{\sqrt{6}}{2}}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)\left[-4\left(x+\dfrac{1}{2}\right)+\dfrac{1}{\sqrt{x+1}+\dfrac{\sqrt{6}}{2}}-\dfrac{3}{\sqrt{3x}+\dfrac{\sqrt{6}}{2}}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\-4\left(x+\dfrac{1}{2}\right)+\dfrac{1}{\sqrt{x+1}+\dfrac{\sqrt{6}}{2}}-\dfrac{3}{\sqrt{3x}+\dfrac{\sqrt{6}}{2}}=0\left(2\right)\end{matrix}\right.\)

\(\left(x\ge0\right)\Rightarrow\left(2\right)< 0\Rightarrow\left(2\right)vô\) \(nghiệm\)

\(\Rightarrow S=\left\{\dfrac{1}{2}\right\}\)

\(\)

ĐKXĐ: x>=-1

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

=>\(4x^2-2\sqrt{x+1}-x-2=0\)

=>\(4x^2+3x-4x-3+1-2\sqrt{x+1}=0\)

=>\(\left(4x+3\right)\left(x-1\right)+1-\sqrt{4x+4}=0\)

=>\(\left(4x+3\right)\left(x-1\right)+\dfrac{1-4x-4}{1+\sqrt{4x+4}}=0\)

=>\(\left(4x+3\right)\left(x-1\right)-\dfrac{4x+3}{1+\sqrt{4x+4}}=0\)

=>\(\left(4x+3\right)\left(x-1-\dfrac{1}{1+\sqrt{4x+4}}\right)=0\)

=>4x+3=0

=>x=-3/4(nhận)

a)

\(x^2-4\sqrt{15}x+19=0\\ < =>x^2-4\sqrt{15}x+60-41=0\\ < =>\left(x-2\sqrt{15}\right)^2-41=0\\ < =>\left(x-2\sqrt{15}-\sqrt{41}\right)\left(x-2\sqrt{15}+\sqrt{41}\right)=0\\ < =>\left[{}\begin{matrix}x-2\sqrt{15}-\sqrt{41}=0\\x-2\sqrt{15}+\sqrt{41}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=2\sqrt{15}+\sqrt{41}\\x=2\sqrt{15}-\sqrt{41}\end{matrix}\right.\)

b)

\(4x^2+4\sqrt{5}x+5=0\\ < =>\left(2x+\sqrt{5}\right)^2=0\\ < =>2x+\sqrt{5}=0\\ < =>2x=-\sqrt{5}\\ < =>-\dfrac{\sqrt{5}}{2}\)

a: Δ=(4căn 15)^2-4*1*19=164>0

Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{5}-2\sqrt{41}}{2}=2\sqrt{5}-\sqrt{41}\\x_2=2\sqrt{5}+\sqrt{41}\end{matrix}\right.\)

b: \(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot\sqrt{5}+5=0\)

=>(2x+căn 5)^2=0

=>2x+căn 5=0

=>x=-1/2*căn 5

bạn ktra lại đề nhé 

\(đk:2\le x\le4\) \(pt\Leftrightarrow\sqrt{x-2}+\sqrt{4-x}=x-2\sqrt{3x}+5\)

\(\left(\sqrt{x-2}+\sqrt{4-x}\right)^2\le2\left(x-2+4-x\right)=4\Rightarrow\sqrt{x-2}+\sqrt{4-x}\le2\)

\(x-2\sqrt{3x}+5=\sqrt{x}^2-2\sqrt{3x}+5=\sqrt{x}^2-2\sqrt{3x}+3+2=\left(\sqrt{x}-\sqrt{3}\right)^2+2\ge2\)

\(\Rightarrow\left\{{}\begin{matrix}VT\le2\\VP\ge2\end{matrix}\right.\) dấu"=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{4-x}=2\\\left(\sqrt{x}-\sqrt{3}\right)^2+2=2\end{matrix}\right.\)

\(\Leftrightarrow x=3\left(tm\right)\)

(ủa đề sai chỗ nào ta?)

Câu 4:

Giả sử điều cần chứng minh là đúng

\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:

\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)

Vậy điều cần chứng minh là đúng

2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)

⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)

⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)

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

⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)

⇔ x = 5

Vậy S = {5}

Lời giải:

Đặt $\sqrt[3]{x^2+3x-5}=a; \sqrt[3]{x+2}=b$. Khi đó pt đã cho tương đương với:

$a+b=\sqrt[3]{a^3+b^3-1}+1$

$\Leftrightarrow a+b-1=\sqrt[3]{a^3+b^3-1}$

$\Leftrightarrow (a+b-1)^3=a^3+b^3-1$

$\Leftrightarrow (a+b)^3-3(a+b)^2+3(a+b)-1=a^3+b^3-1$

$\Leftrightarrow 3ab(a+b)-3(a+b)^2+3(a+b)=0$

$\Leftrightarrow ab(a+b)-(a+b)^2+(a+b)=0$

$\Leftrightarrow (a+b)(ab-a-b+1)=0$

$\Leftrightarrow (a+b)(a-1)(b-1)=0$

Nếu $a+b=0\Leftrightarrow \sqrt[3]{x^2+3x-5}=-\sqrt[3]{x+2}$

$\Leftrightarrow x^2+3x-5=-(x+2)$

$\Leftrightarrow x^2+4x-3=0$

$\Leftrightarrow x=-2\pm \sqrt{7}$

Nếu $a-1=0\Leftrightarrow \sqrt[3]{x^2+3x-5}=1$

$\Leftrightarrow x^2+3x-6=0$

$\Leftrightarrow x=\frac{-3\pm \sqrt{33}}{2}$

Nếu $b-1=0\Leftrightarrow \sqrt[3]{x+2}=1$

$\Leftrightarrow x=-1$

\(ĐKXĐ:\dfrac{74}{27}\le x\le\dfrac{10}{3}\)

PT đã cho tương đương với:

\(4-3\sqrt{10-3x}=x^2-4x+4\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-3\left(1-\sqrt{10-3x}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-3.\dfrac{3\left(x-3\right)}{1+\sqrt{10-3x}}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1-\dfrac{9}{1+\sqrt{10-3x}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\left(1\right)\\\left(x-1\right)\left(1+\sqrt{10-3x}\right)=9\left(2\right)\end{matrix}\right.\)

Ta có:

\(pt\left(1\right)\Leftrightarrow x=3\left(tm\right)\)

\(pt\left(2\right):\left(x-1\right)\left(1+\sqrt{10-3x}\right)=9\)

mà \(\left(x-1\right)\left(1+\sqrt{10-3x}\right)\le\dfrac{7}{3}.\dfrac{7}{3}\) nên \(pt\left(2\right)\) vô nghiệm

Vậy pt đã cho có tập nghiệm \(S=\left\{3\right\}\)

ĐKXĐ: \(-\dfrac{1}{3}\le x\le4\)

\(\Leftrightarrow x+5=\sqrt{3x+1}+2\sqrt{4-x}\)

Ta có: 

\(VP=1.\sqrt{3x+1}+2.\sqrt{4-x}\le\dfrac{1}{2}\left(1+3x+1\right)+\dfrac{1}{2}\left(4+4-x\right)=x+5\)

\(\Rightarrow VP\le VT\)

Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\sqrt{3x+1}=1\\\sqrt{4-x}=2\end{matrix}\right.\) \(\Leftrightarrow x=0\)