Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

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

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

điều kiện : \(x\ge1\)

ta có : \(P=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

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

\(=\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\)

\(\Rightarrow\left[{}\begin{matrix}P=2\sqrt{x-1}\left(x\ge2\right)\\P=2\left(1\le x< 2\right)\end{matrix}\right.\)

vậy .....................................................................................................

tks ạ!

a) \(\sqrt{4\left(1+6x+9x^2\right)^2}\) = \(\sqrt{\left(2\left(1+6x+9x^2\right)\right)^2}\)

= \(\sqrt{\left(2\left(1-6\sqrt{2}+18\right)\right)^2}\) = \(2\left(1-6\sqrt{2}+18\right)\) = \(2\left(3\sqrt{2}-1\right)^2\)

= \(21,029\)

b) \(\sqrt{9a^2\left(b^2+4-4b\right)}\) = \(\sqrt{\left(3a\left(b-2\right)\right)^2}\) = \(\sqrt{\left(-6\left(-\sqrt{3}-2\right)\right)^2}\)

= \(\sqrt{\left(6\sqrt{3}+12\right)^2}\) = \(6\sqrt{3}+12\) = \(22,392\)

a) = 225 

b)  49

c) = 1 

d) 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 

k nha

a) \(\sqrt{x}=15\)

=> x = 15²

 => x = 225

b) \(2\sqrt{x}=14\)

<=> \(\sqrt{x}=7\)

=> x = 7²

=> x = 49

c) \(\sqrt{x}< \sqrt{2}\)

<=> x < 2

mà \(x\ge0\)

=> x= {0;1}

d) \(\sqrt{2x}< 4\)

=> 2x < 16

<=> x < 8

mà \(x\ge0\)

=> x = {0;1;2;3;4;5;6;7}

ok mk nhé!!!!!! 53654645756876969251353253434645655435436464556756252345345634

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

Em mới lớp 7 nên em chỉ làm những câu em biết thôi nhé:

\(a,\sqrt{x}=15\)

\(\Rightarrow x=15^2\)

\(\Rightarrow x=225\)

\(b,2\sqrt{x}=14\)

\(\sqrt{x}=14:2\)

\(\sqrt{x}=7\)

\(x=7^2\)

\(x=49\)

\(c,\sqrt{x}< \sqrt{2}\)

\(\Rightarrow x< 2\)

Còn ý d em không biết làm ạ ! 

\(a)\sqrt{x}=15\)

Vì \(x\ge0\) nên bình phương hai vế ta được:

\(x=15^2\Leftrightarrow x=225\)

Vậy \(x=225\)

\(b)2\sqrt{x}=14\Leftrightarrow\sqrt{x}=7\)

Vì  \(x\ge0\) nên bình phương hai vế ta được:

\(x=7^2\Leftrightarrow x=49\)

Vậy \(x=49\)

\(c)\sqrt{x}< \sqrt{2}\)

Vì \(x\ge0\) nên bình phương hai vế ta được: \(x< 2\)

Vậy \(0\le x\le2\)

\(d)\sqrt{2x}< 4\)

Vì \(x\ge0\)nên bình phương hai vế ta được:

\(2x< 16\Leftrightarrow x< 8\)

Vậy \(0\le x< 8\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

\(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{x-9}\)

\(=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

b) Ta có: \(x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)

\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{\sqrt{3}-\left|\sqrt{3}-1\right|}\)

\(=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\)( thỏa mãn ĐKXĐ )

Thay \(x=1\)vào M ta được:

\(M=\frac{3\sqrt{1}}{\sqrt{1}-3}=\frac{3}{1-3}=\frac{-3}{2}\)

c) \(M=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\sqrt{x}-9+9}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)

Vì \(x\inℕ\)\(\Rightarrow\)Để M là số tự nhiên thì \(\frac{9}{\sqrt{x}-3}\inℕ\)

\(\Rightarrow9⋮\left(\sqrt{x}-3\right)\)\(\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\)(1)

Vì \(x\ge0\)\(\Rightarrow\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}-3\ge-3\)(2)

Từ (1) và (2) \(\Rightarrow\sqrt{x}-3\in\left\{-3;-1;1;3;9\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\)\(\Rightarrow x\in\left\{0;4;16;36;144\right\}\)( thỏa mãn ĐKXĐ )

Thử lại với \(x=4\)ta thấy M không là số tự nhiên

Vậy \(x\in\left\{0;16;36;144\right\}\)

\(\sqrt{-3x^3+5x+14}+\sqrt{-5x^3+6x+28}=\left(4-2x-x^2\right)\sqrt{2-x}\) (ĐKXĐ: \(x\in R,x\le2\))

\(\Leftrightarrow\sqrt{\left(2-x\right)\left(3x^2+6x+7\right)}+\sqrt{\left(2-x\right)\left(5x^2+10x+14\right)}-\left(4-2x-x^2\right)\sqrt{2-x}=0\)

\(\Leftrightarrow\sqrt{2-x}\left(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}-4+2x+x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\left(1\right)\end{cases}}\)

Pt \(\left(1\right)\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-\left(x+1\right)^2+5\left(2\right)\)

Ta có: \(\left(x+1\right)^2\ge0\Rightarrow\sqrt{2\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

Tương tự: \(\sqrt{5\left(x+1\right)^2+9}\ge3\). Từ đó: \(VT_{\left(2\right)}\)\(\ge2+3=5\)

Mà \(VP_{\left(2\right)}=-\left(x+1\right)^2+5\le5\) nên dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)(tm)

Vậy tập nghiệm của pt cho là \(S=\left\{2;-1\right\}.\)