Bài 1:

a) \(2\left(x-\sqrt{12}\right)^2=6\Rightarrow\left(x-\sqrt{12}\right)^2=3\)

TH1l \(x-\sqrt{12}=\sqrt{3}\Rightarrow x=\sqrt{3}+\sqrt{12}=3\sqrt{3}\)

TH2: \(x-\sqrt{12}=-\sqrt{3}\Rightarrow x=-\sqrt{3}+\sqrt{12}=\sqrt{3}\)

b)  \(2x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\2\sqrt{x}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

c) \(|2x+\sqrt{\frac{9}{16}}|-x=\left(\frac{1}{\sqrt{2}}\right)^2\Leftrightarrow\left|2x+\frac{3}{4}\right|-x=\frac{1}{2}\)

TH1: \(2x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{8}\)

Ta có \(2x+\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow x=-\frac{1}{4}\left(tm\right)\)

TH2: \(x< -\frac{3}{8}\)

Ta có \(-2x-\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow-3x=\frac{5}{4}\Leftrightarrow x=-\frac{5}{12}\left(tm\right)\)

Bài 2:  Để \(A=\frac{2\sqrt{x}+3}{\sqrt{x}-2}\) là số nguyên thì \(\frac{2\sqrt{x}+3}{\sqrt{x}-2}\in Z\)

Ta có \(\frac{2\left(\sqrt{x}-2\right)+7}{\sqrt{x}-2}=2+\frac{7}{\sqrt{x}-2}\)

Để \(\frac{2\sqrt{x}+3}{\sqrt{x}-2}\in Z\) thì \(\frac{7}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2\inƯ\left(7\right)\)

Do \(\sqrt{x}-2\ge-2\Rightarrow\sqrt{x}-2\in\left\{-1;1;7\right\}\)

\(\Rightarrow x\in\left\{1;9;81\right\}\)

 Bài 1 :

\(2\left(x-\sqrt{12}\right)^2=6\)

\(\Rightarrow\left(x-\sqrt{12}\right)^2=6:2=3\)

\(\Rightarrow x-\sqrt{12}=\sqrt{3}\)

\(\Rightarrow x=3\sqrt{3}\)

a)

\(3(2x-\frac{1}{2})+2(\frac{3}{8}-x)=2,75\)

\(\Leftrightarrow 6x-\frac{3}{2}+\frac{3}{4}-2x=2,75\)

\(\Leftrightarrow 4x=\frac{7}{2}\Rightarrow x=\frac{7}{8}\)

b)

\(x-\frac{1}{3}(5-3x)=1\frac{1}{2}x+5\frac{1}{2}\)

\(\Leftrightarrow x-\frac{5}{3}+x=x+\frac{1}{2}x+\frac{11}{2}\)

\(\Leftrightarrow \frac{1}{2}x=\frac{43}{6}\) \(\Rightarrow x=\frac{43}{3}\)

c) \(\sqrt{x-1}=4\Rightarrow x-1=4^2\Rightarrow x=4^2+1=17\)

d)

\(|x|-5\frac{3}{7}|-x|-\frac{3}{4}=2|x|-1\frac{1}{7}\)

\(\Leftrightarrow |x|-\frac{38}{7}|x|-\frac{3}{4}=2|x|-\frac{8}{7}\)

\(\Leftrightarrow |x|(1-\frac{38}{7}-2)=\frac{3}{4}-\frac{8}{7}\)

\(\Leftrightarrow |x|.\frac{-45}{7}=\frac{-11}{28}\)

\(\Leftrightarrow |x|=\frac{11}{180}\Rightarrow \left[\begin{matrix} x=\frac{11}{180}\\ x=-\frac{11}{180}\end{matrix}\right.\)

CÁC câu này cứ bình phương 2 vế là ra ấy mà 

Câu a:

|\(\sqrt2\) - \(x\)| = \(\sqrt2\)

\(\left[\begin{array}{l}\sqrt2-x=\sqrt2\\ \sqrt2-x=-\sqrt2\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=2\sqrt2\end{array}\right.\)

Vậy \(x\in\) {0; \(2\sqrt2\)}

Câu b:

|\(x-1\)| = \(\sqrt3\) + 2

\(\left[\begin{array}{l}x-1=\sqrt3+2\\ x-1=-\sqrt{3-2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\sqrt3+2+1\\ x=-\sqrt3-2+1\end{array}\right.\)

\(\left[\begin{array}{l}x=\sqrt3+\left(2+1\right)\\ x=-\sqrt3-\left(2-1\right)\end{array}\right.\)

\(\left[\begin{array}{l}x=\sqrt3+3\\ x=-\sqrt3-1\end{array}\right.\)

Vậy \(x\in\) {- \(\sqrt3\) - 1; \(\sqrt3\) + 3}

|2x-1|=x+3

=> 2x-1=x+3 hoặc 2x-1=-(x+3)

2x-x=1+4 2x-1=-x-3

x=5 2x+x= 1-3

3x=-2

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

|4x+7|=2x+5

=> 4x+7=2x+5

4x-2x=5-7

-2x=-2

x=1

=>4x+7=-(2x+5)

4x+7=-2x-5

4x+2x=-5-7

6x=-12

x=-2

a) \(2\sqrt{x}-10=20\left(ĐKXD:x\ge0\right)\)

\(\Leftrightarrow2\sqrt{x}=30\Leftrightarrow\sqrt{x}=15\)

\(\Leftrightarrow x=225\)

b) \(2x-\sqrt{x}=0\left(ĐKXĐ:x\ge0\right)\)

\(\Leftrightarrow2x=\sqrt{x}\Leftrightarrow4x^2=x\Leftrightarrow4x^2-x=0\Leftrightarrow x\left(4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)

Vậy ....

c) \(x+3\sqrt{x}=0\left(ĐKXĐ:x\ge0\right)\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}}\)

Vậy x = 0

d) \(\left(x-1\right)\left(x^2+1\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x\in\varnothing\end{cases}}}\)

Vậy x = 1

a.\(2\sqrt{x}=20+10\)

\(2\sqrt{x}=30\)

\(\sqrt{x}=30:2\)

\(\sqrt{x}=15\)

\(x=15^2\)

x=225

a) x=0;y=1/10

b) x=10;y=1/2

a) Vì \(x^2\ge0;\left(y-\frac{1}{10}\right)^2\ge0\)

Mà theo đề bài: \(x^2+\left(y-\frac{1}{10}\right)^2=0\)

=> \(\begin{cases}x^2=0\\\left(y-\frac{1}{10}\right)^2=0\end{cases}\) => \(\begin{cases}x=0\\y-\frac{1}{10}=0\end{cases}\) => \(\begin{cases}x=0\\y=\frac{1}{10}\end{cases}\)

Vậy \(x=0;y=\frac{1}{10}\)

b) Vì \(\left(\frac{1}{2}x-5\right)^{26}\ge0;\left(y^2-\frac{1}{4}\right)^{10}\ge0\)

Mà theo đề bài: \(\left(\frac{1}{2}x-5\right)^{26}+\left(y^2-\frac{1}{4}\right)^{10}=0\)

=> \(\begin{cases}\left(\frac{1}{2}x-5\right)^{26}=0\\\left(y^2-\frac{1}{4}\right)^{10}=0\end{cases}\)=> \(\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}\)=> \(\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\)=> \(\begin{cases}x=10\\y\in\left\{\frac{1}{2};\frac{-1}{2}\right\}\end{cases}\)

Vậy \(x=10;y\in\left\{\frac{1}{2};\frac{-1}{2}\right\}\)

Bài 1: 

a: \(\left(2x-1\right)^4=16\)

=>2x-1=2 hoặc 2x-1=-2

=>2x=3 hoặc 2x=-1

=>x=3/2 hoặc x=-1/2

b: \(\left(2x-y+7\right)^{2012}+\left|x-3\right|^{2013}< =0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x+7=y=2\cdot3+7=13\end{matrix}\right.\)

c: \(10800=2^4\cdot3^3\cdot5^2\)

mà \(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\)

nên \(\left\{{}\begin{matrix}x+2=4\\x+1=3\\x=2\end{matrix}\right.\Leftrightarrow x=2\)