Câu a: 

TH1: \(x+\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow x+2x-1=2\Leftrightarrow x=1\)

TH2:\(x+\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow x-2x+1=2\Leftrightarrow x=-1\)

ĐK:  \(x\le2\)

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

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

\(\Leftrightarrow\)\(4x^2-4x+1=4-4x+x^2\)

\(\Leftrightarrow\)\(3x^2=3\)

\(\Leftrightarrow\)\(x=\pm1\)(t/m)

Vậy...

\(1-\sqrt{4x^2-20x+25}=0\)

\(\Leftrightarrow\)\(\sqrt{4x^2-20x+25}=1\)

\(\Leftrightarrow\)\(4x^2-20x+24=0\)

\(\Leftrightarrow\)\(x^2-5x+6=0\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

Vậy...

a) \(\sqrt{4x^2-4x+1}=2-x\left(x\le2\right)\)

<=> \(4x^2-4x+1=\left(2-x\right)^2\)

<=> \(3x^2-3=0\)
<=> \(x_1=1_{ }\)(TM)          ;       \(x_2=-1_{ }_{ }\)(TM)

b)\(1-\sqrt{4x^4-20x^2+25}=0\)

Đặt x² = a ( a\(\ge\)0)

=> pt có dạng :

\(1-\sqrt{4a^2-20a+25}=0\)

<=> \(\sqrt{4a^2-20a+25}=1\)

<=> \(4a^2-20a+25=1\)

=> \(a_1=3\)\(\left(TM\right)\)                  ;      \(a_2=2\left(TM\right)\)

\(a_1=3\Rightarrow x=\pm\sqrt{3}\)

\(a_2=2\Rightarrow x=\pm\sqrt{2}\)

ĐKXĐ : \(4x^2+5x+1\ge0\Leftrightarrow\left(4x+1\right)\left(x+1\right)\ge0\Rightarrow\orbr{\begin{cases}x\le-1\\x\ge-\frac{1}{4}\end{cases}}\)

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

\(\Leftrightarrow\sqrt{4x^2+5x+1}-\frac{2\sqrt{7}}{3}-2\sqrt{x^2-x+1}+\frac{2\sqrt{7}}{3}-9x+3=0\)

\(\Leftrightarrow\frac{4x^2+5x+1-\frac{28}{9}}{\sqrt{4x^2+5x+1}+\frac{2\sqrt{7}}{3}}-2\left(\frac{x^2-x+1-\frac{7}{9}}{\sqrt{x^2-x+1}+\frac{\sqrt{7}}{3}}\right)+3\left(3x-1\right)=0\)

\(\Leftrightarrow\frac{4x^2+5x-\frac{19}{9}}{\sqrt{4x^2+5x+1}+\frac{2\sqrt{7}}{3}}-2.\frac{x^2-x+\frac{2}{9}}{\sqrt{x^2-x+1}+\frac{\sqrt{7}}{3}}+3\left(3x-1\right)=0\)

\(\Leftrightarrow\frac{\left(x-\frac{1}{3}\right)\left(4x+\frac{19}{3}\right)}{\sqrt{4x^2+5x+1}+\frac{2\sqrt{7}}{3}}-\frac{2\left(x-\frac{2}{3}\right)\left(x-\frac{1}{3}\right)}{\sqrt{x^2-x+1}+\frac{\sqrt{7}}{3}}+9\left(x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)\left(\frac{4x+\frac{19}{3}}{\frac{2\sqrt{7}}{3}}-\frac{2x-\frac{4}{3}}{\sqrt{x^2-x+1}+\frac{\sqrt{7}}{2}}+9\right)=0\)

\(\Rightarrow x=\frac{1}{3}\)(TMĐKXĐ)

dùng liên hợp nhé bn 

hỏi chấm

....

- giải

- giải 

- giải

=> x =1 

- bằng mấy nx thì không biết ...

\(\sqrt{8x+1}+\sqrt{46-10x}=-x^3+5x^2+4x+1<=>\sqrt{8x+1}-3+\sqrt{46-10x}-6=-x^3+5x^2+4x+1-3-6\)

\(<=> (x-1)(\frac{8}{\sqrt{8x+1}+3}-5 +x^2-4x-3-\frac{10}{\sqrt{46-10x}+6})=0\)

Xét : \((\frac{8}{\sqrt{8x+1}+3}-5 +x^2-4x-3-\frac{10}{\sqrt{46-10x}+6}) (*)\) ( với điều kiện \(\frac{23}{5}\geq x\geq- \frac{1}{8}\))

\((*)= \frac{8-5(\sqrt{8x+1}+3)}{\sqrt{8x+1}+3} +(x^2-4x-3)-\frac{10}{\sqrt{46-10x}+6})\)

\(= \frac{-7-5(\sqrt{8x+1})}{\sqrt{8x+1}+3} +(x^2-4x-3)-\frac{10}{\sqrt{46-10x}+6}) <0\)

\(=> x=1\)

giúp với mọi người

Đk : \(x\ge\frac{3}{4}\)

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

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

\(\Rightarrow\left(x-2\right)^2=\left(\sqrt{4x-3}\right)^2\)

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

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

\(\Delta=36\Rightarrow\sqrt{\Delta}=6\)

\(\Rightarrow\)Phương trình có hai nghiệm phân biệt :

\(x_1=1\left(tm\right)\)

\(x_2=7\left(tm\right)\)

\(\sqrt{5x^2-2x\sqrt{5}+1}=\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow\)\(5x^2-2x\sqrt{5}+1=6-2\sqrt{5}\)

\(\Leftrightarrow\)\(\left(x\sqrt{5}-1\right)^2=\left(\sqrt{5}-1\right)^2\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x\sqrt{5}-1=\sqrt{5}-1\\x\sqrt{5}-1=1-\sqrt{5}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=\frac{2-\sqrt{5}}{\sqrt{5}}\end{cases}}\)

Vậy...

ĐK:  \(x\ge\frac{3}{4}\)

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

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

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

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

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

đến đây tự làm

Mình làm câu còn lại nha :

ĐK : \(x\ge\frac{\sqrt{5}}{5}\)

\(\sqrt{\left(x\sqrt{5}-1\right)^2}=\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(x\sqrt{5}-1=\sqrt{5}-1\)

\(x\sqrt{5}=\sqrt{5}\)

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

\(\sqrt{4x^2-4x+1}=3-x\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\\ \Leftrightarrow2x-1=3-x\\ \Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\\ \sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\right)\left(x\ge-1\right)\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{9}+1+\sqrt{4}\right)=2\left(x+1\right)\\ \Leftrightarrow6\sqrt{x+1}=2\left(x+1\right)\\ \Leftrightarrow3\sqrt{x+1}=x+1\\ \Leftrightarrow\sqrt{x+1}\left(3-\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{x+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)

a, ĐK: \(x\in R\)

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

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

\(\Leftrightarrow\left|2x-1\right|=3-x\)

TH1: \(\left\{{}\begin{matrix}2x-1\ge0\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{4}{3}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\1-2x=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x=-2\end{matrix}\right.\Leftrightarrow x=-2\)

b, ĐK: \(x\ge-1\)

\(\sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\text{​​}\right)\)

\(\Leftrightarrow3\sqrt{x+1}+\sqrt{x+1}-2\sqrt{x+1}=2\left(x+1\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

a) pt<=> \(\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)

     <=>\(\left|x-2\right|+\left|x-3\right|=1\)

đến đây chia 3 trường hợp để phá trị tuyệt đối là ra 

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

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

câu này cũng tương tự câu a nha

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`