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

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

....

- 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 \(\frac{23}{5}\geq x\geq- \frac{1}{8}\))

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

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

\(=> x=1\)

Đặt x^2+3x=a

=>\(a+2=3\sqrt{a}\)

=>a-3 căn a+2=0

=>(căn a-1)(căn a-2)=0

=>a=1 hoặc a=4

=>x^2+3x=1 hoặc x^2+3x=4

=>(x+4)(x-1)=0 và x^2+3x-1=0

=>\(x\in\left\{1;-4;\dfrac{-3+\sqrt{13}}{2};\dfrac{-3-\sqrt{13}}{2}\right\}\)

\(a,ĐK:x\ge\dfrac{1}{5}\\ PT\Leftrightarrow5x-1=64\\ \Leftrightarrow x=13\left(tm\right)\\ b,ĐK:x\ge\dfrac{2}{5}\\ BPT\Leftrightarrow5x-2< 16\\ \Leftrightarrow x< \dfrac{18}{5}\\ \Leftrightarrow\dfrac{2}{5}\le x< \dfrac{18}{5}\\ c,ĐK:x\ge3\\ PT\Leftrightarrow\left|x-1\right|-\left|x-2\right|=x-3\\ \Leftrightarrow\left[{}\begin{matrix}1-x-\left(2-x\right)=x-3\left(x< 1\right)\\x-1-\left(2-x\right)=x-3\left(1\le x< 2\right)\\x-1-\left(x-2\right)=x-3\left(x\ge2\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=0\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

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 : \(x\ge\frac{\sqrt{5}}{5}\)

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

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

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

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

a.

ĐKXĐ: \(x\ne-1\)

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

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

Đặt \(\sqrt{x^2+x+2}=t>0\)

\(\Rightarrow t^2-2\left(x+1\right)t+4x=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2\\\sqrt{x^2+x+2}=2x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4\\x^2+x+2=4x^2\left(x\ge0\right)\end{matrix}\right.\)

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

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-5x+14-4\sqrt{x+1}=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)

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

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

\(\Leftrightarrow x=3\)