sai đề

Cách liên hợp 

ĐK \(x\ge-2\)

PT <=> \(\sqrt{x+2}+5x+2\ne0\)

\(25x^2+19x+2+2\left(x+1\right)\left(\sqrt{x+2}-5x-2\right)=0\)

Xét \(\sqrt{x+2}+5x+2=0\)=> \(x=\frac{-19-\sqrt{161}}{50}\)

Thay vào ta thấy nó không phải là nghiệm của PT

=> \(\sqrt{x+2}+5x+2\ne0\)

<=> \(25x^2+19x+2+2\left(x+1\right).\frac{x+2-\left(5x+2\right)^2}{\sqrt{x+2}+5x+2}=0\)

<=> \(25x^2+19x+2+2\left(x+1\right).\frac{-25x^2-19x-2}{\sqrt{x+2}+5x+2}=0\)

<=> \(\orbr{\begin{cases}25x^2+19x+2=0\\1-\frac{2\left(x+1\right)}{\sqrt{x+2}+5x+2}=0\left(2\right)\end{cases}}\)

Pt (2)

<=> \(\sqrt{x+2}=-3x\)

<=> \(\hept{\begin{cases}x\le0\\9x^2-x-2=0\end{cases}}\)=> \(x=\frac{1-\sqrt{73}}{18}\)(TM ĐKXĐ)

Pt (1) có nghiệm \(x=\frac{-19+\sqrt{161}}{50}\)(Tm ĐKXĐ)

Vậy Pt có nghiệm \(S=\left\{\frac{1-\sqrt{73}}{18};\frac{-19+\sqrt{161}}{50}\right\}\)

Cách đặt ẩn phụ không hoàn toàn 

ĐK\(x\ge-2\)

PT 

<=> \(15x^2+6x+2\left(x+1\right)\sqrt{x+2}-\left(x+2\right)=0\)

Đặt \(\sqrt{x+2}=a\left(a\ge0\right)\)

=> \(15x^2+6x+2\left(x+1\right).a-a^2=0\)

<=> \(\left(15x^2+2ax-a^2\right)+\left(6x+2a\right)=0\)

<=> \(\left(5x-a\right)\left(3x+a\right)+2\left(3x+a\right)=0\)

<=> \(\left(3x+a\right)\left(5x-a+2\right)=0\)

<=> \(\orbr{\begin{cases}3x+a=0\\5x-a+2=0\end{cases}}\)

+ 3x+a=0

=> \(3x+\sqrt{2+x}=0\)

=> \(\hept{\begin{cases}x\le0\\9x^2-x-2=0\end{cases}}\)=> \(x=\frac{1-\sqrt{73}}{18}\)(TM ĐKXĐ)

+ 5x-a+2=0

=> \(5x+2=\sqrt{x+2}\)

=> \(\hept{\begin{cases}x\ge-\frac{2}{5}\\25x^2+19x+2=0\end{cases}}\)=> \(x=\frac{-19+\sqrt{161}}{50}\)(TM ĐKXĐ)

vậy \(S=\left\{\frac{-19+\sqrt{161}}{50};\frac{1-\sqrt{73}}{18}\right\}\)

cảm ơn bạn nha lúc nào mk đăng câu hỏi bạn cx trả lời sớm nhất , kb nhé cho mk hỏi bạn học lớp mấy và có bí kíp nào giúp mk học tốt dạng này ko

`a,3x^2+7x+2=0`

`<=>3x^2+6x+x+2=0`

`<=>3x(x+2)+x+2=0`

`<=>(x+2)(3x+1)=0`

`<=>x=-2\or\x=-1/3`

d) Ta có: (x-1)(x+2)=70

\(\Leftrightarrow x^2+2x-x-2-70=0\)

\(\Leftrightarrow x^2+x-72=0\)

\(\Leftrightarrow x^2+9x-8x-72=0\)

\(\Leftrightarrow x\left(x+9\right)-8\left(x+9\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=8\end{matrix}\right.\)

Vậy: S={8;-9}

`d,(x+1)(x+2)=70`

`<=>x^2+3x+2=70`

`<=>x^2+3x-68=0`

`<=>(x+3/2)^2=281/4`

`<=>x=(+-\sqrt{281}-3)/2`

Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

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

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

ĐKXĐ:

\(\left(2x+2-2\sqrt{5x-1}\right)+\left(\sqrt{5x^2+x+3}-\left(2x+1\right)\right)+x^2-3x+2=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+2\right)}{x+1+\sqrt{5x-1}}+\dfrac{x^2-3x+2}{\sqrt{5x^2+x+3}+2x+1}+x^2-3x+2=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{2}{x+1+\sqrt{5x-1}}+\dfrac{1}{\sqrt{5x^2+x+3}+2x+1}+1\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\)

ĐKXĐ: \(x>\dfrac{1}{5}\)

\(1-3x^2< \left(x+2\right)\sqrt[]{5x-1}+5x-1\)

\(\Leftrightarrow3x^2+5x-2+\left(x+2\right)\sqrt{5x-1}\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-1\right)+\left(x+2\right)\sqrt{5x-1}>0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-1+\sqrt{5x-1}\right)>0\)

\(\Leftrightarrow3x-1+\sqrt{5x-1}>0\)

\(\Leftrightarrow\sqrt{5x-1}>1-3x\)

TH1: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{5}\\1-3x< 0\end{matrix}\right.\) \(\Leftrightarrow x>\dfrac{1}{3}\)

TH2: \(\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\5x-1>9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\9x^2-11x+2< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{2}{9}< x\le\dfrac{1}{3}\)

Kết luận: \(x>\dfrac{2}{9}\)