=>|x^2+2|=x^2+2x+5

=>x^2+2=x^2+2x+5(Do x^2+2>=2>0 với mọi x)

=>2x+5=2

=>2x=-3

=>x=-3/2

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

\(\Leftrightarrow\left|x^2+2\right|=x^2+2x+5\)  

Mà: \(x^2+2\ge2>0\forall x\)

\(\Leftrightarrow x^2+2=x^2+2x+5\)

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

\(\Leftrightarrow2x+3=0\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=-\dfrac{3}{2}\)

\(x^4-2x^3+3x^2-2x+1=0\)

Chia cả hai vé cho \(x^2\)

\(\Leftrightarrow x^2-2x+3-\dfrac{2}{x}+\dfrac{1}{x^2}\)

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

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

Đặt x+1/x = a, ta có:

\(a^2-2a+1=0\)

\(\Leftrightarrow\left(a-1\right)^2=0\)

\(\Leftrightarrow a=1\)

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

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

\(\Leftrightarrow x^2-x+1=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+3>0\)

Do đó phương trình vô nghiệm

a,\(11-2x=x-1\Leftrightarrow-2x-x=-1-11\Leftrightarrow-3x=-12\Leftrightarrow x=-4\)

b,\(\text{5(3x+2)=4x+1}\Leftrightarrow15x+10=4x+1\Leftrightarrow15x-4x=1-10\Leftrightarrow11x=-9\Leftrightarrow x=\dfrac{-9}{11}\)

c,\(x^2-4-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x+2\right)\left(x-2\right)-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x-2\right)[\left(x+2\right)-\left(x-5\right)]\Leftrightarrow\left(x-2\right)\left[x+2-x+5\right]\Leftrightarrow\left(x-2\right)7\Leftrightarrow7x-14\)

ĐKXĐ: \(x\ge1;x\le-3;x=-1\)

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

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

\(\left(1\right)\Leftrightarrow x+1=0\Rightarrow x=-1\)

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

\(\Leftrightarrow2x+6=x-1+4\sqrt{\left(x-1\right)\left(x+1\right)}+4x+4\)

\(\Leftrightarrow4\sqrt{x^2-1}=3-3x\) \(\Leftrightarrow\left\{{}\begin{matrix}3-3x\ge0\\16\left(x^2-1\right)=\left(3-3x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\7x^2+18x-25=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-25}{7}\end{matrix}\right.\)

Vậy pt có 3 nghiệm: \(x=-1;1;\dfrac{-25}{7}\)

thank

Thực hiện các phép đổi tương đương , ta đưa ( 1 ) về dạng :

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

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

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

\(\Leftrightarrow\left(x+4\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{1}{2}\end{array}\right.\)

Thữ vào mẫu thức : Với \(x=\frac{1}{2}\) thì \(2x^2-5x+2=0\)

Với \(x=-4\) thì \(\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)\ne0\)

Vậy phương trình ( 1 ) là cho nghiệm duy nhất là \(x=-4\)

\(\left\{{}\begin{matrix}x^2y+xy^2=0\left(1\right)\\2x^2+3xy+2y^2=1\left(2\right)\end{matrix}\right.\)

\(pt\left(1\right)\Leftrightarrow xy\left(x+y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\x=-y\end{matrix}\right.\)

Với \(x=0\) thế vào pt(2) ta được\(2.0^2+3.0.y+2y^2=1\Rightarrow2y^2=1\Rightarrow y^2=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{\sqrt{2}}\)

Với \(y=0\) thế vào pt(2) ta được

\(2x^2+3.x.0+2.0^2=1\Rightarrow2x^2=1\Rightarrow x^2=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{\sqrt{2}}\)

Với \(x=-y\) thế vào pt(2) ta được

\(2\left(-y\right)^2+3\left(-y\right).y+2y^2=1\Rightarrow2y^2-3y^2+2y^2=1\Rightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=1\\y=1\Rightarrow x=-1\end{matrix}\right.\)

vậy ...

a) Ta có : (2x + 5)² = (x + 2)²

<=> 4x² + 25 = x² + 4

<=> 4x² - x² = 4 - 25

<=> 3x² = -21

<=> x² = -21 : 3

<=> x² = -7

Đề sao sao 

a) \(\left(2x+5\right)^2=\left(x+2\right)^2\)

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

\(\Leftrightarrow\left(2x+5+x+2\right)\left(2x+5-x-2\right)=0\)

\(\Leftrightarrow\left(3x+7\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{3}\\x=-3\end{cases}}\)

vậy.............

b) \(x^2-5x+6=0\)

\(\Leftrightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)

vậy.................

c) hình như sai đề