Trả lời:

\(1\left(x-2\right)+2=x\left(x+2\right)\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

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

Vậy\(x\in\left\{0;-1\right\}\)

Hok tốt!

Good girl

\(\left(x-2\right)+2=x\left(x+2\right)\)

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

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

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

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

KL

Câu 1 :

Đặt \(n^2+2n+4=a^2\)

\(\Leftrightarrow\left(n+1\right)^2+3=a^2\)

\(\Leftrightarrow a^2-\left(n+1\right)^2=3\)

\(\Leftrightarrow\left(a-n-1\right)\left(a+n+1\right)=3\)

TH1 \(\hept{\begin{cases}a+n+1=3\\a-n-1=1\end{cases}}\)

TH2 : \(\hept{\begin{cases}a+n+1=-3\\a-n-1=-1\end{cases}}\)

TH3 : \(\hept{\begin{cases}a+n+1=-1\\a-n-1=-3\end{cases}}\)

TH4 : \(\hept{\begin{cases}a+n+1=1\\a-n-1=3\end{cases}}\)

Bạn tính ra trong từng TH nhé !

Câu 1 :

Giả sử : \(n^2+2n+4=k^2\left(k\inℤ\right)\)

\(\Rightarrow k^2-\left(n^2+2n+1\right)=3\)

\(\Rightarrow k^2-\left(n+1\right)^2=3\)

\(\Rightarrow\left(k+n+1\right)\left(k-n-1\right)=3\)

Do k + n + 1 > k - n - 1 ( với k;n thuộc Z )

\(\Rightarrow\hept{\begin{cases}k+n+1=3\\k-n-1=1\end{cases}}\Rightarrow\hept{\begin{cases}k+n=2\\k-n=2\end{cases}}\Rightarrow\hept{\begin{cases}k=2\\n=0\end{cases}}\) 

Vậy n = 0 

|x+3|+|x+5|-|x+1/2|+|7-x|

TH1: -x-3-x-5+x-1/2+7-x

    = -2x-3/2

TH2: -x-3-x-5+x-1/2+x-7

   = 0-31/2

=-31/2

Câu 1:

Ta có \(x^3+3x-5=x^3+2x+x-5=\left(x^2+2\right)x+x-5\)

để giá trị của đa thức \(x^3+3x-5\)chia hết cho giá trị của đa thức \(x^2+2\)

thì \(x-5⋮x^2+2\Rightarrow\left(x-5\right)\left(x+5\right)⋮x^2+2\Rightarrow x^2-25⋮x^2+2\)

\(\Leftrightarrow x^2+2-27⋮x^2+2\Rightarrow27⋮x^2+2\)

\(\Leftrightarrow x^2+2\inƯ\left(27\right)\)do \(x^2+2\inℤ,\forall x\inℤ\)

mà \(x^2+2\ge2,\forall x\inℤ\)

\(\Rightarrow x^2+2\in\left\{3;9;27\right\}\)\(\Leftrightarrow x^2\in\left\{1;7;25\right\}\)

mà \(x^2\)là số chính phương \(\forall x\inℤ\)

\(\Rightarrow x^2\in\left\{1;25\right\}\Leftrightarrow x\in\left\{\pm1;\pm5\right\}\)

**bạn nhớ thử lại nhé
\(KL...\)

Bạn Minh Tâm ơi giá trị \(\pm1\)sai rồi

2x²+2y²=5xy

<=>2x²-5xy+2y²=0

<=>(2x²-4xy)-(xy-2y²)=0

<=>2x(x-2y)-y(x-2y)=0

<=>(x-2y).(2x-y)=0

<=> (x-2y)=0 hoặc 2x-y=0

Nếu x-2y=0 =>x=2y

=>E=\(\frac{x+y}{x-y}\)=\(\frac{2y+y}{2y-y}\)=\(\frac{3y}{y}\)=3

Nếu 2x-y=0 =>2x=y

=>E=\(\frac{x+y}{x-y}\)=\(\frac{x+2x}{x-2x}\)=\(\frac{3x}{-1x}\)= -3

2x^2 + 2y^2 = 5xy

<=> 2x^2 + 2y^2 - 5xy = 0

<=> 2x^2  - 4xy + 2y^2 - xy  = 0

<=> 2x(x - 2y) - y(x - 2y) = 0

<=> (2x - y)(x - 2y) = 0

<=> 2x = y hoặc x = 2y

thay vào là xong

Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\cdot\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=3\cdot\frac{1}{abc}\)

( Do \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\) )

Khi đó : \(P=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)

\(5+4x-x+2=\left(5x+4\right)\left(7+5x\right)\)

\(\Leftrightarrow5+4x-x+2=35+28x+25x+20x^2\)

\(\Leftrightarrow x^2+50x+28=0\)

Ta có \(\Delta=50^2-4.1.28=2388,\sqrt{\Delta}=2\sqrt{597}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-50+2\sqrt{597}}{2}=-25+\sqrt{597}\\x=\frac{-50-2\sqrt{597}}{2}=-25-\sqrt{597}\end{cases}}\)

\(5+4x-x+2=\left(5+4x\right)\left(7+5x\right)\)

\(7+3x=\left(5+4x\right)\left(7+5x\right)\)

\(7+3x=35+28x+25x+20x^2\)

\(7+3x-35-28x-25x-20x^2=0\)

\(-28-50x-20x^2=0\)

\(-28-50x-20x^2=0\)

\(x=-\frac{25+\sqrt{65}}{20};-\frac{25-\sqrt{65}}{20}\)