\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

\(x-\frac{2}{8}=\frac{-1}{4}\)

\(\Leftrightarrow x=\frac{-1}{4}+\frac{2}{8}\)

\(\Leftrightarrow x=\frac{-1}{4}+\frac{1}{4}\)

\(\Leftrightarrow x=0\)

Vậy ....

=> 4x^2  - 12x + 4 = 2x^2 - 2x - 2 - 2x^2 - 2x - 13

=> 4x^2 - 12x + 4 = - 4x - 15

=> 4x^2 - 12x + 4x + 4 + 15 = 0 

=> 4x^2 - 8x + 19 = 0

Đề sai 

=> 2x-2 + 3x-6 = x-4

=> 5x-8 = x-4

=> 5x-8-(x-4) = 0

=> 5x-8-x+4=0

=> 4x-4=0

=> 4x=4

=> x=4:4=1

Vậy x=1

Tk mk nha

2(x-1)+3(x-2)=x-4

<=>2x-2+3x-6=x-4

<=>5x-8=x-4

<=>5x-x=8-4

<=>4x=4

<=>x=1

ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ge0\\x-2>0\\x+2>0\\x\ge0\end{matrix}\right.\)  và \(4-x\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x>2\\x>-2\\x\ge0\end{matrix}\right.\) và \(x\ne4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)

=>3(x-2)=4(x+5)

=>4x+20=3x-6

=>x=-26

`(x-2):4 = (5+x) :3`

`(x-2)*3 = 4*(5+x)`

`3x -6 = 20 +4x`

`3x -4x = 20 +6`

`-x =26`

`x=26`

Ta có: \(\frac{3}{1\cdot2}+\frac{3}{3\cdot4}+\cdots+\frac{3}{299\cdot300}\)

\(=3\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{299\cdot300}\right)\)

\(=3\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{299}-\frac{1}{300}\right)\)

\(=3\cdot\left\lbrack1+\frac12+\frac13+\frac14+\cdots+\frac{1}{299}+\frac{1}{300}-2\left(\frac12+\frac14+\cdots+\frac{1}{300}\right)\right\rbrack\)

\(=3\left(1+\frac12+\frac13+\cdots+\frac{1}{300}-1-\frac12-\cdots-\frac{1}{150}\right)\)

\(=3\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\)

Ta có: \(\left(\frac{3}{1\cdot2}+\frac{3}{3\cdot4}+\cdots+\frac{3}{299\cdot300}\right)\cdot\left(x+\frac23\right)=\frac{2}{151}+\frac{2}{152}+\cdots+\frac{2}{300}\)

=>\(3\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\left(x+\frac23\right)=2\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\)

=>\(x+\frac23=\frac23\)

=>x=0

ta có:

1-1/2+1/2-1/3+1/3-1/4+....+1/x -1/x+1 =499/500

1-1/x+1 =499/500

1/x+1 =1/500 

x+1=500

x=499

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{X\times\left(X+1\right)}=\frac{499}{500}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{X}-\frac{1}{X+1}=\frac{499}{500}\)

\(\Leftrightarrow1-\frac{1}{X+1}=\frac{499}{500}\)

\(\Leftrightarrow\frac{1}{X+1}=\frac{1}{500}\)

\(\Leftrightarrow X+1=500\)

\(\Leftrightarrow X=499\)