( Tham khảo nhé  !)

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ai biet

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

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 ...

Đk:\(x\ge1\)

Pt \(\Leftrightarrow\sqrt{x-1}=\sqrt{3x-2}+\sqrt{5x-1}\)

\(\Leftrightarrow x-1=8x-3+2\sqrt{15x^2-13x+2}\)

\(\Leftrightarrow2-7x=2\sqrt{15x^2-13x+2}\) (1)

Với \(x\ge1\Rightarrow\)\(\left\{{}\begin{matrix}2-7x\le2-7.1=-5< 0\\2\sqrt{15x^2-13x+2}=4>0\end{matrix}\right.\)

Từ (1) => Dấu "=" không xảy ra

Vậy pt vô nghiệm.

bai toan nay quá hó

x=10 , ủng hộ mk nha

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

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

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

\(\Leftrightarrow x=0\)

Vậy ....

câu 1:-18/7

9/2

còn câu 2 tui chịu

))= 

khó nhìn quá

Đề là như này à bạn ?

\(\frac{-1}{7}\) . \(2^3\)- \(2^x\) : \(\frac{7}{3}\)= \(2^x\) - 1 

....

=> 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 

\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{1}{4}\)

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

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{4}\)

\(\Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)

\(\Leftrightarrow\dfrac{1}{4}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{\left(x-1\right)x}\right)=\dfrac{1}{8}\)   ( đk x khác 0 , x khác 1)

\(\Leftrightarrow\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{x-1}-\dfrac{1}{x}\right)=\dfrac{1}{8}\)

\(\Leftrightarrow1-\dfrac{1}{x}=\dfrac{1}{2}\)

=> x =2 ( tm)

đặt A=(x+1)+(X+2)+(x+3)+....+(x+99)

=> A= x+1+x+2+x+3+....+x+100

=x+x+x+x+...+x+(1+2+3+4+..+99)( có 99x)

=> 99x+4950=0

=> 99x=-4950

=> x=-50