~ Ko có đề bài -_- ~ 

Ta có : 

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\)\(1-\frac{1}{100}\)

\(=\)\(\frac{99}{100}\)

Chúc bạn học tốt ~ 

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{99x100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/19-1/20

=1/2-1/20

=10/20-1/20

=9/20

\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{19\times20}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=\dfrac{1}{2}-\dfrac{1}{20}\)

\(=\dfrac{9}{20}\)

\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{19\times20}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=\dfrac{1}{2}-\dfrac{1}{20}\)

\(=\dfrac{10}{20}-\dfrac{1}{20}\)

\(=\dfrac{9}{20}\)

\(\frac{2}{3}\times\frac{4}{5}+\frac{1}{3}\times\frac{4}{5}\)

\(=\left(\frac{2}{3}+\frac{1}{3}\right)\times\frac{4}{5}\)

\(=1\times\frac{4}{5}\)

\(=\frac{4}{5}\)

2/3 x 4/5 + 1/3 x 4/5

= 4/5 x (2/3 + 1/3 )

= 4/5 x 1

= 4/5 

mong các bạn nha

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

\(\Rightarrow x-\dfrac{1}{2}=0\)

\(\Rightarrow x=\dfrac{1}{2}\)

b) \(\left(x-2\right)^2=1\)

\(\Rightarrow x-2=1\)

\(\Rightarrow x=3\)

c) \(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-1\)

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

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\).

a , \(\left(x-\dfrac{1}{2}\right)^2=0\)

<=> \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

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

c , \(\left(2x-1\right)^3=-8\Rightarrow2x-1=-2\Rightarrow x=\dfrac{-1}{2}\)

d , \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4^2}\)

<=> \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

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

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

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=0+\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(TM\right)\)

Vậy \(x=\dfrac{1}{2}\) là giá trị cần tìm

b) \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=\left(-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) \(\left(TM\right)\)

Vậy \(x\in\left\{3;1\right\}\)

c) \(\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-2+1\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=\dfrac{-1}{2}\left(TM\right)\)

Vậy \(x=\dfrac{-1}{2}\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\\left(x+\dfrac{1}{2}\right)^2=\dfrac{-1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-1}{4};\dfrac{-3}{4}\right\}\) là giá trị cần tìm

Ta có : \(\left(3x-1\right)^2-\left(x+7\right)^2=0\)

\(\left(3x-1+x+7\right)\left(3x-1-x-7\right)=0\)

\(\left(4x+6\right)\left(2x-8\right)=0\)

Nên : 4x + 6 = 0 hoặc 2x - 8 = 0

4x = -6 hoặc 2x = 8

x = \(\frac{-3}{2}\) hoặc x = 4

Vậy x = \(\frac{-3}{2}\) hoặc x = 4

Đặt \(A=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\)

\(A=\frac{3-2}{2x3}+\frac{4-3}{3x4}+\frac{5-4}{4x5}+\frac{6-5}{5x6}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(A=\frac{1}{2}-\frac{1}{6}\)

\(A=\frac{3}{6}-\frac{1}{6}\)

\(A=\frac{2}{6}=\frac{1}{3}\)

Bạn tham khảo bài giải dưới nhé

Cre: Olm

Hc tốt:)

cái này là tìm x phải ko bn

a)27³-72x=0

-72x=27³

^=>x=27³/-72

x=-273,375

b)=>(x+2)²=6.(x+2)

(x+2)²/(x+2)=6

(x+2)=6

vậy x =6-2=4