Ta có : \(\dfrac{x^2}{5.6}\text{=}\dfrac{x^2}{5}-\dfrac{x^2}{6}\)

\(\dfrac{x^2}{6.7}\text{=}\dfrac{x^2}{6}-\dfrac{x^2}{7}\)

\(...\)

\(\dfrac{x^2}{24.25}\text{=}\dfrac{x^2}{24}-\dfrac{x^2}{25}\)

\(\Rightarrow\) biểu thức chỉ còn :

\(\dfrac{x^2}{5}-\dfrac{x^2}{25}\text{=}\dfrac{5x^2-x^2}{25}\text{=}\dfrac{4x^2}{25}\)

\(a)A\text{=}-5^{22}+22+112-100+5^{22}+2022\text{=}\left(-5^{22}+5^{22}\right)+\left(22+112-100+2022\right)\text{=}0+2056\text{=}2056\)

\(b)B\text{=}\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\text{=}\dfrac{2^{10}.\left(13+65\right)}{2^{10}.26}\text{=}\dfrac{78}{26}\text{=}3\)

Ta có : \(A\text{=}\dfrac{2013.2014-1}{2013.2014}\text{=}\dfrac{2013.2014}{2013.2014}-\dfrac{1}{2013.2014}\text{=}1-\dfrac{1}{2013.2014}\)

\(B\text{=}\dfrac{2014.2015-1}{2014.2015}\text{=}\dfrac{2014.2015}{2014.2015}-\dfrac{1}{2014.2015}\text{=}1-\dfrac{1}{2014.2015}\)

\(Ta\) có : \(\dfrac{1}{2013.2014}>\dfrac{1}{2014.2015}\)

\(\Rightarrow A< B\)

\(\dfrac{7}{8}+\dfrac{7}{8}:\dfrac{1}{8}-\dfrac{1}{2}\\ =\dfrac{7}{8}+\dfrac{7}{8}\cdot\dfrac{8}{1}-\dfrac{1}{2}\\ =\dfrac{7}{8}+\dfrac{7}{1}-\dfrac{1}{2}\\ =\dfrac{7}{8}+\dfrac{56}{8}-\dfrac{4}{8}\\ =\dfrac{59}{8}\)

`7/8 + 7/8 : 1/8 - 1/2`

`= 7/8 + 7/8 xx 8 -1/2`

`= 7/8 + 56/8 -1/2`

`= 7/8+ 7 -1/2`

`= 7/8 + 56/8 -1/2`

`= 63/8 -1/2`

`= 63/8 - 4/8`

`= 59/8`

\(\left(x-3\right)^6=\left(x-1\right)^2\)

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

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

a)

ta có 2n-1=2(n-3)+7

2(n-3)⋮n-3;để 2n-1⋮n-3

<=>7⋮n-3

=>n-3ϵƯ(7)={1;-1;7;-7}

lập bảng

vậy nϵ{4;2;10;-4}

a)

\(\dfrac{6}{13}\cdot\dfrac{8}{7}\cdot\dfrac{-26}{3}\cdot\dfrac{-7}{8}\)

\(=\dfrac{6}{13}\cdot\dfrac{-26}{3}\cdot\dfrac{8}{7}\cdot\dfrac{-7}{8}\)

\(=-4\cdot\left(-1\right)\\ =4\)

b)

\(\dfrac{6}{11}+\dfrac{11}{3}\cdot\dfrac{3}{22}\)

\(=\dfrac{6}{11}+\dfrac{1}{2}\\ =\dfrac{12}{22}+\dfrac{11}{22}\\ =\dfrac{23}{22}\)

dấu chấm là dấu nhân

Nhanh hộ mình nha