mình sửa đề số cuối nhé

\(\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{90}-\dfrac{1}{93}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{93}\right)=\dfrac{176}{1395}\)

\(xy-x+y=4\\ \Rightarrow x\left(y-1\right)+\left(y-1\right)=3\\ \Rightarrow\left(x+1\right)\left(y-1\right)=3\)

Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x+1,y-1\in Z\\x+1,y-1\inƯ\left(3\right)\end{matrix}\right.\)

Ta có bảng:
 

Vậy \(\left(x,y\right)\in\left\{\left(-4;0\right);\left(-2;-2\right);\left(0;4\right);\left(2;2\right)\right\}\)

báo cáo

\(2^x-2^{x-2}=96\\ \Rightarrow2^x-2^x:4=96\\ \Leftrightarrow2^x-2^x.0,25=96\\ \Rightarrow2^x\left(1-0,25\right)=96\\ \Rightarrow2^x.0,75=96\\ \Rightarrow2^x=128\\ \Rightarrow2^x=2^7\\ \Rightarrow x=7\)

báo cáo

\(\left(3x+6\right)-35=-2\\ \Rightarrow3x+6=-2+35\\ \Rightarrow3x+6=33\\ \Rightarrow3x=33-6\\ \Rightarrow3x=27\\ \Rightarrow x=27:3\\ \Rightarrow x=9\)

TL :

( 3x + 6 ) - 35 = -2

3x + 6 = -2 + 35

3x + 6 = 33

3x = 33 - 6

3x = 27

x = 27 : 3

x = 9

Vậy x = 9

HT

\(27-\left(2x+1\right)=4\)

\(2x+1=27-4\)

\(2x+1=23\)

\(2x=23-1\)

\(2x=22\)

\(x=22:2\)

\(x=11\)

Vậy x = 11

#HQX

\(27-\left(2x+1\right)=4\)

\(2x+1=23\)

\(2x=22\)

\(x=11\)

\(a,\dfrac{2}{5}-\dfrac{3}{8}+\dfrac{7}{6}=\dfrac{48}{120}-\dfrac{45}{120}+\dfrac{140}{120}=\dfrac{143}{120}\\ b,\dfrac{2}{3}+\left(\dfrac{5}{7}+\dfrac{-2}{3}\right)=\left(\dfrac{2}{3}+\dfrac{-2}{3}\right)+\dfrac{5}{7}=0+\dfrac{5}{7}=\dfrac{5}{7}\\ c,\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}=\dfrac{-5}{7}\)

\(d,\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}=\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}+\dfrac{1}{11\times13}=\dfrac{1}{2}\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}+\dfrac{2}{11\times13}\right)=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)