\(\frac27-\frac89x=\frac23\)

\(\frac89x=\frac27-\frac23=-\frac{8}{21}\)

\(x=-\frac{8}{21}:\frac89\)

\(x=-\frac{8}{21}\cdot\frac98=-\frac37\)

8/9.x=2/7-2/3

8/9.x=14/21-6/21

8/9.x=8/21

x=8/9;8/21

x=8/9.21/8

x=21/9

x=7/3

dấu ; cho thành dấu chia nhé

\(\frac{2}{x-3}\) ≤ \(\frac23\)

\(\frac{1}{x-3}\) ≤ \(\frac13\)

\(\frac{1}{x-3}-\frac13\) ≤ 0

\(\frac{3-x+3}{3\left(x-3\right)}\) ≤ 0

\(\frac{\left(3+3\right)-x}{3\left(x-3\right)}\) ≤ 0

\(\frac{6-x}{3\left(x-3\right)}\) ≤ 0

6 - \(x\) = 0 ⇒ \(x=6\); \(x-3=0\) ⇒ \(x=3\)

Lập bảng xét dấu ta có:

Theo bảng trên ta có: \(x\) ≥ 6 hoặc \(x\) < 3

SOS!!!

ra kq luôn đko, lười đánh máy

viết cách lm vs kq ra cho mik dc khum

=33X

a: \(\frac{7-x}{2}+\frac23\left(x-7\right)\left(x-3\right)=0\)

=>\(\frac{-\left(x-7\right)}{2}+\frac23\left(x-7\right)\left(x-3\right)=0\)

=>\(\left(x-7\right)\left(-\frac12+\frac23x-2\right)=0\)

=>\(\left(x-7\right)\left(\frac23x-\frac52\right)=0\)

=>\(\left[\begin{array}{l}x-7=0\\ \frac23x-\frac52=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ \frac23x=\frac52\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac52:\frac23=\frac{15}{4}\end{array}\right.\)

b: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)

=>\(\left(3x+6\right)^2=\left(2x+7\right)^2\)

=>\(\left(3x+6\right)^2-\left(2x+7\right)^2=0\)

=>(3x+6-2x-7)(3x+6+2x+7)=0

=>(x-1)(5x+13)=0

=>\(\left[\begin{array}{l}x-1=0\\ 5x+13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-\frac{13}{5}\end{array}\right.\)

a: \(\frac23\times\frac45+\frac23\times\frac15+\frac23\)

\(=\frac23\times\left(\frac45+\frac15\right)+\frac23\)

\(=\frac23+\frac23=\frac43\)

b: \(\frac{2022\times2023+2000}{2023\times2023-23}\)

\(=\frac{2022\times2023+2000}{2022\times2023+2023-23}\)

\(=\frac{2022\times2023+2000}{2022\times2023+2000}\)

=1

\(\begin{cases}y=x+20\left(1\right)\\ \frac{y}{50}-\frac{x}{40}=\frac14\left(2\right)\end{cases}\)

thay (1) vào (2) ta được:

\(\frac{x+20}{50}-\frac{x}{40}=\frac14\)

\(\Leftrightarrow4\left(x+20\right)-5x=50\)

4x + 80 - 5x = 50

-x = -30

⇒ x = 30

⇒ y = 30 + 20 = 50

vậy (x; y) = (30; 50)

\(\frac14-2x=5\)

\(2x=\frac14-5\)

\(2x=\frac{-19}{4}\)

\(x=-\frac{19}{4}:2\)

\(x=\frac{-19}{8}\)

\(\frac12x-\frac13=25\%\)

\(\frac12x-\frac13=\frac14\)

\(\frac12x=\frac14+\frac13\)

\(\frac12x=\frac{7}{12}\)

\(x=\frac{7}{12}:\frac12\)

\(x=\frac76\)

- 19/8;7/6

Câu 1:

c: \(\frac19+\frac28+\frac37+\cdots+\frac91\)

\(=\left(\frac19+1\right)+\left(\frac28+1\right)+\cdots+\left(\frac82+1\right)+1\)

\(=\frac{10}{2}+\frac{10}{3}+\cdots+\frac{10}{10}=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

Ta có: \(\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\right)\cdot x=\frac19+\frac28+\frac37+\cdots+\frac91\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

=>x=10

Câu 2:

d: \(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023\cdot2024}\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

Hẹ hẹ

\(P=\dfrac{\dfrac{8}{12}-\dfrac{3}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{12}{12}-\dfrac{7}{11}}=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{17}{12}-\dfrac{7}{11}}=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{103}\)