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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{a}{b}=\frac{3}{4}\)(1)
\(\frac{a+15}{b}=\frac{7}{6}\)
\(\Rightarrow\frac{a}{b}+\frac{15}{b}=\frac{7}{6}\)(2)
Thay (1) vào (2) ta có :
\(\frac{3}{4}+\frac{15}{b}=\frac{7}{6}\)
\(\Rightarrow\frac{15}{b}=\frac{7}{6}-\frac{3}{4}\)
\(\Rightarrow\frac{15}{b}=\frac{5}{12}\)
\(\Rightarrow5b=12.15\)
\(\Rightarrow b=12.15:5\)
\(\Rightarrow b=36\)
Thay b\(\Rightarrow a=27\) vào (1) ta có :
\(\frac{a}{b}=\frac{a}{36}=\frac{3}{4}\)
\(\Rightarrow4a=36.3\)
\(\Rightarrow a=36.3:4\)
\(\Rightarrow a=27\)
\(\Rightarrow\frac{a}{b}=\frac{27}{36}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,-2022.(x+8)=0`
`=>x+8=0:(-2022)`
`=>x+8=0`
`=>x=0-8`
`=>x=-8`
`b,(7-x)(x+3)=0`
`@ TH1`
`7-x=0`
`=>x=7-0`
`=>x=7`
`@ TH2`
`x+3=0`
`=>x=0-3`
`=>x=-3`
`c,2023x .(14-x)=0`
`@ TH1`
`2023x=0`
`=>x=0:2023`
`=>x=0`
`@ TH2`
`14-x=0`
`=>x=14-0`
`=>x=14`
`d,x^2-x=0`
`=>x(x-1)=0`
`@ TH1`
`x=0`
`@ TH2`
`x-1=0`
`=>x=0+1`
`=>x=1`
\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{7^2.3}{2.3.5^2}\)
\(=\frac{7^2.1}{2.1.5^2}\)
\(=\frac{7^2}{2.5^2}\)
\(=\frac{49}{2.25}\)
\(=\frac{49}{50}\)
\(\frac{7^2.3}{2.3.5^2}=\frac{7^2.3:3}{2.5^2.3:3}=\frac{7^2}{2.5^2}=\frac{49}{50}\)
Tk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải :
-7/12 + 11/8 - 5/9
= 17/72
Học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
lấy tổng của 7/12 và 11/8 trừ đi 5/9 bằng 17/72
#hok tốt
\(\dfrac{2.3.5.\left(-7\right).12}{3.\left(-5\right).7.14}=\dfrac{2.3.5.\left(-7\right).12}{3.5.\left(-7\right).2.7}=\dfrac{12}{7}\)
\(\dfrac{2.3.5.\left(-7\right).12}{3.\left(-5\right).7.14}=\dfrac{1.1.1.\left(-1\right).12}{1.\left(-1\right).1.7}=\dfrac{-12}{-7}=\dfrac{12}{7}\)