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\(=\dfrac{-2}{3}\cdot\dfrac{12}{19}-\dfrac{2}{3}\cdot\dfrac{7}{19}+1\)
=-2/3(12/19+7/19)+1
=1-2/3
=1/3
\(=\dfrac{2021}{2022}\left(\dfrac{6}{17}-\dfrac{23}{17}\right)+\dfrac{2021}{2022}=\dfrac{-2021}{2022}+\dfrac{2021}{2022}=0\)
Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
\(1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\left(đk:x>2\right)\)
đặt
\(A=1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\)
\(\left(x-1\right)^2A=\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\)
\(\left(x-1\right)^2A-A=\left[\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\right]-\left[1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\right]\)
\(\left[\left(x-1\right)^2-1\right]A=\left(x-1\right)^{2022}-1\)
\(A=\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}\)
\(=>\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\\ =>\left(x-1\right)^{2022}-1=17^{2022}-1\\ =>\left(x-1\right)^{2022}=17^{2022}\\ =>x-1=17\\ =>x=18\left(tm\right)\)
a)\(27.65+27.35+300=27.\left(65+35\right)+300\)
\(=27.100+300=2700+300=3000\)
b)\(3838:\left[\left(190-6.5^2\right):4+3\right]\)
\(=3838:\left[\left(190-6.25\right):4+3\right]\)
\(=3838:\left[\left(190-150\right):4+3\right]\)
\(=3838:\left[40:4+3\right]=3838:\left[10+3\right]\)
\(=3838:13=\dfrac{3838}{13}\)
c)\(2022-x=2021\)
\(x=2022-2021=1\)
d)\(26+14:\left(x-5\right)=33\)
\(14:\left(x-5\right)=33-26=7\)
\(x-5=14+7=2\)
\(x=2+5=7\)
e)đề hỏi làm j thế bạn
`2x-15=-25`
`2x=-10`
`x=-5`
___________
`3/5<x/10<4/5`
`3/5=(3xx10)/(5xx10)=30/50`
`x/10=(5x)/(10xx5)=(5x)/50`
`4/5=(4xx10)/(5xx10)=40/50`
`=>30/50<(5x)/50<40/50`
`=>30<5x<40`
`=>x=7`
b: \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{505}{1011}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1010}{1011}\)
=>1/x+1=-1009/2022
=>x+1=-2022/1009
hay x=-3031/1009
ta có
\(\left(x-6\right)^6=\left(x-6\right)^{2022}\Leftrightarrow\left(x-6\right)^6\left[\left(x-6\right)^{2016}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-6=\pm1\end{cases}}\Leftrightarrow x=6\text{ hoặc }x=5\text{ hoặc }x=7\)