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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.
\(a;343-4\times\left(x+1\right)=143\)
\(4\times\left(x+1\right)=343-143\)
\(4\times\left(x+1\right)=200\)
\(x+1=200:4\)
\(x+1=50\)
\(x=50-1=49\)
Mấy câu hỏi trẻ trâu thì tự tính nhé
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\)
\(\Leftrightarrow x=305\)
Vậy x=305
`(11-x):(-5)=15`
`=> 11-x=-75`
`=> x=86`
Vậy `x = 86`
`((2x+3)^2022) . (-7x+84)=0`
`=> (2x+3)^2022 = 0` hoặc `-7x + 84 = 0`
`=> 2x+3=0` hoặc `-7x = -84`
`=> x = -3/2` hoặc `x = 12`
Vậy `x = -3/2` hoặc `x = 12`
`(x-3)(x+1) < 0`
Ta có: `x - 3 < x + 1`
nên: `x - 3 < 0` và `x + 1 > 0`
`=> x < 3 và x > -1`
`=> -1 < x < 3`
Vậy `-1 <x < 3`
#\(N\)
`a, (11-x)`\(\div\)`(-5)=15`
`11-x=15.-5`
`11-x=-75`
`x=11-(-75)`
`x=11+75=86`
`b, (2x-3)^2022.(-7x+84)=0`
`=>`\(\left\{{}\begin{matrix}\left(2x-3\right)^{2022}=0\\\left(-7x+84\right)=0\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}2x-3=0\\-7x+84=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=3+0\\-7x=0-84\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=3\\-7x=-84\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\div2\\x=-84\div-7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=12\end{matrix}\right.\)
`c, (x-3) (x+1) <0`
`-> (x-3) < x+1`
`-> (x-3)<0 , (x+1)>0`
`-> x < 3 , x> (-1)`
`-> 3 > x > -1`
\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+........+\(\frac{1}{x.\left(x+3\right)}\)=\(\frac{101}{1540}\)
3(.\(\frac{1}{5.8}+\frac{1}{8.11}\)+\(\frac{1}{11.14}+.......+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}.3=\frac{303}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=>\(x+3=308\)
\(x=308-3=305\)
Vậy \(x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
=> \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 5
x = 303
\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(3.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{4620}\)
\(\frac{1}{x+3}=...\) (tự làm tiếp)
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+1\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+1}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)
\(\frac{1}{x+1}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=> x + 1 = 308
=> x = 308 - 1
=> x = 307