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1) Ta có: \(3\left(x-1\right)-5\left(x-2\right)=4\left(x+1\right)\)
\(\Leftrightarrow3x-5-5x+10-4x-4=0\)
\(\Leftrightarrow-6x+1=0\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\dfrac{1}{6}\)
2) Ta có: \(-2\left(x-2\right)-4\left(x+1\right)=-3\left(x+3\right)\)
\(\Leftrightarrow-2x+4-4x-4+3x+9=0\)
\(\Leftrightarrow-3x=-9\)
hay x=3
3) Ta có: \(3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
4) Ta có: \(x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) Ta có: \(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
6) Ta có: \(\left(5x-1\right)^3=125\)
\(\Leftrightarrow5x-1=5\)
\(\Leftrightarrow5x=6\)
hay \(x=\dfrac{6}{5}\)
7) Ta có: \(3^{x+1}=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
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`@` `\text {Ans}`
`\downarrow`
`1)`
\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(2x=\dfrac{7}{6}\)
`\Rightarrow`\(x=\dfrac{7}{6}\div2\)
`\Rightarrow`\(x=\dfrac{7}{12}\)
Vậy, `x = 7/12`
`2)`
\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)
`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)
`\Rightarrow`\(x=\dfrac{40}{21}\)
Vậy, `x = 40/21`
`3)`
\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)
`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)
`\Rightarrow`\(x=\dfrac{16}{21}\)
Vậy, `x = 16/21`
`4)`
\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{1}{12}\)
`\Rightarrow`\(x=\dfrac{1}{12}\div3\)
`\Rightarrow`\(x=\dfrac{1}{36}\)
Vậy, `x = 1/36`
`5)`
\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)
`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{52}{21}\)
Vậy, `x = 52/21`
`6)`
\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)
`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(5x=\dfrac{1}{6}\)
`\Rightarrow`\(x=\dfrac{1}{6}\div5\)
`\Rightarrow`\(x=\dfrac{1}{30}\)
Vậy, `x = 1/30.`
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a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-36\right):18=12\)
\(x-36=12.18\)
\(x-36=216\)
\(x=252\)
vậy \(x=252\)
\(5x-3=3x-11\)
\(5x-3x=-11+3\)
\(2x=-8\)
\(x=-4\)
vậy \(x=-4\)
\(3x+5x=16\)
\(8x=16\)
\(x=2\)
vậy \(x=2\)
\(2\left(x+1\right)=10\)
\(x+1=5\)
\(x=4\)
vậy \(x=4\)
k nha Pham Tuyet Nhung
1.=> x-36 = 216 = > x = 216 +36 =252
2. =>2x = -8 = > x= -4
3.8 x = 16 = > x = 2
4 . => x= 10 -1 = 9
ok có j ko hiểu hỏi riêng nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A.3x-\frac{3}{2}-5x+\frac{10}{3}=1\)
\(3x-\frac{3}{2}-5x=-\frac{7}{3}\)
\(3x-5x=-\frac{7}{3}+\frac{3}{2}\)
\(-2x=-\frac{5}{6}\)
\(x=-\frac{5}{6}:\left(-2\right)\)
\(x=\frac{5}{12}\)
\(\left(1-3x\right)+\left(x-3\right)=\left(5x-1\right)-\left(5x+3\right)\)
\(\Rightarrow1-3x+x-3=5x-1-5x-3\)
\(\Rightarrow\left(1-3\right)-\left(3x-x\right)=\left(5x-5x\right)-\left(1+3\right)\)
\(\Rightarrow-2-2x=-4\)
\(\Rightarrow2x=-2+4\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)