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Trl:
\(\left(3x-2^4\right).7^{10}=2.7^{10}\)
\(\Rightarrow\left(3x-16\right)=2.7^{10}:7^{10}\)
\(\Rightarrow\left(3x-16\right)=2.1\)
\(\Rightarrow\left(3x-16\right)=2\)
\(\Rightarrow3x=16+2\)
\(\Rightarrow3x=18\)
\(\Rightarrow x=18:3\)
\(\Rightarrow x=6\)
Hc tốt
\(\left(3x-2^4\right)\times7^{10}=2\times7^{10}\)
\(3x-2^4=2\times7^{10}:7^{10}\)
\(3x-2^4=2\times1\)
\(3x-16=2\)
\(3x=18\)
\(x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
2) \(\Rightarrow5\left(x-2\right).3\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow2\left(x-4\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=x^2+2+3x^2+2-2x^2-2=2x^2+2\)
b: \(B=2x-3-\left(3x-2\right)-\left(2x-4\right)\)
\(=2x-3-3x+2-2x+4=-3x+3\)
c: \(C=6-3x+3x+10=16\)
d: \(D=\left|3x-8\right|+\left|x+2\right|\)
\(=3x-8+x+2=4x-6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
xy-3x+y=10
(=) x.(y-3)+y=10
(=) x.(y-3)+(y-3)=7
(=) (y-3).(x+1)=7
Ta có bảng sau
y-3 | 1 | 7 | -1 | -7 |
y | 4 | 11 | 2 | -4 |
x+1 | 7 | 1 | -7 | -1 |
x | 6 | 0 | -8 | -2 |
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{3x}{-4}=\frac{-9}{6}\)
=> \(\frac{-3x}{4}=\frac{-9}{6}\)
=> \(\frac{-3x}{4}=\frac{-3}{2}\)
=> \(\frac{-3x}{4}=\frac{-6}{4}\)
=> \(-3x=-6\)
=> \(x=\left(-6\right):\left(-3\right)=2\)
b) \(\frac{2}{x}=\frac{y}{10}=\frac{4}{-5}\)
=> \(\frac{2}{x}=\frac{y}{10}=\frac{-4}{5}\)
+) \(\frac{2}{x}=\frac{-4}{5}\)
=> \(\left(-4\right)\cdot x=2\cdot5\)
=> \(\left(-4\right)\cdot x=10\)
=> \(x=\frac{10}{-4}=\frac{-10}{4}=\frac{-5}{2}\)
+) \(\frac{y}{10}=\frac{-4}{5}\)
=> \(5\cdot y=-40\)
=> \(y=\left(-40\right):5=-8\)
Vậy \(x=-\frac{5}{2},y=-8\)
c) \(\frac{7}{-x}=\frac{5}{y}=\frac{2}{34}\)
=> \(\frac{-7}{x}=\frac{5}{y}=\frac{1}{17}\)
+) \(\frac{-7}{x}=\frac{1}{17}\Rightarrow x=\left(-7\right)\cdot17=-119\)
+) \(\frac{5}{y}=\frac{1}{17}\Rightarrow y=5\cdot17=85\)
Vậy \(x=-119,y=85\)
d) \(\frac{x^2}{4}=16\)
=> \(x^2=16\cdot4=64\)
=> \(x\in\left\{8;-8\right\}\)
e) \(\frac{3}{x^3}=\frac{4}{-36}\)
=> \(\frac{3}{x^3}=\frac{1}{-9}=\frac{-1}{9}\)
=> \(x^3=3:\left(-\frac{1}{9}\right)=-27\)
=> \(x=-3\)
Vậy x = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x-15=40-(3x+10)
<=> 2x-15=40-3x-10
<=> 2x-15-3x+30=0
<=> -x+15=0
<=> -x=-15
<=> x=15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^x\cdot4=324\)
\(\Leftrightarrow3^x=\dfrac{324}{4}=81\)
\(\Leftrightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(3^x\cdot4=324\)
\(\Rightarrow3^x=324:4\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
`10-3x=4`
`3x=10-4`
`3x=6`
`x=6:3`
`x=2`
Vậy `x=2`
3x= 10-4
3x= 6
X=6:3
X=2