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Lời giải:
$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$
$5^x(1+5+5^2+5^3)=88.89:2-16$
$5^x.156=3900$
$5^x=3900:156=25=5^2$
$\Rightarrow x=2$
![](https://rs.olm.vn/images/avt/0.png?1311)
Cậu ơi đề bài của cậu có phải có vấn đề không, tìm x với bt trên thì x nào thuộc Z cũng được mà?
Phần phân số cậu có thể viết rõ ra không?
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\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+15\right):3=7\)
\(x+15=7.3\)
\(x+15=21\)
\(x=21-15\)
\(x=6\)
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\(4.\left(6-x\right)=280:36\)
\(4.\left(6-x\right)=\dfrac{70}{9}\)
\(6-x=\dfrac{70}{9}:4\)
\(6-x=\dfrac{35}{18}\)
\(x=6-\dfrac{35}{18}\)
\(x=\dfrac{73}{18}\)
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\(5x-3x-12=8\)
\(2x=8+12\)
\(2x=20\)
\(x=\dfrac{20}{2}\)
\(x=10\)
\(\left(x+15\right):3=7\\ \Rightarrow x+15=7.3=21\\ \Rightarrow x=21-15=6\)
\(4\left(6-x\right)=280:36\\ \Rightarrow4\left(6-x\right)=\dfrac{70}{9}\\ \Rightarrow6-x=\dfrac{70}{9}:4\\ \Rightarrow6-x=\dfrac{35}{18}\\ \Rightarrow x=6-\dfrac{35}{18}=\dfrac{73}{18}\)
\(5x-3x-12=8\\ \Rightarrow2x=8+12=20\\ \Rightarrow x=\dfrac{20}{2}=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{5}x-\dfrac{1}{2}x=\dfrac{3}{2}-0\)
\(\left(\dfrac{3}{5}-\dfrac{1}{2}\right)x=\dfrac{3}{2}\)
\(\dfrac{1}{10}\cdot x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}:\dfrac{1}{10}\)
\(x=15\)
\(\dfrac{3}{5x}-\dfrac{1}{2x}=\dfrac{3}{2-0}\)
\(\dfrac{3}{5x}+\dfrac{-1}{2x}=\dfrac{3}{2}\)
\(\dfrac{3}{5}+\dfrac{-1}{2}=\dfrac{3}{2}\cdot x\)
\(\dfrac{1}{10}=\dfrac{3}{2}\cdot x\)
\(\dfrac{3}{2}\cdot x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}\div\dfrac{3}{2}\)
\(x=\dfrac{2}{30}\)
\(x=\dfrac{1}{15}\)
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`(x-1)/3+(3x-5)/2+(2x)/9+(-5x)/9`
`=(x-1)/3+(3x-5)/2+x/3`
`=(2x-2+9x-15+2x)/6`
`=(13x-17)/6`
3\(x^2\).(5\(x\) + 1) + 6\(x^3\).(5\(x\) + 2) = 9\(x^3\) .(5\(x\) + 3)
15\(x^3\) + 3\(x^2\) + 30\(x^4\) + 12\(x^3\) = 45\(x^4\) + 27\(x^3\)
(15\(x^3\) + 12\(x^3\)) + 3\(x^2\) + 30\(x^4\) - 45\(x^4\) - 27\(x^3\) = 0
27\(x^3\) + 3\(x^2\) - 15\(x^4\) - 27\(x^3\) = 0
3\(x^2\) - 15\(x^4\) = 0
3\(x^2\).(1 - 5\(x^2\)) = 0
\(\left[{}\begin{matrix}x^2=0\\1-5x^2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\5x^2=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=\mp\dfrac{\sqrt{5}}{5}\end{matrix}\right.\)