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a, Đặt A = \(\frac{9}{5.32}+\frac{9}{8.44}+\frac{9}{11.56}+\frac{9}{14.68}+\frac{9}{17.80}\)
\(=\frac{1}{4}\left(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+\frac{9}{14.17}+\frac{9}{17.20}\right)\)
\(=\frac{3}{4}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{3}{4}\cdot\frac{3}{20}=\frac{9}{80}\)
b, Đặt B = \(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}\)
\(=\left(1+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{6}+\frac{1}{7}\right)\)
\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}=\frac{6}{7}\)
\(5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2\)
=>\(5^x+5^x\cdot5+5^x\cdot25+5^x\cdot125=88\cdot\dfrac{\left(88+1\right)}{2}-16\)
=>\(156\cdot5^x=44\cdot89-16=3900\)
=>\(5^x=\dfrac{3900}{156}=25\)
=>x=2
Lời giải:
$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$
$5^x(1+5+5^2+5^3)=88.89:2-16$
$5^x.156=3900$
$5^x=3900:156=25=5^2$
$\Rightarrow x=2$
2⁵ˣ⁺¹ - 2⁵ˣ = 32
2⁵ˣ.(2 - 1) = 2⁵
2⁵ˣ = 2⁵
5x = 5
x = 5 : 5
x = 1
\(2^{5x+1}-2^{5x}=32\)
\(\Rightarrow2^{5x+1}-2^{5x}=2^5\)
\(\Rightarrow2^{5x}\cdot2-2^{5x}\cdot1=2^5\)
\(\Rightarrow2^{5x}\cdot\left(2-1\right)=2^5\)
\(\Rightarrow2^{5x}\cdot1=2^5\)
\(\Rightarrow2^{5x}=2^5\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=\dfrac{5}{5}\)
\(\Rightarrow x=1\)
`@` `\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.`
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
5x+5x+1+5x+2=31
5x + 5x + 5x = 31 - 2 - 1
15x = 28
x= 28/15
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
\(\left[{}\begin{matrix}x^2=0\\1-5x^2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\5x^2=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=\mp\dfrac{\sqrt{5}}{5}\end{matrix}\right.\)
a, 42x - 6 = 1
=> 42 x = 7
=> x = 6
b, 5x + 5x + 1 +5x + 2 = 775
=> 15 x + 3 = 775
=> 15 x = 772
=> x = 772/ 15
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