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\(M=\frac{2.2^{12}.3^6+2^2.2^9.3^9}{2^5.2^7.3^7+2^7.2^3.3^{10}}\)
\(=\frac{2^{11}.3^6\left(2^2+3^3\right)}{2^{10}.3^7\left(2^2+3^3\right)}\)
\(=\frac{2}{3}\)
\(M=\frac{2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9}{2^5.\left(2.3\right)^7+2^7.2^3.\left(3^2\right)^5}\)
\(M=\frac{2.2^{12}.3^6+2^2.2^9.3^9}{2^5.2^7.3^7+2^7.2^3.3^{10}}\)
\(M=\frac{2^{13}.3^6+2^{11}.3^9}{2^{12}.3^7+2^{10}.3^{10}}\)
\(M=\frac{2^{11}.3^6\left(2^2.1+1.3^3\right)}{2^{10}.3^7\left(2^2.1+1.3^3\right)}\)
\(M=\frac{2.31}{3.31}\)
\(M=\frac{2}{3}\)
Study well
\(6^x+4\cdot6^x=180\)
\(6^x\cdot\left(1+4\right)=36\cdot5\)
\(6^x\cdot5=36\cdot5\)
\(6^x=36\)
\(6^x=6^2\)
\(x=2\)
\(\frac{1}{3.4}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{3}-\frac{1}{x+1}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+1}\)
\(=\frac{1}{3}-\frac{1}{x+1}\)
1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)=3/10
<=> \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(x+1\right)x}=\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
<=> \(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)=> x+1=30=>x=29
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1=29\)
a)
4,3. x + (-1,3). x + 1,6=8,2
[4,3 + (-1,3)].x = 8,2 - 1,6
3. x = 6,6
x = 6,6:3
x = 2,2
Vậy x = 2,2.
b) -5,6. x + 16,2 = 3,4. x
-5,6. x + 16,2 = 0 + 3,4. x
-5,6. x - 3,4. x = 0 - 16,2
(-5,6 - 3,4). x = -16,2
-9. x = -16,2
x = -16,2 : -9
x = 1,8
Vậy x = 1,8.
|5.6 - \(x\)| = 4.6
|30 - \(x\)| = 24
\(x< 30\) ⇒ 30 - \(x\) = 24 ⇒ \(x\) = 30 - 24 = -6
\(x>30\) ⇒ -(30 - \(x\)) = 24 ⇒ -30 + \(x\) = 24 ⇒ \(x\) = 24 + 30 = 54
Vậy \(x\) \(\in\) { -6; 54}
`@` `\text {Ans}`
`\downarrow`
\(\left|5\cdot6-x\right|=4\cdot6\)
`\Rightarrow`\(\left|5\cdot6-x\right|=24\)
`\Rightarrow`\(\left[{}\begin{matrix}5\cdot6-x=24\\5\cdot6-x=-24\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}30-x=24\\30-x=-24\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=30-24\\x=30-\left(-24\right)\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=6\\x=54\end{matrix}\right.\)
Vậy, `x={6; 54}.`