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Có: 1024=2^10
=> 2.2^2.2^3......2^x=2^10
=> 1+2+3+...+x=10
1+2+3+...+x=1+2+3+4
=>x=4
Vậy x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 )
a ) \(2.2^2.2^3.....2^x=1024\Leftrightarrow2^{1+2+....+x}=2^{10}\Leftrightarrow1+2+....+x=10\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=10\Leftrightarrow\left(x+1\right)x=20=4.5\Rightarrow x=4\)
b ) \(\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow3x+39=259-7x\Leftrightarrow3x+7x=259-39\Leftrightarrow10x=220\Rightarrow x=22\)
Bài 2 ) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(\frac{50^2-15.125}{5^4}\right)^{2014}=\frac{1}{2}.8-\frac{2}{5}+\left(\frac{5^4.2^2-3.5^4}{5^4}\right)^{2014}\)
\(=4-\frac{2}{5}+\left[\frac{5^4\left(4-3\right)}{5^4}\right]^{2014}=\frac{18}{5}+1=\frac{23}{5}\)
Mình làm bài 1 thui nha, còn bài 2 thì còn tự tính là được thôi mừ !!!
Bài 1:
a) \(2.2^2.2^3...2^x=1024\)
\(=>2^{1+2+3+...+x}=2^{10}\)
\(< =>1+2+3+...+x=10\)
\(=>6+x=10\)
\(=>x=10-6\)
\(=>x=4.\)
Nếu đúng thì k cho mình nhá
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(\frac{2}{5}\right)^{2014}:\left(\frac{4}{25}\right)^{1007}=\left[\left(\frac{2}{5}\right)^2\right]^{1007}:\left(\frac{4}{25}\right)^{1007}\)
\(=\left(\frac{4}{25}\right)^{1007}:\left(\frac{4}{25}\right)^{1007}\)
\(=1\)
b)\(3^{n+1}:9=3^{n+1}:3^2\)
\(=3^{n+1-2}\)
\(=3^{n-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3-2 . 34 . 3x = 37
<=> 32 + x = 37
<=> 2 + x = 7
<=> x = 7 - 2
<=> x = 5
\(2^{-2}\cdot2^x+2\cdot2^x=9\cdot2^6\\ \Rightarrow2^{x-2}+2^{x+1}=9\cdot2^6\\ \Rightarrow2^{x-2}\left(1+2^3\right)=9\cdot2^6\\ \Rightarrow2^{x-2}\cdot9=9\cdot2^6\Rightarrow2^{x-2}=2^6\\ \Rightarrow x-2=6\Rightarrow x=8\)
\(3^{-2}\cdot3^4\cdot3^x=3^7\\ \Rightarrow3^{x+4-2}=3^7\\ \Rightarrow x+2=7\Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
$(3^2-2^3)x+3^2.2^2=4^2.3$
$\Leftrightarrow x+36=48$
$\Leftrightarrow x=48-36=12$
2.
$x^5-x^3=0$
$\Leftrightarrow x^3(x^2-1)=0$
$\Leftrightarrow x^3(x-1)(x+1)=0$
$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$
$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.
$(x-1)^2+(-3)^2=5^2(-1)^{100}$
$\Leftrightarrow (x-1)^2+9=25$
$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$
$\Rightarrow x-1=4$ hoặc $x-1=-4$
$\Leftrightarrow x=5$ hoặc $x=-3$
4.
$(2x-1)^2-(2x-1)=0$
$\Leftrightarrow (2x-1)(2x-1-1)=0$
$\Leftrightarrow (2x-1)(2x-2)=0$
$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$
$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$
$\Lef
`@` `\text {Ans}`
`\downarrow`
\((3^2-2^3)x+3^2.2^2=4^2.3\)
`=> x + (3*2)^2 = 48`
`=> x+6^2 = 48`
`=> x + 36 = 48`
`=> x = 48 - 36`
`=> x=12`
Vậy, `x=12`
\(x^5-x^3=0\)
`=> x^3(x^2 - 1)=0`
`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
Vậy, `x \in {0; +- 1 }`
\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)
`=> (x-1)^2 + 9 = 25*1`
`=> (x-1)^2 + 9 = 25`
`=> (x-1)^2 = 25 - 9`
`=> (x-1)^2 = 16`
`=> (x-1)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy, `x \in {5; -3}`
\((2x-1)^2-(2x-1)=0\)
`=> (2x-1)(2x-1) - (2x-1)=0`
`=> (2x-1)(2x-1-1)=0`
`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`3^3 * x^2 - 2^4 * x^2 = 8^2 * 5 - 4^2 * 3^2`
`=> x^2 . (3^3 - 2^4) = 2^6 . 5 - 2^4 . 3^2`
`=> x^2 . 11 = 2^4 . (2^2 . 5 - 3^2)`
`=> x^2 . 11 = 2^4 . 11`
`=> x^2 . 11 - 2^4 . 11 = 0`
`=> 11 . (x^2 - 16) = 0`
`=> x^2 - 16 = 0`
`=> x^2 = 16`
`=> x^2 = (+-4)^2`
`=> x = `\(\pm4\)
Vậy, `x \in`\(\left\{4;-4\right\}\)
_____
\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2\cdot2^2=4^2\cdot3\)
`=>`\(\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+\left(3\cdot2\right)^2=48\)
`=>`\(\dfrac{23}{108}\cdot x+6^2=48\)
`=>`\(\dfrac{23}{108}x=48-6^2\)
`=>`\(\dfrac{23}{108}x=48-36\)
`=>`\(\dfrac{23}{108}x=12\)
`=>`\(x=\dfrac{1296}{23}\)
Vậy, `x = `\(\dfrac{1296}{23}\)
\(3^3.x^2-2^4.x^2=8^2.5-4^3.3^2\)
\(\Leftrightarrow x^2\left(27-16\right)=2^6.5-2^6.9\)
\(\Leftrightarrow11x^2=2^6.\left(5-9\right)=-4.2^6=-2^8\)
\(\Leftrightarrow x^2=-\dfrac{2^6}{11}< 0\)
\(\Rightarrow x\in\varnothing\)
\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2.2^2=4^2.3\)
\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+36=48\)
\(\Leftrightarrow\dfrac{23}{108}x=12\Leftrightarrow x=\dfrac{12.108}{23}=\dfrac{1296}{23}\)
\(2^{1+2+3+...+x}=2014\)
\(1+2+3+...+x=10\)
x=4