(5^2x-3 ) - 3.5^2 = 5^2.2
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314 : ( 39.27 ) + 52.25 - 43
= 314 : ( 39.33) + 52.25 - ( 22)3
= 314 : 312 + 52.25 - 26
= 32 + 25.(25 - 2)
= 9 + 32.23
= 9 + 736
=745
23.5 - 32.4 + 4.6
= 23.5 - 32.22 + 22.2.3
= 22.( 2.5 - 9 + 2.3)
= 4 .( 10 - 9 + 6 )
= 4 . 7
= 28
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =5-78*32
=5-2496
=-2491
b: \(=6\left(9-6\right)=6\cdot3=18\)
c: \(=46\cdot\dfrac{\left(123-42\right)}{81}=46\)
d: \(=181+3-84+8\cdot25\)
=100+200
=300
e: \(=64\cdot35+140\cdot84-1=2240-1+11760\)
=14000-1
=13999
f: \(=3^3+25\cdot8-1=26+200=226\)
g: \(=3+2^4+1=16+4=20\)
h: \(=36:4\cdot3+2\cdot25-1=27+50-1=27+49=76\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. 25 : 5,7 = 250/57
2. 30:2.8.4 = 480
3. 20:2^2.14= 70
4. 125:5^3.170= 170
5. 64:2^5.30.4=240
6. (25:5^2.30): 15.7=14
bạn à! Nhiều quá mình ko làm hết được. sorry nha.^-^
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a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)
b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4^2.2^5+32.125}{2^3.5^2}\) \(=\)\(\frac{4^2.2^5+2^5.5^3}{2^3.5^2}\)
\(=\frac{\left(4^2+5^3\right).2^5}{2^3.5^2}\)
\(=\) \(\frac{\left(4^2+5^3\right).2^2}{5^2}\)
\(=\) \(\frac{564}{25}\)
\(\frac{4^2.2^5+32.125}{2^3.5^2}=\frac{4^2.2^5+2^5.5^3}{2^3.2^5}\)
\(=\frac{(4^2+5^3).2^5}{2^3.2^5}\)
\(=\frac{(4^2+5^3).2^2}{5^2}\)
\(=\frac{564}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(6^2.2^2.5:\left[3.12-\left(2x-6\right)\right]=2^3.5\)
\(\Rightarrow36.4.5:\left[36-\left(2x-6\right)\right]=8.5\)
\(\Rightarrow720:\left[36-\left(2x-6\right)\right]=40\)
\(\Rightarrow36-\left(2x-6\right)=720:40=18\)
\(\Rightarrow2x-6=36-18=18\)
\(\Rightarrow2x=18+6=24\)
\(\Rightarrow x=24:2=12\)
![](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)
Ta có: \(5^2\cdot x-3-2\cdot2^5=5^2\cdot3\)
\(\Leftrightarrow25\cdot x-3-64=75\)
\(\Leftrightarrow25\cdot x=122\)
hay \(x=\dfrac{122}{25}\)
Vậy: \(x=\dfrac{122}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^2.2^4-2^3.4^2-2^3.5^0+4^2.2^2\)
\(=9.16-8.16-8.1+16.4\)
\(=16.\left(9-8\right)-8+64\)
\(=16.1-8+64\)
\(=8+64\)
\(=72\)
5^2x-3=50+75
5^2x=125
=>x=125:25
=>x=5
\(\left(5^2x-3\right)-3.5^2=5^2.2\)
<=>\(\left(25x-3\right)-3.25=25.2\)
<=>\(\left(25x-3\right)-75=50\)
<=>\(\left(25x-3\right)=125\)
<=>\(25x-3=125\)
<=>\(x-3=125:25\)
<=>\(x-3=5\)
<=>\(x=8\)