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`@` `\text {Ans}`
`\downarrow`
`x^3 = 125`
`=> x^3 = 5^3`
`=> x = 5`
Vậy, `x = 5`
_____
`4^5 \div 4^3`
`=`\(4^{5-3}\)
`= 4^2`
______
\(8^{12}\cdot8^3\div8^{15}\)
`=`\(8^{12+3-15}\)
`=`\(8^0=1\)
______
\(7^{15}\div7^{13}\)
`=`\(7^{15-13}\)
`=`\(7^2\)
_______
\(17^{15}\cdot17^{10}\div17^{22}\)
`=`\(17^{15+10-22}\)
`= 17^3`
_______
\(5^{10}\cdot25\div5^{11}\)
`=`\(5^{10}\cdot5^2\div5^{11}\)
`=`\(5^{10+2-11}\)
`= 5`
_______
\(15^{10}\cdot225\div15^{11}\)
`=`\(15^{10}\cdot15^2\div15^{11}\)
`=`\(15^{12}\div15^{11}\)
`= 15`
`@` `\text {Kaizuu lv uuu}`
\(x^3=125\)
\(\Rightarrow x^3=5^3\)
\(\Rightarrow x=3\)
___________________
\(4^5:4^3\)
\(=4^{5-3}\)
\(=4^2=16\)
___________________
\(8^{12}.8^3:8^{15}\)
\(=8^{15}:8^{15}\)
\(=1\)
___________________
\(7^{15}:7^{13}\)
\(=7^{15-13}\)
\(=7^2=49\)
___________________
\(17^{15}.17^{10}:17^{22}\)
\(=17^{25}:17^{22}\)
\(=17^3=4913\)
___________________
\(5^{10}.25:5^{11}\)
\(=5^{10}.5^2:5^{11}\)
\(=5^{12}:5^{11}\)
\(=5^1=5\)
___________________
\(15^{10}.225:15^{11}\)
\(=15^{10}.15^2:15^{11}\)
\(=15^{12}:15^{11}\)
\(=15^1=15\)
__________________
* an . am = an+m
an : am = an-m
Chúc bạn học tốt
a) 2x . 4 = 128
<=> 2x = 32
<=> 2x = 25
<=> x = 5
b) x15 = x1
<=> x15 - x = 0
<=> x(x14 - 1) = 0
<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
c) (2x + 1)3 = 125
<=> (2x + 1)3 = 53
<=> 2x + 1 = 5
<=> 2x = 4
<=> x = 2
d) (x - 5)4 = (x - 5)6
<=> (x - 5)6 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)2 - 1] = 0
<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
Khi (x - 5)4 = 0 => x - 5 = 0 => x = 5
Khi (x - 5)2 - 1 = 0 <=> (x - 5)2 = 12 <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
\(x=\left(\frac{33}{5}:6-0,125\cdot8+\frac{32}{15}\cdot0,03\right)\cdot\frac{11}{4}\)
\(x=\left(\frac{33}{5}\cdot\frac{1}{6}-\frac{1}{8}\cdot8+\frac{32}{15}\cdot\frac{3}{100}\right)\cdot\frac{11}{4}\)
\(x=\left(\frac{11}{10}-1+\frac{8}{125}\right)\cdot\frac{11}{4}\)
\(x=\frac{41}{250}\cdot\frac{11}{4}\)
\(x=\frac{451}{100}\)
\(x=\left(\frac{33}{5}:6-0,125.8+\frac{32}{15}.0,03\right).\frac{11}{4}\)
\(< =>x=\left(\frac{33}{30}-1+\frac{32}{15}.\frac{3}{100}\right).\frac{11}{4}\)
\(< =>x=\left(\frac{3}{30}+\frac{32}{500}\right).\frac{11}{4}\)
\(< =>x=\left(\frac{1}{10}+\frac{8}{125}\right).\frac{11}{4}\)
\(< =>x=\frac{41}{250}.\frac{11}{4}=\frac{451}{100}\)
a. 8 + (x - 9) = 125 - 64
8 + (x - 9) = 61
x - 9 = 53
x = 62
b. 5 x (X + 7) - 10 = 8 x 5
5 x (X + 7) - 10 = 40
5 x (X + 7) = 50
X + 7 = 10
X = 3
\(125-5\left(4+x\right)=15\)
\(5\left(4+x\right)=125-15\)
\(5\left(4+x\right)=110\)
\(4+x=110:5\)
\(4+x=22\)
\(x=22-4\)
\(x=18\)
\(\left(x+15\right)-125=5^7:5^5\)
\(\Rightarrow\left(x+15\right)-125=5^2\)
\(\Rightarrow\left(x+15\right)-125=25\)
\(\Rightarrow x+15=25+125\)
=> x+15=150
=> x=150-15
Vậy x=135.
\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
=> 720:[41-(2x-5)]=8.5
=> 720:[41-(2x-5)]=40
=> 41-(2x-5)=720:40
=> 41-(2x-5)=18
=> 2x-5=41-18
=> 2x-5=23
=> 2x=23+5
=> 2x=28
=> x=28:2
Vậy x=14.
\(2^7-3.\left(x+4\right)=23\)
=> 128-3.(x+4)=23
=> 3.(x+4)=128-23
=> 3.(x+4)=105
=> x+4=105:3
=> x+4=35
=> x=35-4
Vậy x=31.
\(2015^x.5^3=125\)
=> 2015x.53=53
=> 2015x=53:53
=> 2015x=1
=> 2015x=20150
Vậy x=0.
\(a.\)\(10x+2^2.5=10\)
\(\Leftrightarrow10x+4.5=10\)
\(\Leftrightarrow10x+20=10\)
\(\Leftrightarrow10x=10-20\)
\(\Leftrightarrow10x=-10\)
\(\Leftrightarrow x=-10:10\)
\(\Leftrightarrow x=-1\)
\(b.\)\(125-5\left(4+x\right)=15\)
\(\Leftrightarrow5\left(4+x\right)=125-15\)
\(\Leftrightarrow5\left(4+x\right)=110\)
\(\Leftrightarrow4+x=110:5\)
\(\Leftrightarrow4+x=22\)
\(\Leftrightarrow x=22-4\)
\(\Leftrightarrow x=18\)
\(c.\)\(2^6+\left(218-x\right)=73\)
\(\Leftrightarrow64+\left(218-x\right)=73\)
\(\Leftrightarrow218-x=73-64\)
\(\Leftrightarrow218-x=9\)
\(\Leftrightarrow x=218-9\)
\(\Leftrightarrow x=209\)
5( 4 + x ) = 125 - 15
5( 4 + x ) = 110
4 + x = 110 : 5
4 + x = 22
x = 22 - 4
x = 8