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\(2^x.2^1=32:2=16\)
\(2^x.2^1=2^4\)
\(2^x=2^{4-1}=2^3\Rightarrow x=3\)
\(7^x.7^{12}=56-7=49\)
\(7^x.7^{12}=7^2\)
\(7^x=7^{2-12}=7^{-10}\Rightarrow x=-10\)
\(3^x.3^1=162:6=27\)
\(3^x.3^1=3^3\)
\(3^x=3^{3-1}=3^2\Rightarrow x=2\)
a)(x+810)/7-40=32*5/2
(x+810)/7-40=80
(x+810)/7=80+20
(x+810)/7=120
x+810=120*7
x+810=840
x=840-810
x=30
b)2x-15=17
2x=17+15=32=25
=>x=5
c)3x-25=56
3x=56+25=81=34
=>x=4
d)x10=1x
x=1;0
g)x10=x
x10-x=0
x(x9-1)=0
=>x=0 hoặc x9-1=0
x9=0+1=1=19
Vậy x=1 hoặc x=-1
\(x^5=32\)
\(\Rightarrow x=2\), vì \(2^5\)= 2.2.2.2.2 = 32
Vậy x = 2
\(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(\Rightarrow x=4\)
Vậy x = 4
\(2^x.7=56\)
\(2^x=56:7\)
\(2^x=8\)
\(\Rightarrow x=3\)
Vậy x = 3
a) \(x^5=32\)
\(\Rightarrow x^5=2^5\)
\(\Rightarrow x=2\)
b) \(3^x.3=243\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\Rightarrow x=4\)
c) \(2^x.7=56\)
\(\Rightarrow2^x=8\)
\(\Rightarrow2^x=2^3\Rightarrow x=3\)
a: \(=-25\cdot4-27=-100-27=-127\)
b: \(=-65:\left(-13\right)-25\cdot4=5-100=-95\)
c: \(=100\cdot\left(-47\right)+53\cdot\left(-100\right)=100\cdot\left(-100\right)=-10000\)
d: \(=\left[\left(-3\right)^{13}:3^{10}-73\right]:\left(-5\right)^2-14\)
\(=\left(-100\right):25-14=-4-14=-18\)
\(125-\left(-75\right)+32-\left(48+32\right)\)
\(=125-\left(-75\right)+32-80\)
\(=125+75+32-80\)
\(=200+32-80\)
\(=232-80\)
\(=152\)
\(-63+415-37-115\)
\(=352-57-115\)
\(=315-115\)
\(=200\)
\(13.\left(-2\right)^2+37.4+4.\left(-150\right)\)
\(=13.2^2+37.4+4.-150\)
\(=13.4+37.4+4.-150\)
\(=4.\left(13+37+-150\right)\)
\(=4.-100\)
\(=-400\)
\(-15.x+4=79\)
\(\Leftrightarrow-15.x=79-4\)
\(\Leftrightarrow-15.x=75\)
\(\Leftrightarrow x=75:-15\)
\(\Leftrightarrow x=-5\)
Vậy \(x=-5\)
\(15-\left(2+x\right)=23\)
\(\Leftrightarrow2+x=15-23\)
\(\Leftrightarrow2+x=-8\)
\(\Leftrightarrow x=-8-2\)
\(\Leftrightarrow x=-10\)
Vậy \(x=-10\)
\(\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1^2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow x=\pm1\)
Vậy \(x\in\left\{\pm1\right\}\)
\(56:\left|x-4\right|=7\)
\(\Leftrightarrow\left|x-4\right|=56:7\)
\(\Leftrightarrow\left|x-4\right|=8\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=8\\x-4=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8+4\\x=-8+4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=12\\x=-4\end{cases}}}\)
Vậy \(x\in\left\{12;-4\right\}\)
\(32^x+2^{x+2}=56\)
\(\Leftrightarrow\left(2^5\right)^x+2^{x+2}=56\)
\(\Leftrightarrow2^{5x}+2^{x+2}=56\)
\(\Leftrightarrow2^{x+2}\left(4^{2x-1}+1\right)=56\)
Ta có \(56=2^3\cdot7\)
Mà \(4^{2x-1}+1\)là số lẻ nên \(2^{x+2}=2^3\)
\(\Leftrightarrow x+2=3\)
\(\Leftrightarrow x=1\)
Thanks you :))