tìm x :
a. 2x = 16
b. 2x = 64
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c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
a) \(2^x\cdot4=16\)
\(\Rightarrow2^x=16:4\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
b) \(3^x\cdot3=243\)
\(\Rightarrow3^x=243:3\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)
\(\Rightarrow2x^2+x-6-4x^2+22x-10=-16\)
\(\Rightarrow2x^2-23x=0\Rightarrow x\left(2x-23\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)
b) \(7x^2-7=x^2-2x+1\)
\(\Rightarrow7\left(x^2-1\right)-\left(x^2-2x+1\right)=0\)
\(\Rightarrow7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)\left(7x+7-x+1\right)=0\Rightarrow2\left(x-1\right)\left(3x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{3}\end{matrix}\right.\)
a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)
\(2x^2+x-6-4x^2+22x-10=-16\)
\(-2x^2+23x-16=-16\)
\(23x-2x^2=0\)
\(x\left(23-2x\right)=0\)
⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)
b) \(7x^2-7=x^2-2x+1\)
\(7\left(x^2-1\right)=\left(x-1\right)^2\)
\(7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)
\(\left(7x+7\right)\left(x-1\right)-\left(x-1\right)^2=0\)
\(\left(x-1\right)\left(7x+7-x+1\right)=0\)
\(\left(x-1\right)\left(6x+8\right)=0\)
⇔ \(\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left(4x-1\right)^2=25=5^2=\left(-5\right)^2\\ \Rightarrow\left[{}\begin{matrix}4x-1=5\\4x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\\ b,\Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:9=16=2^4\Rightarrow x=4\\ c,\Rightarrow3^{2x+3}=3^{2\left(x+3\right)}\\ \Rightarrow2x+3=2x+6\Rightarrow0x=3\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)
a)
\(13x+3=16\\ 13x=16-3\\ 13x=13\\ x=13:13\\ x=1\)
b)
\(2x-138=2^4:2^3\\ 2x-138=2\\ 2x=2+138\\ 2x=140\\ x=140:2\\ x=70\)
a)
13x+3=16
13x =16−3
13x =13
x =13:13
x=1
b)
2x−138=2^4:2^3
2x−138=2
2x =2+1382
x =140
x =140:2
x =70
a) \(2^x:4=16\\ \Rightarrow2^x=64\\ \Rightarrow2^x=2^6\\ \Rightarrow x=6\)
b) \(4^{x-3}=256\\ \Rightarrow4^{x-3}=4^4\\ \Rightarrow x-3=4\\ \Rightarrow x=7\)
c) \(\left(2x+1\right)^3=343\\ \Rightarrow\left(2x+1\right)^3=7^3\\ \Rightarrow2x+1=7\\ \Rightarrow x=3\)
d) \(10+2x=4^5:4^3\\ \Rightarrow10+2x=16\\ \Rightarrow x=3\)
a,2^x:4=16
2^x=16.4=64
2^x=2^6
=>x=6
b,4^x-3=256
4^x-3=4^4
=>x-3=4
x=4+3=7
c,(2x+1)^3=343
(2x+1)^3=7^3
=>2x+1=7
2x=7-1=6
x=6:2=3
d,10+2x=4^5:4^3
10+2x=4^2=16
2x=16-10=6
x=6:2=3