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\(2x^2-72=0\Leftrightarrow2\left(x^2-36\right)=0\Leftrightarrow x^2=36\Leftrightarrow x=-6; x=6\)
\(\frac{1}{2}x+\frac{2}{3}\left(x-1\right)=\frac{1}{3}\)
\(\frac{1}{2}x+\frac{2}{3}x-\frac{2}{3}=\frac{1}{3}\)
\(\frac{7}{6}x=\frac{1}{3}+\frac{2}{3}\)
\(x=\frac{6}{7}\)
\(a,\Rightarrow2\left(x+5\right)-12=-20\\ \Rightarrow2\left(x+5\right)=-8\\ \Rightarrow x+5=-4\Rightarrow x=-9\\ b,\Rightarrow x-28+x=-24\\ \Rightarrow2x=4\Rightarrow x=2\\ c,\Rightarrow6\left(x-2\right)^3=-384\\ \Rightarrow\left(x-2\right)^3=-64=\left(-4\right)^3\\ \Rightarrow x-2=-4\Rightarrow x=-2\\ d,\Rightarrow2^x\left(1+2^3\right)=72\\ \Rightarrow2^x\cdot9=72\\ \Rightarrow2^x=8=2^3\Rightarrow x=3\)
a,⇒2(x+5)−12=−20⇒2(x+5)=−8⇒x+5=−4⇒x=−9b,⇒x−28+x=−24⇒2x=4⇒x=2c,⇒6(x−2)3=−384⇒(x−2)3=−64=(−4)3⇒x−2=−4⇒x=−2d,⇒2x(1+23)=72⇒2x⋅9=72⇒2x=8=23⇒x=3a,⇒2(x+5)−12=−20⇒2(x+5)=−8⇒x+5=−4⇒x=−9b,⇒x−28+x=−24⇒2x=4⇒x=2c,⇒6(x−2)3=−384⇒(x−2)3=−64=(−4)3⇒x−2=−4⇒x=−2d,⇒2x(1+23)=72⇒2x⋅9=72⇒2x=8=23⇒x=3
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a) \(\frac{5}{12}=\frac{x}{72}\Rightarrow x=\frac{5.72}{12}=30\)
b)\(\frac{x+3}{15}=\frac{1}{3}\)
\(\Rightarrow x+3=\frac{15.1}{3}=5\)
\(\Rightarrow x=5-3=2\)
c)\(\frac{-2}{9}=\frac{x}{72}\Rightarrow x=\frac{-2.72}{9}=-16\)
\(\Leftrightarrow\left[3x+6-23\right]\cdot\left(9-8+20\cdot0\right)=147\)
=>3x=164
hay x=164/3
\(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.71\)
\(72^{44}-72^{43}=72^{43}.\left(72-1\right)=72^{43}.71\)
Vì \(72^{44}>72^{43}\Rightarrow72^{45}-72^{44}>72^{44}-72^{43}\)
\(2^x+2^{x+3}=72\)
\(\Rightarrow2^x\cdot\left(1+2^3\right)=72\)
\(\Rightarrow2^x\cdot9=72\)
\(\Rightarrow2^x=72:9\)
\(\Rightarrow2^x=8\)
\(\Rightarrow2^x=2^3\)
\(\Rightarrow x=3\)
Lời giải:
$2^x+2^{x+3}=72$
$2^x(1+2^3)=72$
$2^x.9=72$
$2^x=72:9=8=2^3$
$\Rightarrow x=3$
2\(^x\) + 2\(^x\)+3 = 72
2\(^x\) + 2\(^x\).23 = 72
2\(^x\).(1 + 23) = 72
2\(^x\).(1+ 8) = 72
2\(^x\).9 = 72
2\(^x\) = 72 : 9
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)