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![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2^{x+2}-2^x=96\)
\(=>2^x.2^2-2^x=96\)
\(=>2^x.\left(4-1\right)=96\)
\(=>2^x.3=96\)
\(=>2^x=96:3=32\)
\(=>2^x=2^5\)
\(=>x=5\)
\(b,720:\left[41.\left(2x-5\right)\right]=2^3.125:5^2\)
\(=>720:\left[41.\left(2x-5\right)\right]=8.125:25\)
\(=>720:\left[41.\left(2x-5\right)\right]=8.5=40\)
\(=>41.\left(2x-5\right)=720:40=18\)
\(=>2x-5=18:41=\frac{18}{41}\)
\(=>2x=\frac{18}{41}+5=\frac{223}{41}\)
\(=>x=\frac{223}{41}:2=\frac{223}{62}\)
\(c,\left(-2x+7\right)^{19}=\left(-2x+7\right)^{19}.\left(x+1\frac{1}{4}\right)^2\)
\(=>\left(-2x+7\right)^{19}:\left(-2x+7\right)^{19}=\left(x+\frac{5}{4}\right)^2\)
\(=>1=\left(x+\frac{5}{4}\right)^2\)
\(=>1^2=\left(x+\frac{5}{4}\right)^2\)
\(=>1=x+\frac{5}{4}\)
\(=>x=1-\frac{5}{4}=-\frac{1}{4}\)
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