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\(2^1.2^2.2^3.....2^x=1024\Rightarrow2^{1+2+3+...+x}=2^{10}\)
\(\Rightarrow1+2+3+...+x=1024\Rightarrow x=4\)
\(\left(\frac{1}{3}+\frac{1}{6}\right).2^{x+4}-2^x=2^{13}-2^{10}\)
\(\frac{1}{2}.2^x.2^4-2^x=8192-1024\)
\(2^x.8-2^x=7168\)
\(2^x\left(8-1\right)=7168\)
\(2^x.7=7168\)
\(2^x=7168\div7\)
\(2^x=1024\)
\(2^x=2^{10}\)
\(\Rightarrow x=10\)
Vậy \(x=10\).
(1/3+1/6).2^x.2^4-2^x=8192-1024
(1/3+1/6).2^x.2^4-2^x=7168
1/2.2^x.2^4-2^x=7168
1/2.2^x.(2^4-1)=7168
1/2.2^x.(8-1)=7168
1/2.2^x.7=7168
1/2.2^x=7168:7
1/2.2^x=1024
2^x=1024:1/2
2^x=2048
2^x=2^11
x=11
vậy x=11
\(4^{x+1}.2=32\)
\(4^{x+1}=32:2\)
\(4^{x+1}=16\)
\(4^{x+1}=4^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
vậy \(x=1\)
\(\left(x-\frac{2}{3}\right)^2=\frac{25}{81}\)
\(\left(x-\frac{2}{3}\right)^2=\left(\frac{5}{9}\right)^2\)
\(\Rightarrow x-\frac{2}{3}=\frac{5}{9}\)
\(\Rightarrow x=\frac{11}{9}\)
vậy \(x=\frac{11}{9}\)
\(500^{300}=\left(500^3\right)^{100}=125000000^{100}\)
\(300^{500}=\left(300^5\right)^{100}\)
vì \(\left(500^3\right)^{100}< \left(300^3\right)^{100}\)nên\(500^{300}< 300^{500}\)
\(4^{45}=\left(4^9\right)^5=262144^5\)
\(3^{60}=\left(3^{12}\right)^5=531441^5\)
vì \(262144^5< 531441^5\) nên \(4^{45}< 3^{60}\)
\(\frac{x}{6}+\frac{7}{3.2^2}=\frac{17}{18}-\frac{1}{3^2}\)
\(\frac{x}{6}+\frac{7}{12}=\frac{5}{6}\)
\(\frac{x}{6}=\frac{5}{6}-\frac{7}{12}\)
\(\frac{x}{6}=\frac{1}{4}\)
\(x=\frac{1}{4}.6\)
\(x=\frac{3}{2}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\) vậy
\(\frac{11}{45}.x=\frac{22}{45}\)
\(x=\frac{22}{45}\div\frac{11}{45}=2\)
vậy suy ra x =2
mình chắc chắn 100% luôn đó, cái này ở trong violympic toán 7 vòng 12 phải ko
\(\Rightarrow2^x\left(\frac{1}{2}+4\right)=288\Rightarrow2^x.\frac{9}{2}=288\Rightarrow2^x=64=2^6\Rightarrow x=6\)
\(\)\(\left(\frac{1}{2}+4\right).2^x=288\)
\(\frac{9}{2}.2^x=288\)
\(2^x=64\)
\(2^x=2^6\)
=> x=6