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1)Tính:
a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)
b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)
c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)
a
\(4^2.2=16.2=32\)
b\(36^2:6^2=36.36:6.6=36.36:36=36\)
c
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THEO ĐỀ BÀI TA CÓ
1^2+2^2+3^2+...+10^2=385
MÀ 2^2+4^2+....+20^2=2(1^2+2^2+....+10^2)=2.385=770
VẬY 2^2+2^4+....+20^2=770
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a) \(\left(x-\frac{1}{2}\right)^3=27\)
=> \(\left(x-\frac{1}{2}\right)^3=3^3\)
=> \(x-\frac{1}{2}=3\)
=> \(x=3+\frac{1}{2}\)
=> \(x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}.\)
b) \(\left(2x-1\right)^3=-27\)
=> \(\left(2x-1\right)^3=\left(-3\right)^3\)
=> \(2x-1=-3\)
=> \(2x=\left(-3\right)+1\)
=> \(2x=-2\)
=> \(x=\left(-2\right):2\)
=> \(x=-1\)
Vậy \(x=-1.\)
Chúc bạn học tốt!
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Ta có : (x - 1)2 = (1 - x)3
=> (x - 1)2 = -(x - 1)3
=> (x - 1)2 + (x - 1)3 = 0
=> (x - 1)2(x - 1 + 1) = 0
=> (x - 1)2.x = 0
=> \(\orbr{\begin{cases}x-1=0\\x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy \(x\in\left\{1;0\right\}\)là giá trị cần tìm
\(\left(x-1\right)^2=x^2-2x+1\)
\(\left(1-x\right)^3=\left(-x+1\right)^3=-x^3+2x^2-2x+1\)
\(x^2-2x+1=-x^3+2x^2-2x+1\)
\(x^3=x^2\)
\(x=\left\{0;1\right\}\)