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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
a) \(\left(\frac{-1}{3}\right)^4=\frac{\left(-1\right)^4}{3^4}=\frac{1}{81}\)
b) \(\left(-2\frac{1}{4}\right)^3=\left(\frac{-9}{4}\right)^3=\frac{\left(-9\right)^3}{4^3}=\frac{-729}{64}\)
c) \(\left(-0,2\right)^2=\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)
d) \(\left(-5,3\right)^0=1\)
a)\(\left(\frac{-1}{3}\right)^4=\frac{1}{81}\)
b) \(\left(-2\frac{1}{4}\right)^3=\frac{-729}{64}\)
c) \(\left(-0,2\right)^2=\frac{1}{25}\)
d) \(\left(-5,3\right)^0=1\)
Cbht
\(\left(2x-1\right)\left(3x+5\right)+\left(-6x^3+5x\right):x\)
\(=6x^2+10x-3x-5-6x^3+5x:x\)
\(=-6x^3+6x^2+12x-5:x\)
\(=-6x^2+6x+12-\dfrac{5}{x}=-6\left(x^2-x-12\right)-\dfrac{5}{x}\)
A = (2\(x\) - 1)(3\(x\) + 5) + (-6\(x\)3 + 5\(x\)): \(x\)
A = 6\(x^2\) + 10\(x\) - 3\(x\) - 5 + \(x\)(- 6\(x^2\) + 5): \(x\)
A = 6\(x^2\) + 7\(x\) - 5 - 6\(x^2\) + 5
A = (6\(x^2\) - 6\(x^2\)) + 7\(x\) - (5 - 5)
A = 7\(x\)
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!
\(\left(1-2x\right)^2=9\)
\(\Rightarrow\left(1-2x\right)^2=3^2\)
\(\Rightarrow1-2x=3\)
\(\Rightarrow-2x=2\)
\(\Rightarrow x=-1\)