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đáp án
công thức tổng quát là \(\frac{x}{2^x}\)cho x chạy từ 1-10 bằng xích ma là ra 509/256
1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)
\(3.\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^{32}-1\right)\left(2^{32}+1\right)\)
=\(2^{64}-1\)
Bài 1 :
a ) Ta có :
\(\left(x+y\right)^2=x^2+y^2+2xy=20+16=36\)
b ) Ta có :
\(x^2+y^2=\left(x+y\right)^2-2xy=64-30=34\)
3(22+1).(24+1).(28+1)......(232+1)+2
=(22-1)(22+1).(24+1).(28+1)......(232+1)+2
=(24-1).(24+1).(28+1)......(232+1)+2
=(28-1).(28+1)......(232+1)+2
=....
=(232-1)(232+1)+2
=264-1+2
=264+1
3 = 4 -1 = 22 -1 thay vào ta có :
\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{32}+1\right)+2=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{32}+1\right)+2\)
= \(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+2=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+2\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)+2=2^{64}-1+2=2^{64}+1\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
Đặt \(B=\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
=>2B=1/2+1/4+...+1/512
=>B=1/2-1/1024
A=1/2-1/2+1/2024=1/1024