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4)
a) Ta có \(2^{10}+2^{11}+2^{12}\)
\(=2^{10}\left(1+2+4\right)=2^{10}\cdot7⋮7\)
Vậy: \(2^{10}+2^{11}+2^{12}\) chia hết cho 7(đpcm)
b) Ta có: 7*32=224=25+26+27
1: \(4a^2b^4-c^4d^2\)
\(=\left(2ab^2-c^2d\right)\left(2ab^2+c^2d\right)\)
4: \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(3a^2+b^2\right)\)
5: \(\left(a+b\right)^3+\left(a-b\right)^3\)
\(=a^3+b^3+3a^2b+3ab^2+a^3-3a^2b+3ab^2-b^3\)
\(=2a^3+6ab^2\)
\(=2a\left(a^2+3b^2\right)\)
a) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{1}{x^2+x+1}\)
b) \(\dfrac{9}{x^3-9x}-\dfrac{-1}{x+3}\)
\(=\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\)
c) \(\dfrac{x^3-8}{5x+10}.\dfrac{x^2+4x}{x^2+2x+4}\)
\(=\dfrac{x\left(x-2\right)\left(x^2+2x+4\right)\left(x+4\right)}{5\left(x+2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{x\left(x-2\right)\left(x+4\right)}{5\left(x+2\right)}\)
d) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}.\dfrac{2\left(2-x\right)}{x+2}\)
\(=-\dfrac{10\left(x+2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{5}{2}\)
e) \(\dfrac{\left(x-13\right)^2}{2x^5}.\dfrac{-3x^2}{x-13}\)
\(=\dfrac{x-13}{2x^3}.\dfrac{-3}{1}\)
\(=\dfrac{-3\left(x-13\right)}{2x^3}\)
g) \(\dfrac{x^2+6x+9}{1-x}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2}{x-1}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2\left(x-1\right)^2}{2\left(x-1\right)\left(x+3\right)^2}\)
\(=-\dfrac{x-1}{2}\).
Mk mak lm mấy cái con toán này
Thì mk sáng mai cx ko lm xng đôu
Bấm mẹ máy tính cho nó nhanh
K mk
\(a)\)\(\frac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\)\(\frac{2^{12}.5^{12}+2^9.5^{11}-2^8.5^{13}}{2^8.5^{11}}\)
\(=\)\(\frac{2^8.5^{11}\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=\)\(2^4.5+2-5^2\)
\(=\)\(80+2-25\)
\(=\)\(57\)
Chúc bạn học tốt ~
Ta có 99 = 100 - 1 = x - 1
Thay vào A, ta có :
A = -x10 + (x-1)x9 + (x-1)x8 + ...... + (x-1)x + 99
= -x10 + x10 - x9 + x9 - x8 + ........+ x2 - x + 99
= -x + 99
Thay x = 10 vào A ta có :
A = -100 + 99 = -1
1.
= 4x\(^{^{ }2}\)-4x-9x+9
=4x(x-1)-9(x-1)
=(4x-9)(x-1)
a, 5^6 -10^4=5^2. 5^4 -5^4. 2^4
=5^4(5^2 -2^4)
=5^4. 9 \(⋮\) 9
b, (n+3)2- (n -1)2=(n+3- n+1)(n+3+ n- 1)
=4(2n+2)
=8n+ 8\(⋮8\)
\(A=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(A=2^2-1^2+4^2-3^2+...+100^2-99^2\)
\(A=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)
\(A=1\left(1+2\right)+1\left(3+4\right)+....+1\left(99+100\right)\)
\(A=1+2+3+4+....+99+100\)
A=5050
\(B=3^8.7^8-\left(21^4-1\right)\left(21^4+1\right)\)
\(B=\left(3.7\right)^8-\left(21^8-1\right)\)
\(B=21^8-21^8+1\)
B=1
mà A=5050
⇒ A>B
a: =(-8*10^5+10^3):100
=-8*1000+10
=-8000+10=-7990
b: =(2*3^6+3^8-4*3^6):3^4
=2*3^2+3^4-4*3^2
=2*9+81-4*9
=63