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20 tháng 11 2019
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
HN
1 tháng 1 2018
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) \(\dfrac{x^2+6x+9}{1-x}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2}{x-1}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2\left(x-1\right)^2}{2\left(x-1\right)\left(x+3\right)^2}\)
\(=-\dfrac{x-1}{2}\).
24 tháng 5 2022
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: \(\left(a+b\right)^3+\left(a-b\right)^3\)
\(=a^3+b^3+3a^2b+3ab^2+a^3-3a^2b+3ab^2-b^3\)
\(=2a^3+6ab^2\)
\(=2a\left(a^2+3b^2\right)\)
22 tháng 9 2019
1, -3x4y + 6x3y - 3x2y
= -3x2y (x2 - 2x + 1)
= -3x2y(x - 1)2
2, 12x2 - 12xy + 3y2
= 3(4x2 - 4xy + y2)
= 3(2x - y)2
3, 20x4y2 - 20x3y3 + 5x2y4
= 5x2y2(4x2 - 4xy + y2)
= 5x2y2(2x - y)2
4, 16x5y2 - 16x4y3 + 4x3y4
= 4x3y2(4x2 - 4xy + y2)
= 4x3y2(2x - y)2
5, -12x4y + 12x3y2 - 3x2y3
= -3x2y(4x2 - 4xy + y2)
= -3x2y(2x - y)2
6, (a2 + 4)2 - 16a2
= (a2 + 4 - 4a)(a2 + 4 - 4a)
7, (a2 + 9)2 - 36a2
= (a2 + 32)2 - (6a)2
= (a2 + 32 - 6a)(a2 + 32 + 6a)
= (a2 - 6a + 9)(a2 + 6a + 9)
8, (a2 + 4b2)2 - 16a2b2
= (a2 + 4b2 - 4ab)(a2 + 4b2 + 4ab)
= (a2 - 4ab + 4b2)(a2 + 4ab + 4b2)
= (a - 2b)2(a + 2b)2
= (a2 - 4b2)4
Câu này có sai thì bạn thông cảm nhá!!!
9, 36a2 - (a2 + 9)2
= (6a)2 - (a2 + 9)2
=- (a2 - 6a + 9)(a2 + 6a + 9)
= -(a - 3)2(a + 3)2
= -(a2 - 9)4
Câu 10 giống câu 8 bạn nhé
8 tháng 4 2020
a) \(\left(x+2\right)^2-9\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(3x-6\right)^2=0\)
\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x-4\right)=0\\\left(8-2x\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b)\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{23}{7}\end{matrix}\right.\)
c) \(\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-5x+18\right)\left(8x^2+8x+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=3\end{matrix}\right.\)
15 tháng 8 2017
1.
= 4x\(^{^{ }2}\)-4x-9x+9
=4x(x-1)-9(x-1)
=(4x-9)(x-1)
4 tháng 9 2017
\(1.a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
=\(a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
=\(a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
=\(a^4-b^4\)=\(\left(a^2-b^2\right)\left(a^2+b^2\right)\)
HM
29 tháng 5 2018
\(A=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(A=2^2-1^2+4^2-3^2+...+100^2-99^2\)
\(A=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)
\(A=1\left(1+2\right)+1\left(3+4\right)+....+1\left(99+100\right)\)
\(A=1+2+3+4+....+99+100\)
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)2
b)63
☹