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a) 4x^2 - 12xy + 9y^2
=(2x)^2 - 2.2.3xy + (3y)^2
=(2x+3y)^2
b) 27a^3 - 64b^3
=(3a)^3 - (4b)^3
=(3a - 4b) [(3a)^2 +3a.4b +(4B)^2]
d) (2x - 6y)^2 - (3xy - 4)^2
=[ (2x - 6y)+ (3xy - 4) ] [ (2x - 6y)- (3xy - 4) ]
\(1,a,4x^2-12xy+9y^2\)
\(=\left(2x\right)^2-2.3.2xy+\left(3y\right)^2\)
\(=\left(2x-3y\right)^2\)
\(b,27a^3-64b^3\)
\(=\left(3a\right)^3-\left(4b\right)^3\)
\(\left(3a-4b\right)\left(9a^2+12ab+16b^2\right)\)
Áp dụng định lý Bezout ta có:
f(x) chia hết cho x-3 \(\Rightarrow f\left(3\right)=0\)
\(\Leftrightarrow2a+3b=-87\left(1\right)\)
g(x) chia hết cho x-3 \(\Rightarrow g\left(3\right)=0\)
\(\Leftrightarrow-3a+2b=-318\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2a+3b=-87\\-3a+2b=-318\end{cases}\Leftrightarrow}\hept{\begin{cases}a=60\\b=-69\end{cases}}\)
Vậy ...
(3x^3 - 2x^2 + x + 2)(5x^2)
= 15x^5 - 10x^4 + 5x^3 + 10x^2
(3x^2 + 5x - 2)(2x^2 - 4x + 3)
= 3x^4 - 12x^3 + 9x^2 + 10x^3 - 20x^2 + 15x - 4x^2 + 8x - 6
= 6x^4 - 2x^3 - 15x^2 + 23x - 6
a) Ta có: \(A=\left(7-2x\right)\left(7+2x\right)+\left(2x+7\right)^2\)
\(=7-4x^2+4x^2+28x+49\)
\(=28x+56\)
b) Ta có: \(B=\left(4x-5\right)^2-\left(2x-1\right)\left(8x-5\right)\)
\(=16x^2-40x+25-\left(16x^2-10x-8x+5\right)\)
\(=16x^2-40x+25-16x^2+18x-5\)
\(=-22x+20\)
c) Ta có: \(C=\left(5x-3\right)^2-2\left(5x-3\right)\left(5-5x\right)+\left(5x-5\right)^2\)
\(=\left(5x-3\right)^2+2\cdot\left(5x-3\right)\left(5x-5\right)+\left(5x-5\right)^2\)
\(=\left(5x-3+5x-5\right)^2\)
\(=\left(10x-8\right)^2\)
\(=100x^2-160x+64\)
d) Ta có: \(D=\left(2a+3b-c\right)\left(2a-3b+c\right)-\left(4a^2-9b^2-c^2\right)\)
\(=\left[\left(2a+\left(3b-c\right)\right)\left(2a-\left(3b-c\right)\right)\right]-\left(4a^2-9b^2-c^2\right)\)
\(=4a^2-\left(3b-c\right)^2-4a^2+9b^2+c^2\)
\(=-9b^2+6bc-c^2+9b^2+c^2\)
=6bc
mình làm bài 2 trước nha:
a) y.(a-b)+a.(y-b)=a.y-b.y+a.y-b.y
=(a.y+a.y)-(b.y+b.y)
=2.a.y-2.b.y
=2.y.(a-b)
b)x2.(x+y)-y.(x2-y2)=x3+x2.y-x2y+y3=x3+y3