Tính:
a. (x+2y)2(x+2y)2
b. (x−3y)(x+3y)(x−3y)(x+3y)
c. (5−x)2
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a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)
b: \(=-5y-9+xy\)
a) M = (x + 3y)² - (x - 3y)²
= [(x + 3y) - (x - 3y)][(x + 3y) + (x - 3y)]
= (x + 3y - x + 3y)(x + 3y + x - 3y)
= 6y.2x
= 12xy
b) Q = (x - y)² - 4(x - y)(x + 2y) + 4(x + 2y)²
= [(x - y) - 2(x + 2y)]²
= (x - y - 2x - 4y)²
= (-x - 5y)²
a) M = (x + 3y)² - (x - 3y)²
= [(x + 3y) - (x - 3y)][(x + 3y) + (x - 3y)]
= (x + 3y - x + 3y)(x + 3y + x - 3y)
= 6y.2x
= 12xy
b: Q=(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2
=(x-y-2x-4y)^2
=(-x-5y)^2
=x^2+10xy+25y^2
2a) pt <=> (x + 6)^2 = 0
<=> x = -6
b) pt <=> (4x - 1)^2 = 0
<=> x = 1/4
c) pt<=> (x + 1)^3 = 0
<=> x = -1
Bài 1:
a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)
\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)
\(=32x^2+18y^2\)
b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)
\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)
\(=-12x^2-24\)
Bài 2:
a: Ta có: \(x^2+12x+36=0\)
\(\Leftrightarrow x+6=0\)
hay x=-6
b: Ta có: \(16x^2-8x+1=0\)
\(\Leftrightarrow4x-1=0\)
hay \(x=\dfrac{1}{4}\)
Bài 1:
a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)
\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)
\(=32x^2+18y^2\)
b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)
\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)
\(=-12x^2-24\)
c: Ta có: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)
\(=\left(x+2y+x-2y\right)^2\)
\(=4x^2\)
a: \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)
\(=\dfrac{-6x^3y^4}{2x^3y^3}+\dfrac{4x^4y^3}{2x^3y^3}\)
\(=-3y+2x\)
b: \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}=\dfrac{5x^4y^2}{x^3y^2}-\dfrac{x^3y^2}{x^3y^2}\)
\(=5x-1\)
c: \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)
\(=-\dfrac{27x^3y^5}{3x^2y^3}-\dfrac{9x^2y^4}{3x^2y^3}+\dfrac{6x^3y^3}{3x^2y^3}\)
\(=-9xy^2-3y+2x\)
a) \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)
\(=\dfrac{2x^3y^3\cdot\left(-3y+2x\right)}{2x^3y^3}\)
\(=-3y+2x\)
\(=2x-3y\)
b) \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}\)
\(=\dfrac{5x\cdot x^3y^2-x^3y^2\cdot1}{x^3y^2}\)
\(=\dfrac{x^3y^2\cdot\left(5x-1\right)}{x^3y^2}\)
\(=5x-1\)
c) \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)
\(=\dfrac{-3x^2y^3\cdot-9xy^2+-3x^2y^3\cdot-3y+-3x^2y^3\cdot2x}{-3x^2y^3}\)
\(=\dfrac{-3x^2y^3\cdot\left(-9xy^2-3y+2x\right)}{-3x^2y^3}\)
\(=-9xy^2-3x+2x\)
Bài 12:
a) \(\left(\dfrac{1}{2}x+4\right)^2\)
\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)
\(=\dfrac{1}{4}x^2+4x+16\)
b) \(\left(7x-5y\right)^2\)
\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)
\(=49x^2-70xy+25y^2\)
c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)
\(=\left(y^2+6x^2\right)\left(y^2-6x^2\right)\)
\(=y^4-36x^4\)
d) \(\left(x+2y\right)^2\)
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)
\(=x^2+4xy+4y^2\)
e) \(\left(x-3y\right)\left(x+3y\right)\)
\(=x^2-\left(3y\right)^2\)
\(=x^2-9y^2\)
f) \(\left(5-x\right)^2\)
\(=5^2-2\cdot5\cdot x+x^2\)
\(=25-10x+x^2\)
a) (x-2)(x+2)-x(x-1)+8
= x2-4-x2+x+8
= (x2-x2)+(-4+8)+x
= 4+x
b) bn viết lại đề đi:v
đọc khó quá.
a. (x+2y)2)(x+2y)2) =x2+4xy+4y2=x2+4xy+4y2
b. (x−3y)(x+3y)(x−3y)(x+3y) =x2−(3y)2=x2−9y2=x2−(3y)2=x2−9y2
c. (5−x)2(5−x)2 =52−10x+x2=25−10x+x2
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