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a) \(A=x^2-6x+9-9y^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3-3y\right)\left(x-3+3y\right)\)
b) \(B=x^3-3x^2+3x-1+2\left(x^2-1\right)\)
\(=\left(x-1\right)^3+\left(2x+2\right)\left(x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)^2+2x+2\right]\)
\(=\left(x-1\right).\left(x^2+3\right)\)
a, \(A=\left(x-3\right)^2-9y^2=\left(x-3-3y\right)\left(x-3+3y\right)\)
b, \(B=\left(x-1\right)^3+2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left[\left(x-1\right)^2+2\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2-2x+1+2x+2\right)=\left(x-1\right)\left(x^2+3\right)\)
a) Áp dụng HĐT 1 thu được ( 2 x + y ) 2 .
b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được
[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).
c) Ta có: 9 - 6x + x 2 - y 2 = ( 3 - x ) 2 - y 2 = (3 - x - y)(3 -x + y).
d) Ta có: -(x + 2) + 3( x 2 - 4) = -{x + 2) + 3(x + 2)(x - 2)
= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).
a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)
c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)
\(=x^4-14x^2-32\)
\(=\left(x^2-16\right)\left(x^2+2\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)
\(=\left(x^2+5x+1\right)^2\)
1. x2 - 6x + 9=(x-3)2
2. 25 + 10x + x2=(x+5)2
3. \(\dfrac{1}{4}a^2+2ab^2+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
4.\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)
5.x3 + 8y3=(x+8y)(x2-8xy+64y2)
6.8y3 -125=(2y-5)(4y2+10y+25)
7.a6-b3=(a2-b)(a4+a2b+b2)
8 x2 - 10x + 25=(x-2)2
1) \(x^2-6x+9=\left(x-3\right)^2\)
2) \(25+10x+x^2=\left(5+x\right)^2\)
3) \(\dfrac{1}{4}a^2+2ab+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
4) \(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)
5) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
6) \(8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)
7) \(a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
8) \(x^2-10x+25=\left(x-5\right)^2\)
9) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
k cho tui nha
\(x^2-6x+9=x^2-2.3x+3^2=\left(x-3\right)^2\)
\(25+10x+x^2=\left(x+5\right)^2\)
\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2ab^2+\left(2b^2\right)^2=\left(\frac{1}{2}a+2b^2\right)^2\)
\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(y^4-\frac{1}{3}\right)^2\)
\(x^2-10x+25=\left(x-5\right)^2\)
\(x^2+4xy+4y^2=\left(x+2y\right)^2\)
\(\left(3x+2\right)^2-4=\left(3x\right)\left(3x+4\right)\)
\(4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(4x^2-49=\left(2x\right)^2-7^2=\left(2x+7\right)\left(2x-7\right)\)
\(\frac{9}{25}y^4-\frac{1}{4}=\left(\frac{3}{5}y^2\right)^2-\left(\frac{1}{2}\right)^2=\left(\frac{3}{5}y^2-\frac{1}{2}\right)\left(\frac{3}{5}y^2+\frac{1}{2}\right)\)
\(x^{32}-1=\left(x-1\right)\left(x^{31}+x^{30}+...+x+1\right)\)
\(4x^2+4x+1=\left(2x\right)^2+2.2x+1^2=\left(2x+1\right)^2\)
\(x^2-20x+100=\left(x-10\right)^2\)
\(y^4-14y^2+49=\left(y^2\right)^2-2.7.y^2+7^2=\left(y^2-7\right)^2\)