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\(a.9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right)\left(3a+5b^2\right)\)
\(b.\left(2x+y\right)^2-1=\left(2x+y-1\right)\left(2x+y+1\right)\)
\(c.\left(x+y+z\right)^2-\left(x-y-z\right)^2=\left[\left(x+y+z\right)+\left(x-y-z\right)\right]\left[\left(x+y+z\right)\right]-\left(x-y-z\right)\\ =2x.\left(2y+2z\right)\)
a) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right)\left(3a+5b^2\right)\)
b) \(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y+1\right)\left(2x+y-1\right)\)
c) \(\left(x+y+z\right)^2-\left(x-y-z\right)^2=\left(x+y+z+x-y-z\right)\left(x+y+z-x+y+z\right)\)
\(=2x\left(2y+2z\right)\)
Bài làm:
Ta có: \(\frac{9}{4x^2}+\frac{9y^2}{4}-\frac{9y}{2x}\)
\(=\left(\frac{3}{2x}\right)^2-2.\frac{3}{2x}.\frac{3y}{2}+\left(\frac{3y}{2}\right)^2\)
\(=\left(\frac{3}{2x}-\frac{3y}{2}\right)^2\)
BÀi 1 : xem lại đề
bài 2
a) 27 - x^3
= ( 3 -x )( 9 + 3x + x^2)
b) 8x^3 + 0,001
= (2x + 0,1) ( 4x^2 - 0,2x + 0,01)
\(\frac{x^3}{64}-\frac{y^3}{125}=\left(\frac{x}{4}-\frac{y}{5}\right)\left(\frac{x^2}{16}-\frac{xy}{20}+\frac{y^2}{25}\right)\)
a+b=7=>(a+b)2=49
=>a2+2ab+b2=49
Do ab=3
=>2ab=6
=>b2+a2=43
Ta có:a3+b3=(a+b)(a2-ab+b2)
Thay a2+b2=43 ab=3 a+b=7
=> a3+b3=7.(43-3)=7.40=280
a)27-x3=(3-x)(9+3x+x2)
b)8x3+0,001=(2x+0,1)(4x2-0,2x+0,01)
c)x3/64-y3/125=(x/4-y/5)(x2/16+xy/20+y2/25)
a) 2x2 + y2 - 2xy + 10x + 25
= (x2 + y2 - 2xy) + (x2 + 10x + 25)
= (x - y)2 + (x + 5)2
các bn xem đúng ko nhé mk làm bừa nên lên olm hỏi lại mọi người giúp giùm câu b) nha!!
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a) \(27x^3+8^3\)
\(=\left(3x\right)^3+2^3\)
\(=\left(3x+2\right)\left[\left(3x\right)^2+6x+2^2\right]\)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
b) \(8x^3-y^3\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
c) \(x^2+4xy+4y^2\)
\(=\left(x+2y\right)^2\)
\(27x^3+8\)
\(=\left(3x\right)^3+2^3\)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(8x^3-y^3\)
\(=\left(2x\right)^3-y^3\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(x^2+4xy+4y^2\)
\(=x^2+2.x.2y+\left(2y\right)^2\)
\(=\left(x+2y\right)^2\)
_Minh ngụy_
\(4x^2-\frac{1}{9}\left(y+1\right)^2=\left(2x\right)^2-\left(\frac{1}{3}\left(y+1\right)\right)^2\)
\(=\left(2x-\frac{1}{3}\left(y+1\right)\right)\left(2x+\frac{1}{3}\left(y+1\right)\right)\)
\(=\left(2x-\frac{1}{3}y-\frac{1}{3}\right)\left(2x+\frac{1}{3}y+\frac{1}{3}\right)\)
3. \(P=2x^2-8\)
\(P=2x^2-2.4\)
\(P=2\left(x^2-4\right)\)
\(P=2\left(x^2-2^2\right)\)
\(P=2\left(x-2\right)\left(x+2\right)\)
\(\left(2x+3\right)^2-y^2=\left(2x+y+3\right)\left(2x-y+3\right)\)