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Phần a? phải là \(4a^2-4a+1\)chứ
a) \(4a^2-4a+1=\left(2a\right)^2+2.2a+1\)
\(=\left(2a+1\right)^2\)
b) \(9x^2-25y^2=\left(3x\right)^2-\left(5y\right)^2\)
\(=\left(3x-5y\right)\left(3x+5y\right)\)
c) \(1-2x+a^2=\left(1-a\right)^2\)
d) \(\left(2x+1\right)-2.\left(2x+1\right)\left(3x-y\right)+\left(3x-y\right)^2\)
\(=\left[\left(2x+1\right)-\left(3x-y\right)\right]^2\)
nếu có sai thì bn thông cảm
1.
b) nó là hằng đẳng thức rồi bn nhá
c) \(1-2a+a^2\)= \(1^2-2a1+a^2\)=\(\left(1-a\right)^2\)
d)\(\left[\left(2x+1\right)-\left(3x-y\right)\right]^2\)=\(\left(2x+1-3x+y\right)^2\)=\(\left(1-x+y\right)^2\)
2.
a)\(\left(\frac{1}{2}x\right)^2-\left(3y\right)^2\)=\(\left(\frac{x}{2}-3y\right)\left(\frac{x}{2}+3y\right)\)
b) Ko khai triển đc
c) \(4x^2+2xy+\frac{1}{4}y^2\)
e) Sửa đề:
$2x^3-12x^2+17x-2=2x^3-4x^2-8x^2+16x+x-2$
$=2x^2(x-2)-8x(x-2)+(x-2)=(x-2)(2x^2-8x+1)$
f)
$x^3-3x+2=(x^3-x)-(2x-2)=x(x^2-1)-2(x-1)=x(x-1)(x+1)-2(x-1)$
$=(x-1)(x^2+x-2)=(x-1)(x^2-x+2x-2)=(x-1)[x(x-1)+2(x-1)]$
$=(x-1)(x-1)(x+2)=(x-1)^2(x+2)$
g)
$x^3+3x^2=x^2(x+3)$
h)
$x^3+9x^2+26x+24=(x^3+9x^2+27x+27)-x-3$
$=(x+3)^3-(x+3)=(x+3)[(x+3)^2-1]=(x+3)(x+3-1)(x+3+1)$
$=(x+3)(x+2)(x+4)$
a)
$4x^2-3x-1=4x^2-4x+x-1=4x(x-1)+(x-1)=(4x+1)(x-1)$
b)
$6x^2-11x^2=-5x^2$
c)
\(x^2-7xy+12y^2=x^2-4xy-3xy+12y^2\)
\(=x(x-4y)-3y(x-4y)=(x-3y)(x-4y)\)
d)
\(x^2-2xy+y^2+3x-3y=(x^2-2xy+y^2)+(3x-3y)\)
\(=(x-y)^2+3(x-y)=(x-y)(x-y+3)\)
1,
\(\left(\frac{2}{3}x+y\right)^2=\left(\frac{2}{3}x\right)^2+2.\frac{2}{3}x.y+\left(y\right)^2=\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)
\(\left(3a+\frac{1}{2}b\right)^2=\left(3a\right)^2+2.3a.\frac{1}{2}b+\left(\frac{1}{2}b\right)^2=9a^2+3ab+\frac{1}{4}b^2\)
2,
\(25a^2+4b^2+20ab=\left(5a\right)^2+\left(2b\right)^2+2.5a.2b=\left(5a+2b\right)^2\)
\(x^2+2x+1=\left(x\right)^2+2.x.1+\left(1\right)^2=\left(x+1\right)^2\)
\(9x^2+6x+1=\left(3x\right)^2+2.3x.1+\left(1\right)^2=\left(3x+1\right)^2\)
\(\left(2x+3y\right)^2+2.\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
1.
a)\(\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)
b)\(9a^2+3ab+\frac{1}{4}a^2\)
2.
a)\(\left(5x+2b\right)^2\)
b)\(\left(x+1\right)^2\)
c)\(\left(3x+1\right)^2\)
d)\(\left[\left(2x+3y\right)+1\right]^2\)
Bài 12:
a: \(=\left(xy+1+x+y\right)\left(xy+1-x-y\right)\)
\(=\left[x\left(y+1\right)+\left(y+1\right)\right]\left[x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(y+1\right)\left(y-1\right)\)
b: \(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\cdot\left(3x^2+y^2\right)\)
c: \(=3y^2\left(x^4+x^3+x+1\right)\)
\(=3y^2\left[x^3\left(x+1\right)+\left(x+1\right)\right]\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
a: \(\left(3x-2\right)^2=9x^2-12x+4\)
c: \(9x^2-225=9\left(x^2-25\right)=9\left(x-5\right)\left(x+5\right)\)