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\(=\dfrac{4x\left(3x^2+4xy-2\right)}{3x^2+4xy-2}=4x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^2y-12x^3y^2=3x^2y\left(1-4xy\right)\)
\(2x^2\left(x-y\right)-4xy\left(x-y\right)\)
\(=2x\left(x-y\right)\left(x-2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)
\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)
\(=-12x^2-24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)
\(=\left(x+2y+x-2y\right)^2\)
\(=4x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$
b,
$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$
c,
$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$
d,
$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$
$=3x^2+x$
e,
$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$
f,
$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$
$=\frac{23}{4}xy^2$
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề có bị sai k bạn? Mình thấy k chia hết cho đơn thức nào hết
Bài này tìm x,y à bạn?:\(3x^2-4xy+2y^2-12x+36=0\)
\(< =>2x^2-4xy+2y^2+x^2-12x+36\)
\(< =>x^2-2xy+y^2+x^2-2xy+y^2+x^2-12x+36=0\)
\(< =>\left(x-y\right)^2+\left(x-y\right)^2+\left(x-6\right)^2=0\)
\(< =>2\left(x-y\right)^2+\left(x-6\right)^2=0\)
\(=>\left[{}\begin{matrix}2\left(x-y\right)^2=0\\\left(x-6\right)^2=0\end{matrix}\right.\)\(=>x=y=6\)