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1.a)\(20x-5y=5\left(4x-y\right)\)
b)\(5x\left(x-1\right)-3x\left(x-1\right)=\left(5x-3x\right)\left(x-1\right)=2x\left(x-1\right)\)
c)\(x\left(x+y\right)-6x-6y=x\left(x+y\right)-6\left(x+y\right)=\left(x-6\right)\left(x+y\right)\)
d)\(6x^3-9x^2=3x^2\left(2x-3\right)\)
e)\(4x^2y-8xy^2+10x^2y^2=2xy\left(2x-8y+10xy\right)\)
g)\(20x^2y-12x^3=4x^2\left(5y-3x\right)\)
h)\(8x^4+12x^2y-16x^3y^4=4x^2\left(2x^2+12y-16xy^4\right)\)
2.a)\(3x\left(x+1\right)-5y\left(x+1\right)=\left(3x-5y\right)\left(x+1\right)\)
b)\(3x\left(x-6\right)-2\left(x-6\right)=\left(3x-2\right)\left(x-6\right)\)
c)\(4y\left(x-1\right)-\left(1-x\right)=4y\left(x-1\right)+\left(x-1\right)=\left(4y+1\right)\left(x-1\right)\)
d)\(\left(x-3\right)^3+3-x=\left(x-3\right)^3-\left(x-3\right)=\left(x-3\right)\left[\left(x-3\right)^2-1\right]=\left(x-3\right)\left(x-2\right)\left(x-4\right)\)
e)\(7x\left(x-y\right)-\left(y-x\right)=7x\left(x-y\right)+\left(x-y\right)=\left(7x+1\right)\left(x-y\right)\)
h)\(3x^3\left(2y-3z\right)-15x\left(2y-3z\right)^2=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]\)
k)Sai đề: \(3x\left(z+2\right)+5\left(-z-2\right)=3x\left(z+2\right)-5\left(z+2\right)=\left(3x-5\right)\left(z+2\right)\)
l)\(18x^2\left(3+x\right)+3\left(x+3\right)=3\left(x+3\right)\left(6x^2+1\right)\)
m)\(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)
n)\(10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
Bài 1:
a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
c)Đề sai hoàn toàn
d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)
f)\(\left(x^2+y^2\right)^2-4x^2y^2=\left(x-y\right)^2\left(x+y\right)^2\)
g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)
i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)
a, 20x - 5y = 5(4x - y)
b, 4x2 - 8xy2 + 10x2y = 2x(2x - 4y2 + 5xy)
c, 5x (x - 1) - 3x (x - 1) = (5x - 3x) (x - 1) = 2x (x - 1)
d, x (x + y) - 6x - 6y = x (x + y) - (6x + 6y) = x (x + y) - 6 (x + y) = (x - 6) (x + y)
e, x4 - y4 = (x2)2 - (y2)2 = (x2 - y2) (x2 + y2) = [(x + y) (x - y)] (x2 + y2)
f, x2 - 4y2 = x2 - (2y)2 = (x - 2y) (x + 2y)
g, 27x3 - 64 = (3x)3 - 43 = (3x - 4) (9x2 + 12x + 16)
h, (x +1)2 - 16 = (x +1)2 - 42 = (x + 1 + 4) (x + 1 - 4) = (x + 5) (x - 3)
i, (3x + 1)2 - (x - 2)2 = (3x + 1 - x + 2) (3x + 1 + x - 2) = (2x + 3) ( 4x - 1)
\(20x^2y-12x^3=4x^2\left(5y-3x\right)\)
\(8x^4+12x^2y^4-16x^3y^4=4x^2\left(2x^2+3y^4-4xy^4\right)\)
\(6x^3-9x^2=3x^2\left(2x-3\right)\)
\(4xy^2+8xy^2=12xy^2\)
\(3x\left(x+1\right)-5\left(x+1\right)=\left(3x-5\right)\left(x+1\right)\)
\(20x^2y-12x^3\)
\(=4x^2\left(5y-3x\right)\)
\(8x^4+12x^2y^4-16x^3y^4\)
\(=4x^2\left(2x^2+3y^4-4xy^4\right)\)
\(a,x^4+2x^2+1=\left(x^2+1\right)^2\)
\(b,4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
\(c,-x^2-2xy-y^2=-\left(x^2+2xy+y^2\right)=-\left(x+y\right)^2\)
\(d,\left(x+y\right)^2-2\left(x+y\right)-1=\left(x+y\right)\left(x+y-2\right)-1\)
e: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
g: \(=\left(x+2\right)^3\)
h: \(=\left(x+1\right)\left(x^2-x+1\right)+x\left(x+1\right)=\left(x^2+1\right)\left(x+1\right)\)
k: \(=x^3+y^3+3xy\left(x+y\right)-x^3-y^3\)
=3xy(x+y)
f) x2 + 2y2 - 2xy + 2x + 2 - 4y =0
<=>x2 + y2 - 2xy+2x-2y+y2-2y+1+1=0
<=>(x-y)2+2(x-y)+1+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>y=1;x=0
Bạn học thầy Trung phải k nè~~~~
Busted :))))
Bài 2:
a: \(A=\left(2x-y\right)^2=\left(12-2\right)^2=100\)
b: \(=\left(x-3\right)^3=100^3=1000000\)
c: \(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y-3z\right)\left(x-y+3z\right)\)
\(=\left(6+4-90\right)\left(6+4+90\right)=-80\cdot100=-8000\)
where is đề
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