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a) 2xy + 3z + 6y + xz
= (2xy + 6y) + (xz + 3z)
= 2y(x + 3) + z(x + 3)
= (2y + z)(x + 3)
b) 9x - x3
= x(9 - x2)
= x(3 + x)(3 - x)
c) xz + yz + 5.(x + y)
= (xz + yz) + 5(x + y)
= z(x + y) + 5(x + y)
= (z + 5)(x + y)
d) x2 + 4x - y2 + 4
= (x2 + 4x + 4) - y2
= (x + 2)2 - y2
= (x + 2 + y)(x + 2 - y)
có j til mik nha
a) 2xy + 3z + 6y + xz
* Gợi ý : Câu này ta dùng phương pháp nhóm hạng tử và đặt thừ số chung.
Giải :
\(=\left(2xy+6y\right)+\left(3z+xz\right)\)
\(=2y\left(x+3\right)+z\left(x+3\right)\)
\(=\left(2y+z\right)\left(x+3\right)\)
b) 9x - x3
* Gợi ý : Câu này ta dùng phương pháp đặt thừ số chung và dùng hằng đẳng thức.
\(=9.x-x^2.x\)
\(=x\left(9-x^2\right)\)
\(=x\left[\left(3\right)^2-x^2\right]\)
\(=x.\left(3+x\right)\left(3-x\right)\)
1: \(=a\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(a-4\right)\)
2: \(=x\left(x+b\right)+a\left(x+b\right)=\left(x+b\right)\left(x+q\right)\)
3: \(=a\left(x+1\right)-b\left(x+1\right)+c\left(x+1\right)\)
\(=\left(x+1\right)\left(a-b+c\right)\)
6: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
a, \(\left(2x+1\right)^2-2\left(2x+1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1-x+3\right)^2=\left(x+4\right)^2\)
b, \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)
c, \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)=\left(x+1\right)\left(y-z\right)\)
d, \(x^2-xy-8x+8y=\left(x^2-xy\right)-\left(8x-8y\right)\)
\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)
e, \(x^2+2xy+y^2-xz-yz=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y+z\right)\left(x+y\right)\)
f, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)
1,
a, (2x + 1- x + 3)2 = (x+4)2
b,\(x\left(y+z\right)+3\left(y+z\right)=\left(y+z\right)\left(x+3\right)\)
c, \(x\left(y-z\right)+\left(y-z\right)=\left(y-z\right)\left(x+1\right)\)
d,\(x\left(x-y\right)+8\left(y-x\right)\)=\(\left(x-y\right)\left(x-8\right)\)
e,\(\left(x+y\right)^2-z\left(x+y\right)\)=\(\left(x+y\right)\left(x+y-z\right)\)
f,\(25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)
\(=\left(5+2x+y\right)\left(5-2x-y\right)\)
Chúc các bn hc tốt
a) \(6x-6y=6\left(x-y\right)\)
b)\(2xy+3x+6y+xz\)
\(=\left(2xy+xz\right)+\left(6y+3z\right)\)
\(=x\left(2y+z\right)+3\left(2y+z\right)\)
\(=\left(2y+z\right)\left(x+3\right)\)
c)\(x^2+6x+9-y^2\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x-y+3\right)\left(x+y+3\right)\)
d) \(9x-x^3\)
\(=x\left(9-x^2\right)\)
\(=x\left(3-x\right)\left(3+x\right)\)
e)\(x^2-xy+x-y\)
\(=\left(x^2-xy\right)+\left(x-y\right)\)
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+1\right)\)
a, 6x - 6y = 6( x-y )
b, 2xy + 3z + 6y + xz
= ( 2xy + 6y ) + ( 3z + xz )
= 2y( x + 3 ) + z ( 3 + x )
= 2y( 3 + x ) + z ( 3 + x )
= ( 3 + x ) ( 2y + z )
c, x2 + 6x + 9 - y2 = ( x2 + 6x + 9 ) - y2
= ( x + 3 )2 - y2
= ( x + 3 - y ) ( x + 3 + y )
d , 9x - x3 = x ( 9 - x2 )
= x ( 3 - x ) ( 3 + x )
e, x2 - xy + x - y =( x 2 - xy ) + ( x - y )
= x ( x - y ) + ( x - y )
= ( x - y ) ( x + 1 )
\(a,VT=\left(a+b+c\right)\left(a-b+c\right)\)
\(=\left(a+c+b\right)\left(a+c-b\right)\)
\(=\left(a+c\right)^2-b^2\)
\(=a^2+2ac+c^2-b^2=VP\)
\(b,VT=\left(3x+2y\right)\left(3x-2y\right)-\left(4x-2y\right)\left(4x+2y\right)\)
\(=9x^2-4y^2-16x^2+4y^2=-7x^2=VP\)
\(c,VT=x^3-1-x^3-1=-2=VP\)
\(d,VT=8x^3+1-8x^3+1=2=VP\)
\(e,VT=\left(x^2+2xy+4y^2\right)\left(x-2y-2x+1\right)\)
\(=\left(x^2+2xy+4y^2\right)\left(-x-2y+1\right)\)
\(=-x^3-2x^2y+x^2-2x^2y-4xy^2+2xy-4xy^2-8y^3+4y^2\)
( bn kiểm tra lại đề nhé)
\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow x=-2\)
\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)
Dấu \("="\Leftrightarrow x=-5\)
\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(A=x^2+4x+5\)
\(=x^2+4x+4+1\)
\(=\left(x+2\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=-2
\(C=4x^2-4x+5\)
\(=4x^2-4x+1+4\)
\(=\left(2x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
d/ \(x^2+4x-2xy-4y+y^2=\left(x-y\right)^2+4\left(x-y\right)=\left(x-y\right)\left(x-y+4\right)\)
e/ \(x^2+2x+1-16y^2=\left(x+1\right)^2-\left(4y\right)^2=\left(x+1-4y\right)\left(x+1+4y\right)\)