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\(a,VT=\left(a+b+c\right)\left(a-b+c\right)\)
\(=\left(a+c+b\right)\left(a+c-b\right)\)
\(=\left(a+c\right)^2-b^2\)
\(=a^2+2ac+c^2-b^2=VP\)
\(b,VT=\left(3x+2y\right)\left(3x-2y\right)-\left(4x-2y\right)\left(4x+2y\right)\)
\(=9x^2-4y^2-16x^2+4y^2=-7x^2=VP\)
\(c,VT=x^3-1-x^3-1=-2=VP\)
\(d,VT=8x^3+1-8x^3+1=2=VP\)
\(e,VT=\left(x^2+2xy+4y^2\right)\left(x-2y-2x+1\right)\)
\(=\left(x^2+2xy+4y^2\right)\left(-x-2y+1\right)\)
\(=-x^3-2x^2y+x^2-2x^2y-4xy^2+2xy-4xy^2-8y^3+4y^2\)
( bn kiểm tra lại đề nhé)
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
a) \(=x^2+2xy+y^2-x^2+y^2=2xy+2y^2=2y\left(x+y\right)\)
b) \(=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
e) \(=\left(x-3\right)\left(x^2+3x+9\right)-2x\left(x-3\right)=\left(x-3\right)\left(x^2+x+9\right)\)
f) \(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)=\left(x+5\right)\left(x^2-6x+25\right)\)
a) \(\left(x+y\right)^2-\left(x^2-y^2\right)\)
\(=x^2+2xy+y^2-x^2+y^2\)
\(=2y^2+2xy\)
\(=2y\left(x+y\right)\)
c) \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-x^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+\left(y+1\right)\right]\left[2x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(2x+1\right)\left(y+1\right)\left(2x-1\right)\left(y-1\right)\)
\(=\left(4x^2-1\right)\left(y^2-1\right)\)
bình phương tổng chứ
b, B= x^2+ 2xy+y^2 +4y+4
= x^2+2xy+y^2+y^2+4y+4
=(x+y)^2+(y+2)^2
c, C= 2x^2+6xy+9y^2+2x+1
= x^2+6xy+9y^2+x^2+2x+1
= (x+3)^2+(x+1)^2
d, D= x(x+2) +(x+1)(x+3) +2
= x^2+2x+x^2+3x+x+3+2
= x^2+2x+1+x^2+4x+4
= (x+1)^2+(x+2)^2
e, E= x^2-2xy+2y^2+2y+1
= x^2-2xy+y^2+y^2+2y+1
= (x-y)^2+(y+1)^2
f, F= 4x^2-12xy+10y^2+4y+4
=4x^2-12xy+9y^2+y^2+4y+4
=(2x-3y)^2+(y+2)^2
g, G=2x^2+4xy+4y^2+4x+4
=x^2+4xy+4y^2+x^2+4x+4
=(x+2y)^2+(x+2)^2
Xong r.... dài quá...mới hè lớp 7 nên có j bỏ qua ak
Bài làm
a) 4x - 8y
<=> 4( x - 2y )
b) 12x( x - 2y ) - 8y( x - 2y )
<=> ( 12x - 8y )( x - 2y )
<=> 4( 3x - 2y )( x - 2y )
c) 2x + 2y - x2 - xy
= 2( x + y ) - x( x + y )
= ( x + y )( 2 - x )
d) x2 - 4y2
<=> ( x - 2y )( x + 2y )
e) x3 + x2y - 4x - 4y
<=> x2( x + y ) - 4( x + y )
<=> ( x - 2 )( x + 2 )( x + y )
g) 3x2 - 6xy + 3y2 - 12x3
<=>3( x2 - 3xy + y2 - 4x3 )
# Học tốt #
a)4(x-2y)
b)(x-2y)(12x-8y)
=4(x-2y)(3x-2y)
c)2(x+y)-x(x+y)
=(2-x)(x+y)
d)(x-2y)(x+2y)
e)x2(x+y)-4(x+y)
=(x+y)(x2-4)
=(x+y)(x-2)(x+2)
g)3(x2-2xy+y2-4z3)
=3[(x-y)2-4z3]
????????????phải là 4z2chứ nhỉ.....
a: \(A=-\left(x^2-4x\right)=-\left(x^2-4x+4-4\right)\)
\(=-\left(x-2\right)^2+4\le4\)
Dấu '=' xảy ra khi x=2
b: \(B=-2\left(y^2+2y+1-1\right)\)
\(=-2\left(y+1\right)^2+2\le2\)
Dấu '=' xảy ra khi y=-1
c: \(C=-\left(x^2+y^2+2x+y-3\right)\)
\(=-\left(x^2+2x+1+y^2+y+\dfrac{1}{4}-\dfrac{17}{4}\right)\)
\(=-\left(x+1\right)^2-\left(y+\dfrac{1}{2}\right)^2+\dfrac{17}{4}\le\dfrac{17}{4}\)
Dấu '=' xảy ra khi x=-1 và y=-1/2
a. \(2a^2+5ab-3b^2-7b-2\)
\(=\left(2a^2+6ab+2a\right)-\left(ab+3b^2+b\right)-\left(2a+6b+2\right)\)
\(=2a\left(a+3b+1\right)-b\left(a+3b+1\right)-2\left(a+3b+1\right)\)
\(=\left(2a-b-2\right)\left(a+3b+1\right)\)
b. \(2x^2-7xy+x+3y^2-3y\)
\(=\left(2x^2-xy\right)-\left(6xy-3y^2\right)+\left(x-3y\right)\)
\(=x\left(2x-y\right)-3y\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y+1\right)\)
c. \(6x^2-xy-2y^2+3x-2y\)
\(=\left(6x^2+3xy\right)-\left(4xy-2y^2\right)+\left(3x-2y\right)\)
\(=3x\left(2x+y\right)-2y\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y+1\right)\)
\(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left(\left(x+y\right)^2-z^2\right)\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
\(3x^2+6xy+3y^2-3z^2\)
\(\text{Phân tích thành nhân tử}\)
\(\left(-3\right)\left(z-y-x\right)\left(z+y+x\right)\)
\(2x^2+4x+2-2y^2\)
\(\text{Phân tích thành nhân tử}\)
\(\left(-2\right)\left(y-x-1\right)\left(y+x+1\right)\)
\(2x^2-2xy-4x+4y\)
\(\text{Phân tích thành nhân tử}\)
\(\left(-2\right)\left(x-2\right)\left(y-x\right)\)