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\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
- Viết 7 hằng đẳng thức đáng nhớ :
\(\left(A+B\right)^2=A^2+2AB+B^2\)
\(\left(A-B\right)^2=A^2-2AB+B^2\)
\(A^2-B^2=\left(A-B\right)\left(A+B\right)\)
\(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)
\(\left(A-B\right)^3=A^3-3A^2B+3AB^2-B^3\)
\(A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)\)
\(A^3+B^3=\left(A+B\right)\left(A^2-AB+B^2\right)\)
- Áp dụng :
\(a,\left(x+2y\right)^2=x^2+4xy+4y^2\)
\(b,\left(\dfrac{5x-1}{2}\right)^2=\dfrac{\left(5x-1\right)^2}{2^2}=\dfrac{25x^2-10x+1}{4}\)
\(c,\left(\dfrac{1}{3x-3}\right)\left(\dfrac{1}{3x+3}\right)=\dfrac{1.1}{\left(3x-3\right)\left(3x+3\right)}=\dfrac{1}{9x^2-9}\)
\(d,\left(2x+3\right)^3=8x^3+36x^2+54x+27\)
\(e,\left(\dfrac{1}{4y-2x}\right)^2=\dfrac{1}{\left(4y-2x\right)^2}=\dfrac{1}{16y^2-16xy+4x^2}\)
\(f,\left(2x-y\right)\left(4x^2+2xy+y^2\right)=\left(2x\right)^3-y^3=8x^3-y^3\)
\(g,\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
a) Ta có: \(-3x^2\left(2x^2-\frac{1}{3}x+2\right)\)
\(=-6x^4+x^3-6x^2\)
b) Ta có: \(2xy^2\left(x-3y+xy\right)\)
\(=2x^2y^2-6xy^3+2x^2y^3\)
c) Ta có: \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
d) Ta có: \(-\left(2-x\right)\left(2x+3\right)\)
\(=\left(x-2\right)\left(2x+3\right)\)
\(=2x^2+3x-4x-6\)
\(=2x^2-x-6\)
e) Ta có: \(\left(3x^3-2x^2+x\right):\left(-2x\right)\)
\(=\frac{-3}{2}x^2+x-\frac{1}{2}\)
f) Ta có: \(\left(15x^2y^2-21x^3y+2x^2y\right):\left(3x^2y\right)\)
\(=5y-7x+\frac{2}{3}\)
g)
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) x2( x - 1 ) - x + 1
= x2( x - 1 ) - ( x - 1 )
= ( x - 1 )( x2 - 1 )
= ( x - 1 )( x - 1 )( x + 1 )
= ( x - 1 )2( x + 1 )
b) ( a + b )3 - ( a - b )3
= ( a3 + 3a2b + 3ab2 + b3 ) - ( a3 - 3a2b + 3ab2 - b3 )
= a3 + 3a2b + 3ab2 + b3 - a3 + 3a2b - 3ab2 + b3
= 6a2b + 2b3
= 2b( 3a2 + b )
c) 6x( x - 3 ) + 9 - 3x2
= 6x2 - 18x + 9 - 3x2
= 3x2 - 18x + 9
= 3( x2 - 6x + 3 )
d) x( x - y ) - 5x + 5y
= x( x - y ) - ( 5x - 5y )
= x( x - y ) - 5( x - y )
= ( x - y )( x - 5 )
e) 3( x + 4 ) - x2 - 4x
= 3( x + 4 ) - ( x2 + 4x )
= 3( x + 4 ) - x( x + 4 )
= ( x + 4 )( 3 - x )
f) x2 + 4x - y2 + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y )( x + 2 + y )
g) x2 + 5x
= x( x + 5 )
h) -x2 + 2x + 2y + y2
= ( y2 - x2 ) + ( 2x + 2y )
= ( y - x )( y + x ) + 2( x + y )
= ( x + y )( y - x + 2 )
a) 2x2.(5x3-4x2y-7xy +1) =10x5-8x4y-14x3y+2x2 b) (5x -2y)(x2 -xy +1) =5x3-5x2y+5x-2x2y+2xy2-2y =5x3-7x2y+2xy2+5x-2y c) (\(\dfrac{1}{2}\)x -1)(2x -3) =x2-\(\dfrac{3}{2}\)x-2x+3 =x2-\(\dfrac{7}{2}\)x+3 d) (x +3y)2 =x2+6xy+9y2 e) (3x -2y)2 =9x2-12xy+4y2 g) (\(\dfrac{1}{4}\)x - 3y)(\(\dfrac{1}{4}\)x +3y) =\(\dfrac{1}{16}\)x2-9y2 f) (2x +3)3 =8x3+36x2+54x+27 h) (3 -2y)3 =27-54y+36y2-8y3