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a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
Trả lời:
Bài 4:
b, B = ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 )
= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1
= x8 - 1
Thay x = 2 vào biểu thức B, ta có:
28 - 1 = 255
c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 )
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7 + 1
Thay x = 2 vào biểu thức C, ta có:
27 + 1 = 129
d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 )
= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x
= x
Thay x = - 5 vào biểu thức D, ta có:
D = - 5
Bài 5:
a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )
= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4
= x4 - y4
Thay x = 2; y = - 1/2 vào biểu thức A, ta có:
A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16
b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 )
= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5
= a5 + a4b - ab4 - b5
Thay a = 3; b = - 2 vào biểu thức B, ta có:
B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65
c, ( x2 - 2xy + 2y2 ) ( x2 + y2 ) + 2x3y - 3x2y2 + 2xy3
= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y2 + 2xy3
= x4 + 2y4
Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:
( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Bài 1: Thực hiện phép tính
a) 3x(2x2 - 5x + 9) = \(6x^3-15x^2+27x\)
b) 5x(x2-xy+1) = \(5x^3-5xy+5x\)
c) -2/3x2y(3xy-x2+y) = \(-2x^3y^2+\dfrac{2}{3}x^4y-\dfrac{2}{3}x^2y^2\)
2) Thực hiện phép tính
a) (5x-2y) (x2-xy+1) = \(5x^3+5x-7y-2x^3y+2xy^2\)
b) (x+3y)(x2-2xy+y) = \(x^3-x^2y+xy+6xy^2+y^2\)
c) (3x-5y) (4x+ 7y) = \(12x^2-xy-35y^2\)
Bài 3: Rút gọn các biểu thức sau(bằng cách khai triển hằng đẳng thức):
a) (x+y)2+(x-y)2
= \(x^2+2xy+y^2+x^2-2xy+y^2\)
= \(\left(x^2+x^2\right)+\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
= \(2x^2+2y^2=2\left(x^2+y^2\right)\)
b) (x+2)(x-2)-(x-3)(x+1)
= \(x^2-4\) - \(\left(x^2-2x-3\right)\)= \(x^2-4-x^2+2x+3\)
= \(\left(x^2-x^2\right)+2x+\left(-4+3\right)\)=\(2x-1\)
c) (x-2)(x+2)-(x-2)2
=>\(x^2-4-\left(x^2-2.x.2+2^2\right)=x^2-4-x^2-4x+4=\left(x^2-x^2\right)+\left(-4+4\right)-4x=-4x\)
d) (2x+y)(4x2-2xy+y2)-(2x-y)(4x2+2xy+y2)
= \(8x^3+y^3-\left(8x^3-y^3\right)\)
= \(8x^3+y^3-8x^3+y^3\)
= \(\left(8x^3-8x^3\right)+\left(y^3+y^3\right)\)= \(2y^3\)
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
It's khai triển :)
a) \(\left(5x-x^2\right)\left(5x+x^2\right)=25x^2-x^4\)
b) \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)=8x^3-y^3\)
c) \(\left(x+3\right)\left(x^2-3x+9\right)=x^3-27\)
d) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)
e) \(x^2-2x+9=\left(x-1\right)^2+8??\) ko ra gì cả-.-
g) \(\left(x+1\right)\left(x-1\right)=x^2-1\)
h) \(\left(x-2y\right)\left(x+2y\right)=x^2-4y^2\)
i) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)
a: \(x^2+x-2x-2\)
\(=x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2\right)=\left(-1+1\right)\left(-1-2\right)=0\)
b: \(3x^2-2x+9x-6\)
\(=x\left(3x-2\right)+3\left(3x-2\right)\)
\(=\left(3x-2\right)\left(x+3\right)=\left(3\cdot7-2\right)\left(7+3\right)\)
\(=19\cdot10=190\)
c: \(2x^2-3xy-xy^2\)
\(=x\left(2x-3y-y^2\right)\)
\(=2\left(2\cdot2-3\cdot3-9\right)\)
\(=2\cdot\left(4-18\right)=-28\)
a) (x - 1/2x²y)²
= x² - 2x . 1/2 x²y + (1/2x²y)²
= x² - x³y + 1/4 x⁴y²
b) (2xy² - 1)(1 + 2xy²)
= (2xy²)² - 1²
= 4x²y⁴ - 1
c) (x - y + 2)²
= (x - y)² + 2(x - y).2 + 2²
= x² - 2xy + y² + 4x - 4y + 4
= x² + y² - 2xy + 4x - 4y + 4
d) (x + 1/2)(1/2 - x)
= (1/2)² - x²
= 1/4 - x²
e) (x² - 1/3)²
= (x²)² - 2x².1/3 + (1/3)²
= x⁴ - 2/3 x² + 1/9