Cho x-y = 4 ; x2 + y2 = 12. Tính A = x3 - y3
Cho x-y = 4 ; x2 + y2 = 12. Tính A = x3 - y3
có `x-y=4`
`<=>x^2 -2xy+y^2 =16`
`<=>12-2xy+16`
`<=>-2xy=4`
`<=>xy=-2`
`x^3 -y^3`
`=(x-y)(x^2 +xy+y^2)`
`=4(12-2)`
`=4*10`
`=40`
Có: \(x-y=4\)
\(\Rightarrow\left(x-y\right)^2=16\)
\(\Rightarrow x^2-2xy+y^2=16\)
\(\Rightarrow x^2+y^2-2xy=16\)
\(\Rightarrow12-2xy=16\) \(\Leftrightarrow2xy=-4\Leftrightarrow xy=-2\)
Lại có: \(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=4.\left(x^2+y^2+xy\right)\) (do \(x-y=4\))
\(=4.\left(12-2\right)\) (do \(x^2+y^2=12;xy=-2\))
\(=4.10=40\)
Vậy \(A=40\).
rút gọn bt sau: (2x-3)2-(2x+3)2
`@` `\text {Ans}`
`\downarrow`
`(2x - 3)^2 - (2x + 3)^2`
`= 4x^2 - 12x + 9 - (4x^2 + 12x + 9)`
`= 4x^2 - 12x + 9 - 4x^2 - 12x - 9`
`= (4x^2 - 4x^2) + (-12x - 12x) + (9-9)`
`= -24x`
____
`@` CT:
`(A + B)^2 = A^2 + 2AB + B^2`
`(A - B)^2 = A^2 - 2AB + B^2`
\(\left(2x-3\right)^2-\left(2x+3\right)^2\)
\(=\left[\left(2x-3\right)+\left(2x+3\right)\right]\left[\left(2x-3\right)-\left(2x+3\right)\right]\)
\(=\left(2x-3+2x+3\right)\left(2x-3-2x-3\right)\)
\(=4x\cdot-6\)
\(=-24x\)
Bài 4. Tìm x, biết:
a) (2x + 1)^2 - 4(x + 2)^2 = 9
b) (x + 3)^2 - (x - 4)( x + 8) = 1
c) 3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36
\(a,\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\\ \Leftrightarrow4x^2-4x^2+4x-16x+1-16-9=0\\ \Leftrightarrow-12x=24\\ \Leftrightarrow x=\dfrac{24}{-12}=-2\\ b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2-x^2+6x-4x=1-9-32\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\\ c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\\ \Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\\ \Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Leftrightarrow3x^2+4x^2-7x^2+12x-4x=36-12-1-63\\ \Leftrightarrow8x=-40\\ \Leftrightarrow x=\dfrac{-40}{8}=-5\)
1: (4a-b)(4a+b)=(4a)^2-b^2=16a^2-b^2
(x^2y+2y)(x^2y-2y)
=(x^2y)^2-4y^2
=x^4y^2-4y^2
(3/4x+3/5y)(3/5y-3/4x)=(3/5y)^2-(3/4x)^2=9/25y^2-9/16x^2
2: (x+2)(x^2-2x+4)=x^3+8
(3x+2y)(9x^2-6xy+4y^2)=27x^3+8y^3
3: (5-3x)(5+3x+9x^2)=125-27x^3
(1/2x-1/5y)(1/4x^2+1/10xy+1/25y^2)=1/8x^3-1/125y^3
Câu 1 viết các đa thức sau dưới dạng tích
a, 25+10x+x^2
b, 8x^3 - 1/8
c, x^2 - 10x + 25
Giúp em với ạ em cảm ơn
a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)
1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)
2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)
3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)
Viết các đa thức sau thành tích
a, 4x^2 - 25y^2
b, 8x^3 + 27
c, 125x^3 - 64y^3
Giúp em với cảm ơn mọi người
`a, 4x^2 - 25y^2 = (2x-5y)(2x+5y)`.
`b, 8x^3 +27 = (2x+3)(4x^2 - 6x + 9)`.
`c, 125x^3 - 64y^3 = (5x)^3 - (4y)^3 = (5x-4y)(25x^2 + 20xy + 16y^2)`.
\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
a)x^2+y^2-6x+8y+25=0
b)(x+y)^2=(x-1)(y+1)
a: =>x^2-6x+9+y^2+8y+16=0
=>(x-3)^2+(y+4)^2=0
=>x=3 và y=-4
Tính (x-3/2)(x+3/2y)
Sửa đề:(x-3/2y)(x+3/2y)
=x^2-(3/2y)^2
=x^2-9/4y^2
Rút gọn biểu thức (2x-1)^2+(x+3)^2-5(x+7)(x-7)
=4x^2-4x+1+x^2+6x+9-5x^2+245
=2x+255
Rút gọn biểu thức (x+4)^2-(x+1)(x-1)
\(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=x^2+8x+16-\left(x^2-1\right)=x^2+8x+16-x^2+1=8x+17\)