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Giải:
a) \(3x^2y-6xy^2\)
\(=3xy\left(x-2y\right)\)
Vậy ...
b) \(\left(2x-a\right)x^2-\left(2x-a\right)y\)
\(=\left(2x-a\right)\left(x^2-y\right)\)
\(=\left(2x-a\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)
Vậy ...
c) \(25a^2-c^2\)
\(=\left(5a-c\right)\left(5a+c\right)\)
Vậy ...
d) \(4-36x+81x^2\)
\(=2^2-2.2.9x+\left(9x\right)^2\)
\(=\left(2-9x\right)^2\)
Vậy ...
e) \(\left(x+7\right)2-\left(2x-9\right)2\)
\(=2\left[\left(x+7\right)-\left(2x-9\right)\right]\)
\(=2\left(x+7-2x+9\right)\)
\(=2\left(16-x\right)\)
Vậy ...
f) \(x^2-6x+8\)
\(=x^2-6x+9-1\)
\(=\left(x-3\right)^2-1\)
\(=\left(x-4\right)\left(x-2\right)\)
Vậy ...
a) =2x - 3 =0
x = 3/2
b) (5x -1)2 = 0
5x - 1 = 0
x = 1/5
c) = ( x +3)2 = 0
x+3 = 0
x = -3
d) =(13+y)(13-y) = 0
y = 13; -13
e) xem lại đề bài này
\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)
\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)
\(x^2+3x-x-3=x\left(x+3\right)-\left(x+3\right)=\left(x-1\right)\left(x+3\right)\)
\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)
`a)(2x-1)^2-0,25=0`
`<=>(2x-1-0,5)(2x-1+0,5)=0`
`<=>(2x-1,5)(2x-0,5)=0`
`<=>[(x=0,75)(x=0,25):}`
`b)x^2+9=6x`
`<=>(x-3)^2=0`
`<=>x-3=0`
`<=>x=3`
`c)(x^2-4)-3x-6=0`
`<=>(x-2)(x+2)-3(x+2)=0`
`<=>(x+2)(x-2-3)=0`
`<=>(x+2)(x-5)=0`
`<=>[(x=-2),(x=5):}`
a: =>(2x-1-0,5)(2x-1+0,5)=0
=>(2x-1,5)(2x-0,5)=0
=>x=0,25 hoặc x=0,75
b: =>x^2-6x+9=0
=>(x-3)^2=0
=>x-3=0
=>x=3
c: =>(x-2)(x+2)-3(x+2)=0
=>(x+2)(x-5)=0
=>x=5 hoặc x=-2
\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)
\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)
Vậy \(x=\dfrac{-255}{2}\)
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
đáp án: -(x-1)
bạn đổi dấu vế phải, và phân tích hằng đẳng thức vế trái
`#3107`
`a)`
`(6x - 2)^2 + 4(3x - 1)(2 + y) + (y + 2)^2 - (6x + y)^2`
`= [(6x - 2)^2 - (6x + y)^2] + 4(3x - 1)(2 + y) + (2 + y)^2`
`= (6x - 2 - 6x - y)(6x -2 + 6x + y) + (2 + y)*[ 4(3x - 1) + 2 + y]`
`= (2 - y)(12x + y - 2) + (2 + y)*(12x - 4 + 2 + y)`
`= (2 - y)(12x + y - 2) + (2 + y)*(12x + y - 2)`
`= (12x + y - 2)(2 - y + 2 + y)`
`= (12x + y - 2)*4`
`= 48x + 4y - 8`
`b)`
\(5(2x-1)^2+2(x-1)(x+3)-2(5-2x)^2-2x(7x+12)\)
`= 5(4x^2 - 4x + 1) + 2(x^2 + 2x - 3) - 2(25 - 20x + 4x^2) - 14x^2 - 24x`
`= 20x^2 - 20x + 5 + 2x^2 + 4x - 6 - 50 + 40x - 8x^2 - 14x^2 - 24x`
`= - 51`
`c)`
\(2(5x-1)(x^2-5x+1)+(x^2-5x+1)^2+(5x-1)^2-(x^2-1)(x^2+1)\)
`= [ 2(5x - 1) + x^2 - 5x + 1] * (x^2 - 5x + 1) + (5x - 1)^2 - [ (x^2)^2 - 1]`
`= (10x - 2 + x^2 - 5x + 1) * (x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`
`= (x^2 + 5x - 1)(x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`
`= x^4 - (5x - 1)^2 + (5x - 1)^2 - x^4 + 1`
`= 1`
`d)`
\((x^2+4)^2-(x^2+4)(x^2-4)(x^2+16)-8(x-4)(x+4)\)
`= (x^2 + 4)*[x^2 + 4 - (x^2 - 4)(x^2 + 16)] - 8(x^2 - 16)`
`= (x^2 + 4)(x^4 + 12x^2 - 64) - 8x^2 + 128`
`= x^6 + 16x^4 - 16x^2 - 256 - 8x^2 + 128`
`= x^6 + 16x^4 - 24x^2 - 128`
a)-(x-y)(x2+xy-1)=-(x3+x2y-x-x2y-xy2+y)
=-(x3-xy2-x+y)
=-x3+xy2+x-y
b)x2(x-1)-(x3+1)(x-y)=x3-x2-x3+x2y-x+y
=-x2+x2y-x+y
c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x2-3x-4x+2-15x2-10x-3x-2
=-9x2-20x
d) hình như bạn ghi lỗi
Bài 2: C=x(x2-y)-x2(x+y)+y(x2-x)
=x3-xy-x3-x2y+x2y-xy
=-2xy
Thay x=1/2,y=-1 vào C, ta có:
C=-2.1/2.(-1)=1
Vậy C=1 khi x=1/2 và y=-1.
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