Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
a: \(=2x^2-6x+x-3-20x+8x^2\)
\(=10x^2-25x-3\)
b: \(=x^2+4x+4-2\left(x^2-9\right)+10\)
\(=x^2+4x+14-2x^2+18\)
\(=-x^2+4x+32\)
a: \(=\dfrac{\left(x+1\right)\left[\left(3x-2\right)-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)
=3x-2-2x^2+2x-5x+5
=-2x^2+3
b: \(=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\)
c: =x^3-3x^2+3x-1-x^3-1+9x^2-1
=6x^2+3x-3
\(a,\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]:\left(x+1\right)\)
\(=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x-1\right)\left(x+1\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(3x-2-\left(2x+5\left(x-1\right)\right)\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(3x-2-2x^2+2x-5x+5\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(-2x^2+3\right)\right].\dfrac{1}{x+1}\)
\(=-2x^2+3\)
\(b,\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)\)
\(=\left(2x+1\right)\left[\left(2x+1\right)-2\left(3-x\right)\right]\)
\(=\left(2x+1\right)\left(2x+1-6+2x\right)\)
\(=\left(2x+1\right)\left(4x-5\right)\)
\(c,\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(3x+1\right)\left(1-3x\right)\)
\(=x^3-3x^2+3x-1-x^3-1-\left(3x-9x^2+1-3x\right)\)
\(=-3x^2+3x-2-3x+9x^2-1+3x\)
\(=6x^2+3x-3\)
`@` `\text {Ans}`
`\downarrow`
`A= (2x - 3)^2 - (2x + 3)^2`
`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`
`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`
`= -6 * 4x`
`= -24x`
\(a,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)+\left(-6x-5x\right)+24\)
\(=0-11x+24\)
\(=-11x+24\)
\(b,\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x\)
\(=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x\)
\(=\left(14x^2-5x^2-9x^2\right)+\left(7x-6x-20x+2x+17x\right)+\left(-3+8\right)\)
\(=0+0+5\)
\(=5\)
a)
\(2x\left(9-x\right)+\left(2x+5\right)\left(x+1\right)\\ =18x-2x^2+2x^2+5x+2x+5\\ =25x+5\\ =5\left(5x+1\right)\)
b)
\(\left(x-3\right)^2+\left(x+3\right)^2+2\left(3-x\right)\left(3+x\right)\\ =\left(3-x\right)^2+2\left(3-x\right)\left(3+x\right)+\left(3+x\right)^2\\ =\left[\left(3-x\right)+\left(3+x\right)\right]^2\\ =6^2=36\)
a: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
b: \(=3x^2-6x-5x+5x^2-8x^2+24\)
=-11x+24
Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
\(2x.\left(9-x\right)+\left(2x+5\right).\left(x+1\right)\)
\(=2x.9+2x.\left(-x\right)+2x.x+2x.1+5x+5.1\)
\(=18x-2x^2+2x^2+2x+5x+5\)
\(=25x+5\)
\(\left(x-3\right)^2+\left(x+3\right)^2+2.\left(3-x\right).\left(3+x\right)\)
\(=\left(x-3\right)^2-2.\left(x-3\right).\left(x+3\right)+\left(x+3\right)^2\)
\(=[\left(x-3\right)-\left(x+3\right)]^2\)
\(=\left(x-3-x-3\right)^2\)
\(=36\)