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6 tháng 9 2022
a: \(=\dfrac{2x\left(3x^2+2\right)+3x^2+2}{3x^2+2}=2x+1\)
b: \(=\dfrac{2x^3-10x^2-17x^2+85x+30x-150}{x-5}=2x^2-17x+30\)
c: \(=\dfrac{12x^4-8x^3+12x^3-8x^2+8x^2-\dfrac{16}{3}x+\dfrac{43}{3}x-\dfrac{86}{9}+\dfrac{113}{9}}{3x-2}\)
\(=4x^3+4x^2+\dfrac{8}{3}x+\dfrac{43}{9}x+\dfrac{\dfrac{113}{9}}{3x-2}\)
6 tháng 9 2022
a: \(=\dfrac{2x\left(3x^2+2\right)+3x^2+2}{3x^2+2}=2x+1\)
b: \(=\dfrac{2x^3-10x^2-17x^2+85x+30x-150}{x-5}=2x^2-17x+30\)
c: \(=\dfrac{12x^4-8x^3+12x^3-8x^2+8x^2-\dfrac{16}{3}x+\dfrac{43}{3}x-\dfrac{86}{9}+\dfrac{113}{9}}{3x-2}\)
\(=4x^3+4x^2+\dfrac{8}{3}x+\dfrac{43}{9}x+\dfrac{\dfrac{113}{9}}{3x-2}\)
ST
3
PN
10 tháng 8 2020
\(a,\left(2x^3-27x^2+115x-150\right)\left(x-5\right)\)
\(=x\left(2x^3-27x^2+115-150\right)-5\left(2x^3-27x^2+115-150\right)\)
\(=2x^4-27x^3+115x-150x-10x^3+135x^2-575+750\)
\(=2x^4-37x^3+135x^2-35x+175\)
KN
3 tháng 11 2019
a) \(\left(6x^3+3x^2+4x+2\right):\left(3x^2+2\right)\)
\(=\left[3x^2\left(2x+1\right)+2\left(2x+1\right)\right]⋮\left(3x^2+2\right)\)
\(=\left[\left(3x^2+2\right)\left(2x+1\right)\right]⋮\left(3x^2+2\right)\)
\(=2x+1\)
b) \(\left(2x^3-22x^2-5x^2+60x+55x-150\right):\left(x-5\right)\)
\(=\left[\left(2x^3-22x^2+60x\right)-\left(5x^2-55x+150\right)\right]:\left(x-5\right)\)
\(=\left[2x\left(x^2-11x+30\right)-5\left(x^2-11x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x^2-11x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x^2-5x-6x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x-5\right)\left(x-6\right)\right]:\left(x-5\right)\)
\(=\left(2x-5\right)\left(x-6\right)\)
\(=2x^2-17x+30\)
KN
3 tháng 11 2019
d) \(\left(x^5+4x^3+3x^2-5x+15\right):\left(x^3-x+3\right)\)
\(=\left(x^5+5x^3+3x^2-x^3-5x+15\right):\left(x^3-x+3\right)\)
\(=\left[\left(x^5-x^3+3x^2\right)+\left(5x^3-5x^2+15\right)\right]:\left(x^3-x+3\right)\)
\(=\left[x^2\left(x^3-x+3\right)+5\left(x^3-x^2+3\right)\right]:\left(x^3-x+3\right)\)
\(=\left[\left(x^2+5\right)\left(x^3-x+3\right)\right]:\left(x^3-x+3\right)\)
\(=x^2+5\)
12 tháng 1 2023
a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
b: =(1-2x)(1+2x)
c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
d: =(x+3)^3
e: \(=\left(2x-y\right)^3\)
f: =(x+2y)(x^2-2xy+4y^2)
30 tháng 7 2020
Bài 4:
a) Ta có: \(x^3+6x^2+12x+8\)
\(=x^3+2x^2+4x^2+8x+4x+8\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+4x+4\right)\)
\(=\left(x+2\right)^3\)
b) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
c) Ta có: \(1-9x+27x^2-27x^3\)
\(=1-3x-6x+18x^2+9x^2-27x^3\)
\(=\left(1-3x\right)-6x\left(1-3x\right)+9x^2\left(1-3x\right)\)
\(=\left(1-3x\right)\left(1-6x+9x^2\right)\)
\(=\left(1-3x\right)^3\)
d) Ta có: \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=\left(x+\frac{1}{2}\right)^3\)
e) Ta có: \(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(3x-2y\right)^3\)