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a) \(20a^4b^5c^2:\left(-5ab^2c\right)^2=20a^4b^5c^2:\left(25a^2b^4c^2\right)=\dfrac{4}{5}a^2b\)
b) \(\left(-15x^2y^3\right)^7:\left(15xy^3\right)^6-\left(32x^{18}y^5\right):\left(-4x^5y\right)^2=-15x^8y^3-2x^8y^3=-17x^8y^3\)
c) \(-13-13x^5y^2:\left(-2xy\right)-\left(x^2+2x+1\right):\left(x+1\right)=-13+\dfrac{13}{2}x^4y-\left(x+1\right)^2:\left(x+1\right)=-13+\dfrac{13}{2}x^4y-x-1=-14+\dfrac{13}{2}x^4y-x\)
a: \(\dfrac{20a^4b^5c^2}{\left(-5ab^2c\right)^2}=\dfrac{20a^4b^5c^2}{25a^2b^4c^2}=\dfrac{4}{5}a^2b\)
b: \(\dfrac{\left(-15x^2y^3\right)^7}{\left(15xy^3\right)^6}-\dfrac{\left(32x^{18}y^5\right)}{\left(-4x^5y\right)^2}\)
\(=\dfrac{\left(-15\right)^7\cdot x^{14}\cdot y^{21}}{15^6\cdot x^6\cdot y^{18}}-\dfrac{32x^{18}y^5}{16x^{10}y^2}\)
\(=-15x^8y^3-2x^8y^3\)
\(=-17x^8y^3\)
a: \(\dfrac{20a^4b^5c^2}{\left(-5ab^2c\right)^2}\)
\(=\dfrac{20a^4b^5c^2}{25a^2b^4c^2}\)
\(=\dfrac{4}{5}a^2b\)
b: \(\dfrac{\left(-15x^2y^3\right)^7}{\left(15xy^3\right)^6}-\dfrac{32x^{18}y^5}{\left(-4x^5y\right)^2}\)
\(=\dfrac{-15^7\cdot x^{14}\cdot y^{21}}{15^6\cdot x^6\cdot y^{18}}-\dfrac{32x^{18}y^5}{16x^{10}y^2}\)
\(=-15x^8y^3-2x^8y^3\)
c: \(\dfrac{-\dfrac{1}{3}x^5y^2}{-2xy}-\dfrac{x^2+2x+1}{x+1}\)
\(=\dfrac{2}{3}x^3y-x-1\)
Bài 1:
a: \(=\dfrac{2x^4-8x^3+2x^2+2x^3-8x^2+2x+18x^2-72x+18+56x-15}{x^2-4x+1}\)
\(=2x^2+2x+18+\dfrac{56x-15}{x^2-4x+1}\)
b: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
a: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
\(=\left(x^4-x^3+x^2+x^3-x^2+x+3x^2-3x+3+2018\right):\left(x^2-x+1\right)\\ =\left[\left(x^2-x+1\right)\left(x^2+x+3\right)+2018\right]:\left(x^2-x+1\right)\\ =x^2+x+3\left(\text{dư 2018}\right)\)
1: \(=\left(x-1\right)^2\)
2: \(x\in\left\{0;20\right\}\)
Câu 13:
\(1,=\left(x-1\right)^2\\ 2,\Leftrightarrow x\left(x-20\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=20\end{matrix}\right.\\ 3,\text{Đề lỗi}\)
Câu 14:
\(1,ĐK:x\ne-2\\ 2,=\dfrac{\left(x+2\right)^2}{x+2}=x+2\\ 3,\Leftrightarrow x+2=0\Leftrightarrow x=-2\left(ktm\right)\Leftrightarrow x\in\varnothing\)
Câu 16:
\(A=x^2-4x+4+20=\left(x-2\right)^2+20\ge20\)
Dấu \("="\Leftrightarrow x=2\)
`-1/3x^5y^2:(-2xy)-(x^2+2x+1):(x+1)`
`=-1/3:(-2).(x^5:x).(y^2:y)-(x+1)^2:(x+1)`
`=-1/6x^4y-(x+1)`
`=-1/6x^4y-x-1`
\(\dfrac{-1}{3}x^5y^2:\left(-2xy\right)-\left(x^2+2x+1\right):\left(x+1\right)\)
\(=\dfrac{1}{6}x^4y-x-1\)