20x^4y-25x^2y^2+3x
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22 tháng 1 2022
\(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)
\(=20x^4y:5x^2y-25x^2y^2:5x^2y-3x^2y:5x^2y\)
\(=4x^2-5y-\frac{3}{5}\)
15 tháng 7 2021
Bài 1:
a) Ta có: \(A=-x^2-4x-2\)
\(=-\left(x^2+4x+2\right)\)
\(=-\left(x^2+4x+4-2\right)\)
\(=-\left(x+2\right)^2+2\le2\forall x\)
Dấu '=' xảy ra khi x=-2
b) Ta có: \(B=-2x^2-3x+5\)
\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)
\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)
c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)
\(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9\le9\forall x\)
Dấu '=' xảy ra khi x=-1
15 tháng 7 2021
Bài 2:
a) Ta có: \(=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)
b) Ta có: \(B=9x^2-6xy+2y^2+1\)
\(=9x^2-6xy+y^2+y^2+1\)
\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)
c) Ta có: \(E=x^2-2x+y^2-4y+6\)
\(=x^2-2x+1+y^2-4y+4+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)
14 tháng 2 2017
\(\hept{\begin{cases}\frac{25x^2-y^2}{20x-4y-3\left(5x+y\right)}=3\\\frac{25x^2-y^2}{2\left(5x-y\right)+10x+2y}=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{\left(5x-y\right)\left(5x+y\right)}{4\left(5x-y\right)-3\left(5x+y\right)}=3\\\frac{\left(5x-y\right)\left(5x+y\right)}{2\left(5x-y\right)+2\left(5x+y\right)}=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{4\left(5x-y\right)-3\left(5x+y\right)}{\left(5x-y\right)\left(5x+y\right)}=\frac{1}{3}\\\frac{2\left(5x-y\right)+2\left(5x+y\right)}{\left(5x-y\right)\left(5x+y\right)}=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{4}{5x+y}-\frac{3}{5x-y}=\frac{1}{3}\\\frac{2}{5x+y}+\frac{2}{5x-y}=1\end{cases}}\)
Đặt: \(\hept{\begin{cases}\frac{1}{5x+y}=a\\\frac{1}{5x-y}=b\end{cases}}\)thì hệ thành
\(\hept{\begin{cases}4a-3b=\frac{1}{3}\\2a+2b=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=\frac{11}{42}\\b=\frac{5}{21}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{5x+y}=\frac{11}{42}\\\frac{1}{5x-y}=\frac{5}{21}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{441}{550}\\y=-\frac{21}{110}\end{cases}}\)
PS: Bí thì bỏ chứ đăng lên làm gì :3
28 tháng 10 2020
a. \(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)
\(=4x^2-5y-\frac{3}{5}\)
b. \(\left(15xy^2+17xy^3+18y^2\right):6y^2\)
\(=\frac{5}{2}x+\frac{17}{6}xy+3\)
c. \(\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(y-x\right)^2\)
\(=\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(x-y\right)^2\)
\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)
d. \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)
H9
HT.Phong (9A5)
CTVHS
30 tháng 7 2023
1) \(4x^5y^2-8x^4y^2+4x^3y^2\)
\(=4x^3y^2\left(x^2-2x+1\right)\)
\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)
\(=4x^3y^2\left(x-1\right)^2\)
2) \(5x^4y^2-10x^3y^2+5x^2y^2\)
\(=5x^2y^2\left(x^2-2x+1\right)\)
\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)
\(=5x^2y^2\left(x-1\right)^2\)
3) \(12x^2-12xy+3y^2\)
\(=3\left(4x^2-4xy+y^2\right)\)
\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=3\left(2x-y\right)^2\)
4) \(8x^3-8x^2y+2xy^2\)
\(=2x\left(4x^2-4xy+y^2\right)\)
\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=2x\left(2x-y\right)^2\)
5) \(20x^4y^2-20x^3y^3+5x^2y^4\)
\(=5x^2y^2\left(4x^2-4xy+y^2\right)\)
\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=5x^2y^2\left(2x-y\right)^2\)
29 tháng 7 2023
1: 4x^5y^2-8x^4y^2+4x^3y^2
=4x^3y^2(x^2-2x+1)
=4x^3y^2(x-1)^2
2: \(=5x^2y^2\left(x^2-2x+1\right)=5x^2y^2\left(x-1\right)^2\)
3: \(=3\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)^2\)
4: \(=2x\left(4x^2-4xy+y^2\right)=2x\left(2x-y\right)^2\)
5: \(=5x^2y^2\left(4x^2-4xy+y^2\right)=5x^2y^2\left(2x-y\right)^2\)
24 tháng 10 2021
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
GH
21 tháng 10 2017
a) (10x3y - 5x2y2 - 25x4y3) : (- 5xy)
= - 2x2 + xy + 5x3y2
b) (27x3 - y3) : (3x - y)
= (3x - y)(9x2 - 3xy + y2) : (3x - y)
= 9x2 - 3xy + y2
9 tháng 12 2018
a, 15x3y5z : 5x2y3 = 3xy2z.
b, 12x4y2 : ( - 9xy2 ) = \(\frac{3}{4}x^3\).
c, ( 30x4y3 - 25x2y3 - 3x4y4 ) : 5x2y3 = \(6x^2-5-\frac{3}{5}x^2y.\)
d, ( 4x4 - 8x2y2 + 12x5y ) : ( - 4x2 ) = -x2 + 2y2 - 3x3y.
=x*20x^3y-x*25xy^2+3*x
=x(20x^3y-25xy^2+3)