\(\dfrac{3}{2}.ax=3.\dfrac{3x-2y}{2}.a\)
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23 tháng 9 2018
a) ta có : \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2=\dfrac{3x^5y^2}{2x^2y^2}+\dfrac{4x^3y^3}{2x^2y^2}-\dfrac{5x^2y^4}{2x^2y^2}\)
\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)
b) ta có : \(\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right):\dfrac{3}{5}ax^3\)
\(=\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right)\dfrac{5}{3ax^3}\)
\(=\dfrac{3}{5}.\dfrac{5}{3}\dfrac{a^6x^3}{ax^3}+\dfrac{3}{7}.\dfrac{5}{3}\dfrac{a^3x^4}{ax^3}-\dfrac{9}{10}.\dfrac{5}{3}\dfrac{ax^5}{ax^3}\)
\(=a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}a^2\)
30 tháng 5 2023
a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2
b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y
=>A-B=12xy^2-14x^2y
c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2
=>A-B=-5x^2y^3-x^3y^2
d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2
D
19 tháng 1 2022
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)
19 tháng 12 2021
c: \(=\dfrac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}=\dfrac{3x}{x^2+1}\)
7 tháng 8 2021
a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)
\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)
b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)
\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)
c, tương tự
28 tháng 10 2023
a: \(\left(3x+y\right)^2=\left(3x\right)^2+2\cdot3x\cdot y+y^2=9x^2+6xy+y^2\)
b: \(\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)
\(=x^2+4xy+4y^2\)
c: \(\left(\dfrac{2}{3}x-1\right)^2=\left(\dfrac{2}{3}x\right)^2-2\cdot\dfrac{2}{3}x\cdot1+1^2\)
\(=\dfrac{4}{9}x^2-\dfrac{4}{3}x+1\)
d: \(\left(\dfrac{1}{3}x+2\right)^2=\left(\dfrac{1}{3}x\right)^2+2\cdot\dfrac{1}{3}x\cdot2+2^2\)
\(=\dfrac{1}{9}x^2+\dfrac{4}{3}x+4\)
e: \(\left(4x-2y\right)^2=\left(4x\right)^2-2\cdot4x\cdot2y+\left(2y\right)^2\)
\(=16x^2-16xy+4y^2\)
H9
HT.Phong (9A5)
CTVHS
28 tháng 10 2023
a) \(\left(3x+y\right)^2\)
\(=\left(3x\right)^2+2\cdot3x\cdot y+y^2\)
\(=9x^2+6xy+y^2\)
b) \(\left(x+2y\right)^2\)
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)
\(=x^2+4xy+4y^2\)
c) \(\left(\dfrac{2}{3}x-1\right)^2\)
\(=\left(\dfrac{2}{3}x\right)^2-2\cdot\dfrac{2}{3}x\cdot1+1^2\)
\(=\dfrac{4}{9}x^2-\dfrac{4}{3}x+1\)
d) \(\left(\dfrac{1}{3}x+2\right)^2\)
\(=\left(\dfrac{1}{3}x\right)^2+2\cdot\dfrac{1}{3}x\cdot2+2^2\)
\(=\dfrac{x^2}{9}+\dfrac{4}{3}x+4\)
e) \(\left(4x-2y\right)^2\)
\(=\left(4x\right)^2-2\cdot4x\cdot2y+\left(2y\right)^2\)
\(=16x^2-16xy+4y^2\)
18 tháng 10 2021
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
\(\dfrac{3}{2}ax=3a.\dfrac{3x-2y}{2}\\ \Leftrightarrow\dfrac{3}{2}ax-3a.\dfrac{3x-2y}{2}=0\\ \Leftrightarrow\dfrac{3}{2}a\left(X-3x+2y\right)=0\\ \left\{{}\begin{matrix}\dfrac{3}{2}a=0\\-2x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0\\x=y\end{matrix}\right.\)