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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 : \(\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right):\dfrac{3}{5}ax^3\)
\(=\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right)\dfrac{5}{3ax^3}\)
\(=\dfrac{3}{5}.\dfrac{5}{3}\dfrac{a^6x^3}{ax^3}+\dfrac{3}{7}.\dfrac{5}{3}\dfrac{a^3x^4}{ax^3}-\dfrac{9}{10}.\dfrac{5}{3}\dfrac{ax^5}{ax^3}\)
\(=a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}a^2\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
1: =>2x-5=4 hoặc 2x-5=-4
=>2x=9 hoặc 2x=1
=>x=9/2hoặc x=1/2
2: \(\Leftrightarrow\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}=\dfrac{-1}{8}\)(vô lý)
3: \(\Leftrightarrow\left|5x-3\right|=x+5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(5x-3-x-5\right)\left(5x-3+x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(4x-8\right)\left(6x+2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;-\dfrac{1}{3}\right\}\)
a)\(\dfrac{32x^5\left(3y-7\right)^5}{-4x\left(7-3y\right)^4}=\dfrac{-4x.\left(-8x^4\right)\left(3y-7\right)^4\left(3y-7\right)}{-4x\left(3y-7\right)^4}\)
\(=\dfrac{\left(-8x^4\right)\left(3y-7\right)}{1}=\left(-8x^4\right)\left(3y-7\right)\)
\(=-32x^4y+56x^4\)
b) \(\dfrac{12x^3\left(3x-5\right)^2}{4x\left(3x-5\right)^2}-\dfrac{2x\left(x+7\right)}{\left(x+7\right)^3}=\dfrac{12x^3}{4x}-\dfrac{2x}{\left(x+7\right)^2}\)
\(=3x^2-\dfrac{2x}{\left(x+7\right)^2}\)
\(\)
bạn ơi bạn ghi nhầm đề bài câu b r