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a: \(=\dfrac{5x^2y^4}{-10x^2y}=-\dfrac{1}{2}y^3=-\dfrac{1}{2}\cdot8=-4\)

b: \(=\dfrac{15x^4y^2}{5x^3y}+\dfrac{20x^3y^2}{5x^2y}=3xy+4xy=7xy\)

\(=7\cdot\dfrac{1}{7}\cdot2009=2009\)

 

15 tháng 8 2018

1. 20\(x^2y^3\) : 4x\(y^2\) = 5xy

2. \(\dfrac{-1}{2}x^4y^4\) : \(\dfrac{2}{3}x^2y^2\) = \(\dfrac{-3}{4}x^2y^2\)

3. \(\left(-xy\right)^6:\left(-xy\right)^2=\left(-xy\right)^2\) = xy

4. 27\(x^2y^3z^4:\left(-3xyz\right)^2\) = 27\(x^2y^3z^4\) : 9 \(x^2y^2z^2\) = 3y\(z^2\)

5. \(\left(-x\right)^{10}:\left(-x\right)^5=\left(-x\right)^2\) = x

26 tháng 7 2018

1,\(x^4-x=0\\ ->x\left(x-1\right)\left(x^2+x+1\right)=0\\ ->\left(......\right)\)

2\(x^4-x^2=0\\ ->x^2\left(x^2-1\right)\\ ->x^2\left(x-1\right)\left(x+1\right)\\ ->......\)

3,\(x^5+x^2\\ ->x^2\left(x^3+1\right)\\ ->x^2\left(x+1\right)\left(x^2-x+1\right)\\ ->.......\)

4\(3x\left(x-20\right)-x+20=0->\left(3x-1\right)\left(x-20\right)=0->.....\)

1: \(=\dfrac{x^2-1}{x\left(x^2-1\right)}=\dfrac{1}{x}\)

2: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{y\left(x-2\right)}=\dfrac{x+2}{y}\)

3: \(=\dfrac{2x^2+2xy-xy-y^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)\left(2x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2x-y}{x-y}\)

4: \(=\dfrac{x\left(x^2-1\right)}{x\left(x^2-x-2\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{x-1}{x-2}\)

 

13 tháng 9 2018

a) \(\Leftrightarrow\left(-63x^2+78x-15\right)+\left(63x^3+x-20\right)=44\)

\(\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\)

\(\Leftrightarrow79x-35=44\)

\(\Leftrightarrow79x=44+35\)

\(\Leftrightarrow79x=79\)

\(\Leftrightarrow x=1\)

b) \(\Leftrightarrow\left(x^2+3x+2\right).\left(x+5\right)-x^2.\left(x+8\right)=27\)

\(\Leftrightarrow x.\left(x^2+3x+2\right)+5.\left(x^2+3x+2\right)-x^3-8x^2=27\)

\(\Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\)

\(\Leftrightarrow17x+10=27\)

\(\Leftrightarrow17x=17\)

\(\Leftrightarrow x=1\)

4 tháng 12 2018

1) \(16x\left(2-x\right)-\left(4x-5\right)^2=0\)

\(32x-16x^2-16x^2+40x-25=0\)

\(72x-16x^2-25=0\)

Đề sai ko bạn nhonhung

2) \(\left(x-7\right)^2+3=\left(x-2\right)\left(x+2\right)\)

\(\left(x^2-14x+7\right)+3-\left(x-2\right)\left(x+2\right)=0\)

\(x^2-14x+7+3-x^2+4=0\)

\(-14x+14=0\)

\(x=1\)

3) \(\left(2x-3\right)^2-\left(7x-2x\right)^2=2\)

\(\left(2x-3\right)^2-\left(5x\right)^2=2\)

\(\left(2x-3-5x\right)\left(2x-3+5x\right)=2\)

\(\left(-3x-3\right)\left(7x-3\right)=2\)

=> lập bảng tìm x

4) \(\left(5x-7\right)^2-\left(1-3x\right)^2=16x\left(x-3\right)\)

\(25x^2-70x+49-9x^2+6x-1-16x^2+48x=0\)

\(-16x+48=0\)

\(x=3\)

4 tháng 12 2018

đề có vẻ sai nhiều quá :((

8 tháng 8 2018

1) \(-6x^2-x+7=0\)

\(\Leftrightarrow-6x^2+6x-7x+7=0\)

\(\Leftrightarrow\left(-6x^2+6x\right)-\left(7x-7\right)=0\)

\(\Leftrightarrow-6x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(-6x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-6x-7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)

2) \(-4x^2-5x+9=0\)

\(\Leftrightarrow-4x^2+4x-9x+9=0\)

\(\Leftrightarrow\left(-4x^2+4x\right)-\left(9x-9\right)=0\)

\(\Leftrightarrow-4x\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(-4x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-4x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=1\end{matrix}\right.\)

3) \(x^2+3x-4=0\)

\(\Leftrightarrow x^2-x+4x-4=0\)

\(\Leftrightarrow\left(x^2-x\right)+\left(4x-4\right)=0\)

\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

4) \(x^2-6x-7=0\)

\(\Leftrightarrow x^2+x-7x-7=0\)

\(\Leftrightarrow\left(x^2+x\right)-\left(7x+7\right)=0\)

\(\Leftrightarrow x\left(x+1\right)-7\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

5) \(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow\left(x^2+x\right)+\left(4x+4\right)=0\)

\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

a) \(-6x^2-x+7=0\)

\(\Leftrightarrow-6x^2+6x-7x+7=0\)

\(\Leftrightarrow-6x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-6x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-7}{6}\end{matrix}\right.\)

b) \(-4x^2-5x+9=0\)

\(\Leftrightarrow-4x^2+4x-9x+9=0\)

\(\Leftrightarrow-4x\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-4x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2,25\end{matrix}\right.\)

c) \(x^2+3x-4=0\)

\(\Leftrightarrow x^2-x+4x-4=0\)

\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

d) \(x^2-6x-7=0\)

\(\Leftrightarrow x^2+x-7x-7=0\)

\(\Leftrightarrow x\left(x+1\right)-7\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)

e) \(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)