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21 tháng 8 2021

\(\left(x^3-x^2-5x+21\right):\left(x^2-4x+7\right)\)

\(=\left(x^3-4x^2+3x^2+7x-12x+21\right):\left(x^2-4x+7\right)\)

\(=\left[\left(x^3-4x^2+7x\right)+\left(3x^2-12x+21\right)\right]:\left(x^2-4x+7\right)\)

\(=\left[x\left(x^2-4x+7\right)+3\left(x^2-4x+7\right)\right]:\left(x^2-4x+7\right)\)

\(=\left[\left(x^2-4x+7\right)\left(x+3\right)\right]:\left(x^2-4x+7\right)\)

\(=x+3\)

21 tháng 8 2021

đầy đủ giúp em nhé

\(\left(x^3-6x^2+9x+14\right):\left(x-7\right)\)

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

\(=[x^2\left(x-7\right)+x\left(x-7\right)-2\left(x-7\right)]:\left(x-7\right)\)

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

\(=x^2+x-2\)

11 tháng 8 2021

gi 9x46

a ) -36a2 + x2 + 4y2 - 4xy

= ( x2 - 4xy + 4y) - (6a)2

= ( x -2y )2 - (6a)2

= ( x - 2y - 6a ).(x - 2y + 6a )

b ) 10ax - 5ay +2x - y 

= ( 10ax - 5ay ) + ( 2x - y )

= 5a ( 2x - y ) + ( 2x - y )

= ( 2x - y ) . (5a + 1 ) 

c ) 2a2b(x + y) - 4ab2(-x - y )

= 2a2b( x+ y ) + 4ab2(x + y )

= 2ab(x + y ) ( a + 2b )

13 tháng 8 2021

a, \(-36a^2+x^2+4y^2-4xy=\left(x+2y\right)^2-\left(6a\right)^2=\left(x+2y-6a\right)\left(x+2y+6a\right)\)

b, \(10ax-5ay+2x-y=5a\left(2x-y\right)+2x-y=\left(5a+1\right)\left(2x-y\right)\)

c, \(2a^2b\left(x+y\right)-4ab^2\left(-x-y\right)=2a^2b\left(x+y\right)+4ab^2\left(x+y\right)\)

\(=\left(2a^2b+4ab^2\right)\left(x+y\right)=2ab\left(a+2b\right)\left(x+y\right)\)

9 tháng 10 2021

a)=\(3x^3-15x^2+21x\)

b)\(=-2x^4y-10x^2y+2xy\)

c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)

d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)

e)\(=x^2-4y^2\)

f)\(=-2x^2y^3+y-3\)

g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)

h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)

i)\(=x^2-x-3\)

j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)

24 tháng 10 2021

Tại sao ý b có dấu - trước ngoặc đâu mà đổi dấu mong bn giải đáp

13 tháng 8 2023

a) \(\left(x^2+5x-6\right):\left(x-1\right)\)

\(=\left[x\left(x+6\right)-\left(x+6\right)\right]:\left(x-1\right)\)

\(=\left(x-1\right)\left(x+6\right):\left(x-1\right)\)

\(=x+6\)

b) \(\left(x^3-x^2-5x+21\right):\left(x^2-4x+7\right)\)

\(=\left(x+3\right)\left(x^2-4x+7\right):\left(x^2-4x+7\right)\)

\(=x+3\)

2 tháng 9 2021

Đầy đủ giúp em nhé

9 tháng 10 2021

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)