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\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
\(1,x^3-7x+6\)
\(=x^3+3x^2-3x^2-9x+2x+6\)
\(=x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+2\right)\)
\(=\left(x+3\right)\left(x^2-2x-x+2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x-1\right)\)
\(2,x^3-9x^2+6x+16\)
\(=x^3+x^2-10x^2-10x+16x+16\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left(x^2-2x-8x+16\right)\)
\(=\left(x+1\right)\left(x-8\right)\left(x-2\right)\)
mk ms lm hai câu thôi mà đã mệt r , bh mk lm bt mai đi học ,lúc khác lm đ cko bn
a) ( x2 - 5 )( x + 3 ) = x3 + 3x2 - 5x - 15
b) ( x + 4 )( x - x2 ) = x2 - x3 + 4x - 4x2 = -x3 - 3x2 + 4x
c) ( x2 - 6 )( x + 2 ) + ( x + 3 )( x - x2 ) = x3 + 2x2 - 6x - 12 + x2 - x3 + 3x - 3x2 = -3x - 12 = -3( x + 4 )
d) x( x - y ) - y( x - y ) = ( x - y )( x - y ) = ( x - y )2
e) x2( x + y ) - x( x2 - y ) = x3 + x2y - x3 + xy = x2y + xy = xy( x + 1 )
f) 3x( 12x - 4 ) - 9x( 4x - 3 ) = 36x2 - 12x - 36x2 + 27x = 15x
Bài làm
a) ( x2 - 5 )( x + 3 )
= x3 + 3x2 - 5x - 15
b) ( x + 4 )( x - x2 )
= ( x + 4 ) . x( 1 - x )
= x( x + 4 )( 1 - x )
= x( x - x2 + 4 - 4x )
= x( 4 - x2 - 3x )
= 4x - x3 - 3x2
c) ( x2 - 6 )( x + 2 ) + ( x + 3 )( x - x2 )
= ( x - 3 )( x + 3 )( x + 2 ) + ( x + 3 )( x - x2 )
= ( x + 3 )[ ( x - 3 )( x + 2 ) + ( x - x2 )]
= ( x + 3 ) [ x2 + 2x - 3x - 6 + x2 - x2 ]
= ( x + 3 ) ( x2 - x - 6 )
= x3 - x2 - 6x + 3x2 - 3x - 18
= x3 + 2x2 - 9x - 18
d) x( x - y ) - y( x - y )
= ( x - y )( x - y )
= ( x - y )2
= x2 - 2xy + y
e) x2( x + y ) - x( x2 - y )
= x3 + x2y - x3 + xy
= x2y + xy
f) 3x( 12x - 4 ) - 9x( 4x - 3 )
= 3x . 3( 4x - 1 ) - 9x( 4x - 3 )
= 9x( 4x - 1 ) - 9x( 4x - 3 )
= 9x( 4x - 1 - 4x + 3 )
= 9x . 2
= 18x
\(a.x^4-16x^2=0\Leftrightarrow\left(x^2+4x\right)\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+4=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
\(b.\left(x-5\right)^3-x+5=0\)
\(\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
a) x4 - 16x2 = 0
<=> x2 ( x2 - 16 ) = 0
<=> \(\left[{}\begin{matrix}x^2=0\\x^2-16=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
Vậy...
b) ( x - 5)3 - x + 5 = 0
<=> ( x - 5)3 - (x - 5) = 0
<=> (x - 5) [ (x - 5)2 - 1] =0
<=> \(\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Vậy...
c) 5(x - 2) = x2 - 4
<=> 5(x - 2) - (x2 - 4) = 0
<=> (x - 2)( 5 - x - 2) = 0
<=> (x - 2)( 3 - x ) = 0
<=> \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy...
d) x - 3 = (3 - x)2
<=> x - 3 - (x - 3)2 = 0
<=> (x - 3)(1 - x + 3) = 0
<=> (x - 3)( 4 - x ) = 0
<=> \(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy...
e) x2 (x - 5) + 5 - x = 0
<=> x2 (x - 5) - (x - 5) = 0
<=> (x2 - 1)( x - 5) = 0
<=> \(\left[{}\begin{matrix}\left(x-1\right)\left(x+1\right)=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
,
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
a, \(x^6-x^4-9x^3+9x^2\)
= \(x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
=\(x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
= \(\left(x-1\right)\left(x^4\left(x+1\right)-9x^2\right)\)
= \(\left(x-1\right)\left(x^5+x-9x^2\right)\)
b, \(x^4-4x^3+8x^2-16x+16\)
= \(x^4-4x^3+4x^2+4x^2-16x+16\)
\(=x^2\left(x^2-4x+4\right)+4\left(x^2-4x+4\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
c, \(\left(xy+4\right)^2-4\left(x+y\right)^2\)
= \(\left(xy+4\right)^2-\left(2\left(x+y\right)\right)^2\)
= \(\left(xy-2x-2y+4\right)\left(xy+2x+2y+4\right)\)
= \(\left(x\left(y-2\right)-2\left(y-2\right)\right)\left(x\left(y+2\right)+2\left(y+2\right)\right)\)
=\(\left(x-2\right)\left(y-2\right)\left(x+2\right)\left(y+2\right)\)
d, \(\left(a+b+c\right)^2+\left(a-b+c\right)^2-4b^2\)
= \(a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2-2ab+2ac-2bc-4b^2\)
=\(2a^2+2b^2+2c^2+4ac-4b^2\)