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c)(x2+x)2-2(x2+x)-15
đặt x2+x=a ta có
a2-2a-15
=a2+3a-5a-15
=(a2+3a)-(5a+15)
=a(a+3)-5(a+3)
=(a+3)(a-5)
thay a=x2+x
(x2+x+3)(x2+x-5)
a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )
\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )
\(\dfrac{2x^2-x}{x-1}=\dfrac{x-2x^2}{1-x}\)
b)Ta có : \(\dfrac{1}{x-1}=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x+1}{x^3-1}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2-2x}{x^3-1}\)
\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )
c) MTC = ( x+ 2)2(x - 2)2
Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}\)
\(\dfrac{x}{x^2-4}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x^2-2^2\right)}{\left(x+2\right)^2\left(x-2\right)^2}=\dfrac{x^3-4x}{\left(x+2\right)^2\left(x-2\right)^2}\)
d) MTC = xyz( x - y)( y - z)( x - z)
Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Cộng các phân thức lại ta có :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
a) \(x^4-3x^3-x+3=x^4-x^3-2x^3+2x^2-2x^2+2x-3x+3\)
\(=x^3\left(x-1\right)-2x^2\left(x-1\right)-2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x^3-2x^2-2x-3\right)\left(x-1\right)=\left(x^3+x^2+x-3x^2-3x-3\right)\left(x-1\right)\)
\(=\left(x\left(x^2+x+1\right)-3\left(x^2+x+1\right)\right)\left(x+1\right)=\left(x^2+x+1\right)\left(x-3\right)\left(x-1\right)\)
b) \(x^2y^2\left(y-x\right)+y^2z^2\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=-x^2y^2\left(x-y\right)+y^2z^2\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=\left(y^2z^2-x^2y^2\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z^2-x^2\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z+x\right)\left(z-x\right)\left(x-y\right)-z^2x^2\left(z-x\right)\)
\(=\left(y^2\left(z+x\right)\left(x-y\right)-z^2x^2\right)\left(z-x\right)\)
c) câu này đề có sai o bn
hình như đề là : \(4x^2+4x^2y-8y^2\) mới đúng chứ ?? ?
a) 3x+2(x-5)=-x+2
<=> 3x+2x+x=2+10
<=>6x=12
<=>x=2
b) 3x2-2x=0
<=>x(3x-2)=0
<=>\(\left[{}\begin{matrix}x=0\\3x-2=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) \(\dfrac{2x}{3}\)+\(\dfrac{x-4}{6}\)=2-\(\dfrac{x}{2}\)
<=>\(\dfrac{8x+2x-8}{12}\)=\(\dfrac{24-6x}{12}\)
<=> 8x+2x-8=24-6x
<=>8x+2x+6x=24+8
<=>16x=32
<=>x=2
d) \(\dfrac{x-2}{x+2}\)-\(\dfrac{3}{x-2}\)= -\(\dfrac{2\left(x-11\right)}{4-x^2}\) ( ĐKXĐ: x\(\ne\)\(\pm\)2)
<=> \(\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{x^2-4}\)=\(\dfrac{2\left(x-11\right)}{x^2-4}\)
=> (x-2)2-3(x+2)=2(x-11)
<=> x2-4x+4-3x-6=2x-22
<=> x2-4x-3x-2x=-22-4+6
<=> x-9x+20=0
<=> (x-4)(x-5)=0
<=>\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) ( thỏa mãn diều kiện )
d) (x2+1)(x2-4x+4)=0
=> x2-4x+4=0 (x2+1\(\ge\)1 với mọi x)
=>(x-2)2 =0
=>x=2
Giải:
a) \(x^2-2xy+y^2-xz+yz\)
\(=\left(x^2-2xy+y^2\right)-\left(xz-yz\right)\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
b) \(3x\left(x-1\right)+7x^2\left(x-1\right)\)
\(=x\left(x-1\right)\left(3+7x\right)\)
c) \(x^3+2x^2y+xy^2-9x\)
\(=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left[\left(x^2+2xy+y^2\right)-9\right]\)
\(=x\left[\left(x+y\right)^2-3^2\right]\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
Chúc bạn học tốt!
Phân tích đa thức thành nhân tử :
Hướng dẫn câu a : Bạn vận dụng phương pháp dùng hằng đẳng thức và đặt nhân tử chung để phân tích đa thức này thành nhân tử nhé.
a) \(x^2-2xy+y^2-xz+yz\)
\(=\left(x^2-2xy+y^2\right)-\left(xz-yz\right)\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
Hướng dẫn làm câu b) : Bạn vận dụng kiến thức về đặt nhân tử chung để phân tích.
b) \(3x\left(x-1\right)+7x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+7x^2\right)\)
\(=\left(x-1\right)x\left(3+7x\right)\)
c: \(\left(x^2+2x\right)^2+9x^2+18x+20\)
\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)
\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)
d: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8+2x\right)\left(x^2+4x+8+x\right)\)
\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
\(=\left(x^2+5x+8\right)\left(x+4\right)\left(x+2\right)\)
\(A=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)
\(\Leftrightarrow A=\left(25x^2-20x+4\right)-\left(36x^2+12x+1\right)+11\left(x^2-4\right)-\left(48-32x\right)\)
\(\Leftrightarrow A=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)
\(\Leftrightarrow A=-89\)
Vây biểu thức A không phụ thuộc vào biến
a,\(A=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)
\(A=(25x^2-20x+4)-\left(36x^2+12x+1\right)+11\left(x^2-4\right)-48+32x\) \(A=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)
\(A=25x^2-36x^2+11x^2-20x-12x+32x+4-1-44-48\)
\(A=-89\)
Vậy giá trị của biểu thức trên không phụ thuộc vào giá trị của x.
`a)P(x)+Q(x)=3x^4-x^3+4x^2+2x+1-2x^4-x^2+x-2`
`=x^4-x^3+3x^2+3x-1`
`b)Q(x)-H(x)=-2x^4-2`
`=>H(x)=Q(x)-(-2x^4-2)`
`=>H(x)=-2x^4-x^2+x-2+2x^4+2`
`=>H(x)=-x^2+x`
`c)` Cho `H(x)=0`
`=>-x^2+x=0`
`=>-x(x-1)=0`
`@TH1:-x=0=>x=0`
`@TH2:x-1=0=>x=1`
\(a,P\left(x\right)+Q\left(x\right)=x^4-x^3+3x^2+3x-1\)
\(b,H\left(x\right)=Q\left(x\right)+2x^4+2=-2x^4-x^2+x-2+2x^4+2=-x^2+x\)
\(c,H\left(x\right)=-x^2+x=x\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)