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Phân tích đa thức thành nhân tử
4x3 + 38x2y + 72xy2 - 90y3
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a) Ta thực hiện phép chia \(3x^3+13x^2-7x+5\) cho \(3x-2\). Khi đó ta có:
\(A=\frac{3x^3+13x^2-7x+5}{3x-2}=3x^2+5x+1+\frac{7}{3x-2}\)
Nếu x nguyên thì \(3x^2+5x+1\in\text{Z}\) nên để A nguyên thì \(\frac{7}{3x-2}\in Z\)
\(\Rightarrow3x-2\in\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x\in\left\{1;3\right\}\)
b) Ta có: \(B=\frac{2x^5+4x^4-7x^3-44}{2x^2-7}=\left(x^3+2x^2+7\right)+\frac{5}{2x^2-7}\)
Để B nguyên thì \(\frac{5}{2x^2-7}\in Z\Rightarrow2x^2-7\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-1;1;2;-2\right\}\)
a) ( x-3y ) ( x + 1 )
b) ( x+y+5 ) ( x+y-5 )
c) ( x-5 ) ( x+2 )
Hk tốt
a, Để phân thức trên có nghĩa thì:
\(3x^3-19x^2+33x-9\ne0\)
\(\Rightarrow3x^3-9x^2-10x^2+30x+3x-9\ne0\)
\(\Rightarrow3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)\ne0\)
\(\Rightarrow\left(x-3\right)\left(3x^2-10x+3\right)\ne0\)
\(\Rightarrow\left(x-3\right).\left[3x^2-9x-x+3\right]\ne0\)
\(\Rightarrow\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]\ne0\)
\(\Rightarrow\left(x-3\right)^2.\left(3x-1\right)\ne0\)
\(\Rightarrow\hept{\begin{cases}x-3\ne0\\3x-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne3\\x\ne\frac{1}{3}\end{cases}}}\)
a) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
\(=\left[\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3\right]-\left(y^2+z^2\right)^3\)
\(=\left(x^2+y^2+z^2-x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]-\left(y^2+z^2\right)^3\)
\(=\left(y^2+z^2\right)\left(x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4\right)-\left(y^2+z^2\right)^3\)
\(=\left(y^2+z^2\right)\left[x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4-\left(y^2+z^2\right)^2\right]\)
\(=\left(y^2+z^2\right)\left(x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4-y^4-2y^2z^2-z^4\right)\)
\(=\left(y^2+z^2\right)\left(3x^4+3x^2y^2-3x^2z^2-3y^2z^2\right)\)
= 3(y2+z2)(x4+x2y2-x2z2-y2z2)
= 3(y2+z2)[x2(x2+y2)-z2(x2+y2)]
= 3(y2+z2)(x2-z2)(x2+y2)
= 3(y2+z2)(x-z)(x+z)(x2+y2)
b) \(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2=3xy\left(x+y\right)\)
c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2.z+3\left(x+y\right).z^2+z^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2.z+3\left(x+y\right).z^2-\left(x^3+y^3\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right).z+3z^2\right]-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2+3xz+3yz+3z^2-x^2+xy-y^2\right)\)
= (x+y)[3xy+3xz+3yz+3z2 ]
= 3(x+y)(xy+xz+yz+z2)
= 3(x+y)[x(y+z)+z(y+z)]
= 3(x+y)(x+z)(y+z)
a) \(\left(x^2+y^2\right)^3+\left(z^2-x^3\right)-\left(y^2+z^2\right)^3\)
\(=x^6+3x^4y^2+3x^4y^2+y^6+z^2+-x^2+-y^6+-3y^4z+-3y^2z^4+-z^6\)
\(=x^6+3x^4y^2+3x^2y^4+-3y^4z^4+-z^6+-x^2+z^2\)
b) \(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3+-x^3+-y^3\)
\(=\left(x^3+-x^3\right)+\left(3x^2y\right)+\left(3xy^2\right)+\left(y^3+-y^3\right)\)
\(=3x^2y+3xy^2\)
c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3x^2z+3xy^2+6xyz+3xz^2+y^3+3y^2z+3yz^2+z^2-x^3-y^3-z^3\)
\(=3x^2y+3x^2z+3xy^2+3xy^2+6xyz+3xz^2+3y^2z+3yz^2\)
P/s: Ko chắc
Ta có: \(2a^2+2b^2=5ab\Leftrightarrow2\left(a^2+2ab+b^2\right)=9ab\Leftrightarrow\left(a+b\right)^2=\frac{9ab}{2}\)
Mặt khác: \(2a^2+2b^2=5ab\Leftrightarrow2\left(a^2-2ab+b^2\right)=ab\Leftrightarrow\left(a-b\right)^2=\frac{ab}{2}\)
Do đó: \(\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\left(\frac{a+b}{a-b}\right)^2=\frac{\frac{9ab}{2}}{\frac{ab}{2}}=9\Leftrightarrow M=\frac{a+b}{a-b}=\pm3\)
Mà a > b > 0 => M = 3
Ta có: \(2a^2+2b^2=5ab\Leftrightarrow2\left(a^2+2ab+b^2\right)=9ab\Leftrightarrow\left(a+b\right)^2=\frac{9ab}{2}\)
Mặt khác: \(2a^2+2b^2=5ab\Leftrightarrow2\left(a^2-2ab+b^2\right)=ab\Leftrightarrow\left(a-b\right)^2=\frac{ab}{2}\)
Do đó: \(\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\left(\frac{a+b}{a-b}\right)^2=\frac{\frac{9ab}{2}}{\frac{ab}{2}}=9\Leftrightarrow M=\frac{a+b}{a-b}=\pm3\)
Mà \(a>b>0\Rightarrow M=3\)
Ta có: (x-y)^3+(y-z)^3+(z-x)^3
Bạn để ý thấy (x-y)^3+(y-z)^3 là hằng đẳng thức dạng A^3+B^3=(A+B)(A^2-AB+B^2). Vậy ta có thể phân tích (x-y)^3+(y-z)^3 như sau
(x-y+y-z)((x-y)^2-(x-y)(y-z)+(y-z)^2)
(x-z)((x-y)^2-(x-y)(y-z)+(y-z)^2)
-(z-x)((x-y)^2-(x-y)(y-z)+(y-z)^2)
cách khác:
Đặt: \(x-y=a;\)\(y-z=b;\)\(z-x=c\)
suy ra: \(a+b+c=0\)
=> \(a+b=-c\)
=> \(\left(a+b\right)^3=-c^3\)
=> \(a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3\)
<=> \(a^3+b^3+c^3=-3ab\left(a+b\right)\)
<=> \(a^3+b^3+c^3=-3ab\left(-c\right)=3abc\)
Thay trở lại đc: \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
4452424574574
\(4x^3+38x^2y+72xy^2-90y^3=2\left(2x^3+19x^2y+36xy^2-45y^3\right)\)
\(=2\left(2x^3+10x^2y+9x^2y+45xy^2-9xy^2-45y^3\right)\)
\(=2\left[2x^2\left(x+5y\right)+9xy\left(x+5y\right)-9y^2\left(x+5y\right)\right]\)
\(=2\left(x+5y\right)\left(2x^2+9xy-9y^2\right)\)