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a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)
\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
hay P=x+3
\(A=x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left(y^2-z^2\right)+y\left(-y^2+z^2-x^2+y^2\right)+z\left(x^2-y^2\right)=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)=\left(y-z\right)\left(y+z\right)\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(y-z\right)=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)
a )
`VP= (a+b)^3-3ab(a+b)`
`=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`
`=a^3+b^3 =VT (đpcm)`
b)
b) Ta có
`VT=a3+b3+c3−3abc`
`=(a+b)3−3ab(a+b)+c3−3abc`
`=[(a+b)3+c3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)2+c2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a2+b2+2ab+c2−ac−bc−3ab)`
`=(a+b+c)(a2+b2+c2−ab−bc−ca)=VP`
a) Ta có:
`VP= (a+b)^3-3ab(a+b)`
`=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`
`=a^3 + b^3=VT(dpcm)`
b) Ta có
`VT=a^3+b^3+c^3−3abc`
`=(a+b)^3−3ab(a+b)+c^3−3abc`
`=[(a+b)^3+c^3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)^2+c^2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a^2+b^2+2ab+c^2−ac−bc−3ab)`
`=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)=VP`
\(a,VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Rightarrow VT=a^2c^2+b^2c^2+a^2d^2+b^2d^2=VP\left(đpcm\right)\)
b, Tham khảo:Chứng minh hằng đẳng thức:(a+b+c)3= a3 + b3 + c3 + 3(a+b)(b+c)(c+a) - Hoc24
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)