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Trả lời:
a, \(-xy.\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+3xy\)
b, \(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y\)
\(=12x^6y^5:6x^2y^2-3x^3y^4:6x^2y+4x^2y+6x^2y\)
\(=2x^4y^3-\frac{1}{2}xy^3+\frac{2}{3}\)
a.\(\left(-xy\right)\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+6xy\)
b.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)
a) Theo định lí Bezout ta có:
\(f\left(-5\right)=3.\left(-5\right)^2-5a+27=2\)
\(\Leftrightarrow75-5a+27=2\)
\(\Leftrightarrow102-5a=2\)
\(\Rightarrow a=20\)
b) \(x^3+ax^2+x+b=\left(x^2-x+2\right).\left(x+m\right)\)(Trong đó m là số nguyên)
\(\Leftrightarrow x^3+ax^2+x+b=x^3+x^2.\left(m-1\right)-mx+2m\)
Sử dụng phương pháp đồng nhất hệ số ta có:
\(\hept{\begin{cases}ax^2=m-1\\x=-mx\\2m=b\end{cases}}\Leftrightarrow\hept{\begin{cases}a=m-1\\m=-1\\2m=b\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-2\\b=-2\end{cases}}\Leftrightarrow a=b=-2\)
- a, [x^2.(x-3)-(x-3)] :( x-3) = (x-3 ).(x^2-1) : (x-3) =X^2-1
2 b, (x-y-z)^5-3 = (x-y-z)^2
3 c, x^2-1
4 d, 2x^4 + x^2 - 6x^2 + x^3 - 3 - 3x / x^2 - 3
= x^2(2x^2 + x + 1) - 3(2x^2 + x + 1) / x^2 - 3
= (2x^2 + x + 1)(x^2 - 3) / x^2 - 3
= 2x^2 + x + 1
5 e, 2.(x-1)
6 f, (2x3 – 5x2 + 6x – 15) : (2x – 5)
=(2x3−5x2)+(6x−15)=(2x3−5x2)+(6x−15)
=x2(2x−5)+3(2x−5)=x2(2x−5)+3(2x−5)
=(x2+3)(2x−5)=(x2+3)(2x−5)
=(2x3−5x2+6x−15):(2x−5)=x2+3
Bài 209 : đăng tách ra cho mn cùng làm nhé
a,sửa đề : \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)
c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)
\(\left(x+5\right)\left(x^2-5x+25\right)\)
\(=\left(x+5\right)\left(x^2-5.x+5^2\right)\)
\(=x^3+5^3\)
\(=x^3+125\)
3) \(27-y^3\)
\(=3^3-y^3\)
\(=\left(3-y\right)\left(9-3y+y^2\right)\)