Sửa đề:

Tìm a biết P(1)=Q(-2)

Ta có:

\(P\left(1\right)=1^3+3a.1+a^2=a^2+3a+1\)

\(Q\left(-2\right)=2.\left(-2\right)^2-\left(2a+3\right).\left(-2\right)+a^2\)

\(=2.4+2\left(2a+3\right)+a^2\)

\(=8+4a+6+a^2=a^2+4a+14\)

Mà \(P\left(1\right)=Q\left(-2\right)\)

\(\Rightarrow a^2+3a+1=a^2+4a+14\)

\(\Rightarrow3a-4a=14-1\Rightarrow-a=13\Rightarrow a=-13\)
Vậy................

Chúc bạn học tốt!!!

a: Ta có: \(ax-x+1=a^2\)

\(\Leftrightarrow x\left(a-1\right)=a^2-1\)

hay x=a+1

Đặt \(P\left(x\right)=a^2x^3+3ax^2-6x-2a\)

Để \(P\left(x\right)⋮x+1\Leftrightarrow P\left(-1\right)=0\)

\(\Leftrightarrow P\left(-1\right)=-a^2+3a+6-2a=0\)

\(\Leftrightarrow-a^2+a+6=0\)

\(\Leftrightarrow-a^2+3a-2a+6=0\)

\(\Leftrightarrow-a\left(a-3\right)-2\left(a-3\right)=0\)

\(\Leftrightarrow\left(a-3\right)\left(-a-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-3=0\\-a-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\a=-2\end{matrix}\right.\)

Vậy....

Áp dụng định lý Bezout ta có:

\(A\left(x\right)⋮\left(x+1\right)\Rightarrow A\left(-1\right)=0\)

                               \(\Leftrightarrow a^2\left(-1\right)^3+3a\left(-1\right)^2-6.\left(-1\right)-2a=0\)

                                \(\Leftrightarrow-a+3a+6+2a=0\)

                               \(\Leftrightarrow4a+6=0\)

                                \(\Leftrightarrow a=\frac{-3}{2}\)

Vậy \(a=\frac{-3}{2}\)để \(A\left(x\right)⋮\left(x+1\right)\)

em gửi bài qua fb thầy chữa cho nhé, tìm fb của thầy bằng sđt: 0975705122 nhé.

b. Sử dụng các hằng đẳng thức

 \(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)

và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Do (a - b) + (b - c) + (c - a) =  0 nên áp dụng hđt  \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:

\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)

Bài 1 :

\(b,ax^2+3ax+9=a^2\) 

\(\Leftrightarrow a^2x+3ax+9-a^2=0\) 

\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\) 

\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)

Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\) 

\(\Leftrightarrow ax=a-3\) 

Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\) 

c.Ta có \(\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{2}{xz}-\frac{2}{xy}+\frac{2}{yz}=1\)

Do x = y + z nên \(\frac{-2}{xz}-\frac{2}{xy}+\frac{2}{yz}=\frac{-2y-2z+2\left(y+z\right)}{\left(y+z\right)zy}=0\)

Vậy nên \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1.\)