\(P\left(x\right)-Q\left(x\right)=\left(-2x+\frac{1}{2}x^2+3x^4-3x^2-3\right)-\left(3x^4+x^3-4x^2+1,5x^3-3x^4+2x+1\right)\\ P\left(x\right)-Q\left(x\right)=-2x+\frac{1}{2}x^2+3x^4-3x^2-3-3x^4-x^3+4x^2-1,5x^3+3x^4-2x-1\\ P\left(x\right)-Q\left(x\right)=\left(-2x-2x\right)+\left(\frac{1}{2}x^2-3x^2+4x^2\right)+\left(3x^4-3x^4+3x^4\right)+\left(-3-1\right)+\left(-x^3-1,5x^3\right)\\ P\left(x\right)-Q\left(x\right)=-4x+\frac{3}{2}x^2+3x^4-4-\frac{5}{2}x^3\)

\(R\left(x\right)+P\left(x\right)-Q\left(x\right)+x^2=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)+\left(P\left(x\right)-Q\left(x\right)\right)+x^2=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)-4x+\frac{3}{2}x^2+3x^4-4-\frac{5}{2}x^3+x^2=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)-4x+\left(\frac{3}{2}x+x^2\right)+3x^4-4-\frac{5}{2}x^3=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)-4x+\frac{5}{2}x^2+3x^4-4-\frac{5}{2}x^3=2x^3-\frac{3}{2}x+1\\ \Rightarrow R\left(x\right)=2x^3-\frac{3}{2}x+1+4x-\frac{5}{2}x^2-3x^4+4+\frac{5}{2}x^3\\ \Rightarrow R\left(x\right)=\left(2x^3+\frac{5}{2}x^3\right)+\left(\frac{-3}{2}x+4x\right)+\left(1+4\right)-\frac{5}{2}x^2-3x^4\\ \Rightarrow R\left(x\right)=\frac{9}{2}x^3+\frac{5}{2}x+5-\frac{5}{2}x^2-3x^4\)

Bài 1 : k bt làm

Bài 2 :

Ta có : \(\left(x-6\right).P\left(x\right)=\left(x+1\right).P\left(x-4\right)\) với mọi x

+) Với \(x=6\Leftrightarrow\left(6-6\right).P\left(6\right)=\left(6+1\right).P\left(6-4\right)\)

\(\Leftrightarrow0.P\left(6\right)=7.P\left(2\right)\)

\(\Leftrightarrow0=7.P\left(2\right)\)

\(\Leftrightarrow P\left(2\right)=0\)

\(\Leftrightarrow x=2\) là 1 nghiệm của \(P\left(x\right)\left(1\right)\)

+) Với \(x=-1\Leftrightarrow\left(-1-6\right).P\left(-1\right)=\left(-1+1\right).P\left(-1-4\right)\)

\(\Leftrightarrow\left(-7\right).P\left(-1\right)=0.P\left(-5\right)\)

\(\Leftrightarrow\left(-7\right).P\left(-1\right)=0\)

\(\Leftrightarrow P\left(-1\right)=0\)

\(\Leftrightarrow x=-1\) là 1 nghiệm của \(P\left(x\right)\) \(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow P\left(x\right)\) có ót nhất 2 nghiệm

nghiệm của đa thức xác định đa thức đó bằng 0

0 mà k bằng 0. You định làm nên cái nghịch lý ak -.-

a) \(\left(x^2-2x+2\right)\left(x-2\right)\left(x^2-2x+2\right)\left(x+2\right)\)

\(=\left(x^3-2x^2-2x^2+4x+2x-4\right)\left(x^3+2^3\right)\)

\(=\left(x^3-4x^2+6x-4\right)\left(x^3+8\right)\)

\(=x^6+8x^3-4x^5-32x^2+6x^4+48x-4x^3-32\)

\(=x^6-4x^5+4x^3-32x^2+48x-32\)

b) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)

\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]+x^3-3x\left(x^2-1\right)\)

\(=2x\left[\left(x^2+2x+1\right)-\left(x^2-1\right)+\left(x^2-2x+1\right)\right]+x^3-\left(3x^3-3x\right)\)

\(=2x\left(x^2+2x+1-x^2+1+x^2-2x+1\right)+x^3-3x^3+3x\)

\(=2x\left(x^2+3\right)+x^3-3x^3+3x\)

\(=2x^3+6x-2x^3+3x\)

\(=9x\)

2 câu kia đợi tí đã nhé!

c) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)

\(=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(a^2+b^2+c^2+2ab-2bc-2ca\right)+\left(4a^2-4ab+b^2\right)\)

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+4a^2-4ab+b^2\)

\(=6a^2+3b^2+2c^2\)

d) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+2\left(a+b\right)^2\)

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

\(=4a^2+4b^2+2c^2+6ab.\)

Bài 1:

a, Ta có:

\(\left(a+b+c\right)^2-\left(ab+bc+ca\right)=0\Leftrightarrow a^2+b^2+c^2+ab+bc+ca=0\)\(\Leftrightarrow2a^2+2b^2+2c^2+2ab+2bc+2ca=0\)

\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=0\Leftrightarrow a+b=b+c=c+a=0\)

\(\Leftrightarrow a=b=c=0\)

Vậy điều kiện để phân thức M được xác định là a, b, c không đồng thời = 0

b, Ta có:

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

Đặt: \(a^2+b^2+c^2=x,ab+bc+ca=y\)

=> \(\left(a+b+c\right)^2=x+2y\)

Ta cũng có:

\(M=\dfrac{x\left(x+2y\right)+y^2}{x+2y-y}=\dfrac{x^2+2xy+y^2}{x+y}=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)

\(=a^2+b^2+c^2+ab+bc+ca\)

\(F\left(x\right)=3x-6;x=\dfrac{6}{3}=2\)

\(H\left(x\right)=-5x+30;x=-\dfrac{30}{5}=-6\)

\(G\left(x\right)=\left(x-3\right)\left(16-4x\right)\Leftrightarrow\left[{}\begin{matrix}x-3=0;x=3\\16-4x=0;x=4\end{matrix}\right.\)

\(K\left(x\right)=x^2-81=\left(x-9\right)\left(x+9\right)\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\)

\(M\left(x\right)=x^2+7x-8=\left(x-1\right)\left(x+8\right);\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\)

\(N\left(x\right)=5x^2+9x+4\)

\(N\left(x\right)=5x^2+5x+4x+4=5x\left(x+1\right)+4\left(x+1\right)\)

\(N\left(x\right)=\left(x+1\right)\left(5x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Rút gọn:

\(P\left(x\right)=2x^2+4x\)

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

Để \(R\left(x\right)-P\left(x\right)-Q\left(x\right)=0\)

<=> \(R\left(x\right)=P\left(x\right)+Q\left(x\right)\)

= \(\left(2x^2+4x\right)+\left(-x^3+2x^2-x+2\right)\)

= \(-x^3+4x^2+3x+2\)

KL: \(R\left(x\right)=-x^3+4x^2+3x+2\)