\(P\left(x\right)-Q\left(x\right)=7x^2-9x+6\)

Để TMĐK đề bài thì: \(7x^2-9x+6=2x^2-3x+6\)

\(\Leftrightarrow5x^2-6x=0\Leftrightarrow x\left(5x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{5}\end{cases}}\)

a: P(x)=6x^4+5x^3-3x^2+5x-10

Q(x)=5x^4+5x^3+2x^2-4x+4

b: P(x)+Q(x)

=6x^4+5x^3-3x^2+5x-10+5x^4+5x^3+2x^2-4x+4

=11x^4+10x^3-x^2+x-6

P(x)-Q(x)

=6x^4+5x^3-3x^2+5x-10-5x^4-5x^3-2x^2+4x-4

=x^4-5x^2+9x-14

a: P(x)=6x^4+5x^3-3x^2+5x-10

Q(x)=5x^4+5x^3+2x^2-4x+4

b: P(x)+Q(x)

=6x^4+5x^3-3x^2+5x-10+5x^4+5x^3+2x^2-4x+4

=11x^4+10x^3-x^2+x-6

P(x)-Q(x) =6x^4+5x^3-3x^2+5x-10-5x^4-5x^3-2x^2+4x-4

=x^4-5x^2+9x-14

a, \(P\left(x\right)=4x^3+2x-3+2x-2x^2-1\\ =4x^3-2x^2+\left(2x+2x\right)+\left(-3-1\right)\\ =4x^3-2x^2+4x-4\)

Bậc của P(x) là 3

\(Q\left(x\right)=6x^3-3x+5-2x+3x^2\\ =6x^3+3x^2+\left(-3x-2x\right)+5\\ =6x^3+3x^2-5x+5\)

Bậc của Q(x) là 3

b, \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=4x^3-2x^2+4x-4+6x^3+3x^2-5x+5\\ =\left(4x^3+6x^3\right)+\left(-2x^2+3x^2\right)+\left(4x-5x\right)+\left(-4+5\right)\\ =10x^3+x^2-x+1\)

a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)

\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)

2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)

\(\frac{2x}{3y}.\frac{3y}{2x}=1\)

3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)

4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)

5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)

6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)

7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)

Làm rõ lâu.

Chép đề đúng chưa bạn? 2 phân số đầu có ngoặc không vậy?

Nguyễn Công Tỉnh đúng r bạn, mình sửa lại r

Bạn tự tìm ĐKXĐ nhé!

\(B=\left(\frac{x}{x^2-x-6}-\frac{x-1}{3x^2-4x-15}\right):\frac{x^4-2x^2+1}{3x^2+11x+10}.\left(x^2-2x+1\right)\)

\(=\left(\frac{x}{\left(x-3\right)\left(x+2\right)}-\frac{x-1}{\left(x-3\right)\left(3x+5\right)}\right):\frac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}.\left(x-1\right)^2\)

\(=\left(\frac{\left(3x+5\right)x}{\left(x-3\right)\left(x+2\right)\left(3x+5\right)}-\frac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right).\frac{\left(3x+5\right)\left(x+2\right)}{\left(x-1\right)^2\left(x+1\right)^2}.\left(x-1\right)^2\)

\(=\frac{3x^2+5x-\left(x^2+2x-x-2\right)}{\left(x-3\right)\left(x+2\right)\left(3x+5\right)}.\frac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2}\)

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

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

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

\(=\frac{2}{x-3}\)

Vậy...