Thực hiện các phép tính sau:      

    \(\text{a) 3x². (2x³ – x + 5)=}6x^6-3x^3+15x^2\)

   \(\text{ b) (4xy + 3y – 5x). x²y}=4x^3y^2+3x^2y^2-5x^3y\)

\(\text{ d) ( 3x – 5 )( x²– 5x + 7 )}=3x^3+21x-35\)

Học tốt

a)=6x^5-3x^3+15x^2

b)=4x^3y^2+3x^2y^2-5x^3y

c)=3x^3-5x^2-15x^2+25x+21x-35=3x^3-20x^2+46x-35

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

a) P xác định \(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x-2\ne0\\2-x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}}\)

Vậy P xác định khi \(x\ne-3;x\ne2\)

b) \(P=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\left(x\ne-3;x\ne2\right)\)

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

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

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

\(\Leftrightarrow P=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)

Vậy \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)

c) \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)

Để P\(=-\frac{3}{4}\Rightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)

\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow4x+3x=6+16\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\frac{22}{7}\left(tm\right)\)

Vậy \(=\frac{22}{7}\)thì \(P=\frac{-3}{4}\)

d) \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)

\(\Leftrightarrow P=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)

Để P nhận giá trị nguyên thì \(1-\frac{2}{x-2}\)nhận giá trị nguyên 

\(\Leftrightarrow\frac{2}{x-2}\)nhận giá trị nguyên (1)

\(x\inℤ\Rightarrow x-2\inℤ\)(2)

\(\left(1\right)\left(2\right)\Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Ta có bảng giá trị

Vậy \(x\in\left\{0;-1;3;4\right\}\)

\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\left(x^2+x+1-3x^2-2x^2+2x\right)=0\)

\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)

\(\Leftrightarrow-4x^2+4x-x+1=0\)

\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)

Vậy \(x=\frac{-1}{4}\)