Ôn tập: Phân thức đại số - Hỏi đáp

ta có

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)=4\) (vì a+b=c=abc)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)

\(\Leftrightarrow M=2\)

\(\left(2x+1\right)^2-\left(2x-3\right)\left(2x+3\right)=3\left(x+2\right)\\ \Rightarrow4x^2+4x+1-4x^2+9=3x+6\\ \Rightarrow4x+10=3x+6\\ \Rightarrow4x-3x=-10+6\\ \Rightarrow x=-4\)

a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)

Ta có :

\(a^3+b^3+c^3-3abc\)

\(=a^3+b^3+c^3+3a^2b+3ab^2-3a^2b-3ab^2-3abc\)

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

\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)

\(=\left\{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\right\}-3ab\left(a+b+c\right)\)

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

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

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

\(\Rightarrow P=\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(=a+b+c\)

\(=2014\)

Vậy P = 2014

\(C=\dfrac{2}{x^2-6x+15}=\dfrac{2}{x^2-6x+9+6}=\dfrac{2}{\left(x-3\right)^2+6}\le\dfrac{2}{6}=\dfrac{1}{3}\)

Dấu "=" xảy ra khi: \(x=3\)

Vậy \(max_C=\dfrac{1}{3}\) khi \(x=3\)

A=-

A=\(\dfrac{x+2}{x+3}+\left(\dfrac{-5}{\left(x-2\right)\left(x+3\right)}\right)\)

A=\(\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}+\dfrac{-5}{\left(x-2\right)\left(x+3\right)}\)

A=\(\dfrac{x^2-9}{\left(x-2\left(x+3\right)\right)}\)

A=\(\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

A=\(\dfrac{x-3}{x-2}\)

\(2018A=\dfrac{2018x^2+2.2018.x+2018^2}{x^2}=\dfrac{2017x^2}{x^2}+\dfrac{x^2+2.2018+2018^2}{x^2}=2017+\dfrac{\left(x+2018\right)^2}{x^2}\ge2017\Rightarrow A\ge\dfrac{2017}{2018}\)

Dấu "=" xảy ra <=> x = -2018

a)

Biểu thức được xác định khi

\(\left\{{}\begin{matrix}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\\x\ne-1\end{matrix}\right.\)

b)

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

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

\(=\dfrac{\left(x+1\right)\left(x+1\right)+3.2-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}.\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}.\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

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

\(=4\)

Câu hỏi của My Trần - Toán lớp 8 | Học trực tuyến

\(E=\dfrac{5x+5}{2x^2+2x}\)

a) ĐKXĐ \(x\ne-1;0\)

b) \(E=1\Leftrightarrow\dfrac{5x+5}{2x^2+2x}=1\)

\(\Leftrightarrow5x+5=2x^2+2x\)

\(\Leftrightarrow5\left(x+1\right)=2x\left(x+1\right)\)

\(\Leftrightarrow5=2x\rightarrow x=\dfrac{5}{2}\) (TMĐK)

Kết luận : ...................

a)Để phân thức E đc xđ thì:2x²+2x\(\ne\)0

\(\Leftrightarrow\)2x(x+1)\(\ne\)0

\(\Leftrightarrow\left\{{}\begin{matrix}\\\end{matrix}\right.\)2x\(\ne\)0;x+1\(\ne\)0

\(\Leftrightarrow\)x\(\ne\)0;-1

b)với E=1 thì:\(\dfrac{5x+5}{2x^2+2x}=1\)

\(\Rightarrow\)5x+5=2x²+2x

\(\Rightarrow\)5(x+1)=2x(x+1)

\(\Rightarrow\)5=2x\(\Rightarrow\)x=2:5

\(\Rightarrow\)x=\(\dfrac{2}{5}\)(TMĐK)

Bài 1 Phân tích đa thức thành nhân tử

a, 2x3 - 50x

=2x\(\left(x^2-25\right)\)

=2x(x-5)(x+5)

b, x2- 6x + 9- 4y2

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

=(x-3-4y)(x-3+4y)