Phân thức đại số - Hỏi đáp

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)

Ta có: MTC:\(2\left(x-1\right)\left(x+1\right)\)

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

CHÚC BẠN HỌC TỐT

a ) Rút gọn : \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)

\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2}{x-5}\)

\(\Leftrightarrow A=\dfrac{1}{x+5}.\)

Khi \(A=-3\),thì :

\(\dfrac{1}{x+5}=-3\Leftrightarrow x=-\dfrac{16}{3}\)

Ta có : \(9x^2-42x+49\)

\(=\left(3x\right)^2-2.3x.7+49\)

\(=\left(3x-7\right)^2\)

Thay \(x=-\dfrac{16}{3},\) ta có :

\(\left(3.\dfrac{-16}{3}-7\right)^2=\left(-16-7\right)^2=\left(-23\right)^2=529\)

Vì x + y = 2 và x, y nguyên dương nên \(x=y=1\)

Khi đó A = 4.

Vậy A_min = 4

Lời giải:

a) Để A được xác định

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\ne0\)

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

Vậy ...

b) Ta có:

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

\(\Leftrightarrow A=\dfrac{x+3}{x-1}\)

Vậy ...

c) Để \(A=0\)

\(\Leftrightarrow\dfrac{x+3}{x-1}=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\) (Thỏa mãn điều kiện xác định)

Vậy ...

d) Để \(A=2\)

\(\Leftrightarrow\dfrac{x+3}{x-1}=2\)

\(\Leftrightarrow2\left(x-1\right)=x+3\)

\(\Leftrightarrow2x-2=x+3\)

\(\Leftrightarrow2x-x=3+2\)

\(\Leftrightarrow x=5\) (Thỏa mãn điều kiện xác định)

Vậy ...

e) Để A có giá trị nguyên

\(\Leftrightarrow A\in Z\)

\(\Leftrightarrow\dfrac{x+3}{x-1}\in Z\)

\(\Leftrightarrow x+3⋮x-1\)

\(\Leftrightarrow x-1+4⋮x-1\)

Vì \(x-1⋮x-1\)

\(\Leftrightarrow4⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(4\right)\in\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow x\in\left\{0;2;-1;3;-3;5\right\}\) (Thỏa mãn điều kiện xác định)

Vậy ...