\(\frac{x^2-25}{x^2-3x}\):\(\frac{x^2+5x}{x-3}\)
a. Điều kiện xác định
b. Rút gọn cho A
c. Tìm x khi A=4
K
Khách
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10 tháng 12 2016
a)Đk:\(\begin{cases}x^2-25\ne0\\x^2+5x\ne0\end{cases}\)\(\Leftrightarrow\begin{cases}\left(x-5\right)\left(x+5\right)\ne0\\x\left(x+5\right)\ne0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ne5\\x\ne-5\\x\ne0\end{cases}\)
\(A=\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}=\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\)
\(=\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x+5\right)^2}{x\left(x-5\right)\left(x+5\right)}=\frac{x^2-\left(x^2+10x+25\right)}{x\left(x-5\right)\left(x+5\right)}\)\(=\frac{10x-25}{x^3-25x}\)
12 tháng 1 2019
a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2x+1-3x-1+x^2}{3x}\)
\(A=\frac{x^2-x}{3x}\)
\(A=\frac{x\left(x-1\right)}{3x}\)
\(A=\frac{x-1}{3}\)
b) Thay x = 4 ta có :
\(A=\frac{4-1}{3}=\frac{3}{3}=1\)
c) Để A thuộc Z thì \(x-1⋮3\)
\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)
\(\Rightarrow x\in\left\{1;4;7;...\right\}\)
Vậy.....
10 tháng 12 2018
a) Phân thức M xác định khi :
+) \(x\ne0\)
+) \(x-2\ne0\Leftrightarrow x\ne2\)
b) \(M=\left(\frac{2}{x}-\frac{2}{x-2}\right):\frac{3x}{x-2}\)
\(M=\left(\frac{2\left(x-2\right)}{x\left(x-2\right)}-\frac{2x}{x\left(x-2\right)}\right)\cdot\frac{x-2}{3x}\)
\(M=\left(\frac{2x-4-2x}{x\left(x-2\right)}\right)\cdot\frac{x-2}{3x}\)
\(M=\frac{-4\cdot\left(x-2\right)}{x\left(x-2\right)\cdot3x}\)
\(M=\frac{-4}{3x^2}\)
c) Thay x = -2 ta có :
\(M=\frac{-4}{3\cdot\left(-2\right)^2}=\frac{-1\cdot4}{3\cdot4}=\frac{-1}{3}\)
Vậy........
H
30 tháng 1 2019
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
H
30 tháng 1 2019
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
12 tháng 12 2018
a, ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)
b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)
c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)
\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)
H
12 tháng 12 2018
a, ĐỂ A xác định :
\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)
\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)
\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)
\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{x-4}{x+3}\)
b
a, x khác 3,0,-5
b, A= (x-5)(x+5) / x(x-3) . (x-3)/x(x+5)
A= x-5/x^2
c, khi A=4
<=> x-5 / x2 =4
=>4x2 -x +5 =0
=> ko có giá trị x để A=4 (câu này ko bt đúng hay sai, hoặc ghi đề sai )