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12 tháng 7 2017
\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)
c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)
\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)
\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)
\(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x\left(x^2+2-2x\right)-\left(x^2+2-2x\right)}\)
\(=\dfrac{x^2+2+2x}{x-1}\)
Bài 2:
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{10}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{1}{x+1}\)
c) Trong ngoặc giữa hai phân số là dấu gì vậy ?
13 tháng 11 2018
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
TA
21 tháng 2 2018
\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{9}{\left(x-3\right)\left(x+6\right)}=\dfrac{4}{3}\)
=> \(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{1}{x-3}-\dfrac{1}{x+6}=\dfrac{4}{3}\)
=> \(\dfrac{1}{x+1}-\dfrac{1}{x+6}-\dfrac{1}{x-3}+\dfrac{1}{x+6}=0\)
=> \(\dfrac{1}{x+1}-\dfrac{1}{x-3}=0\)
Ma \(\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\)
=> pt vo nghiem
TA
21 tháng 2 2018
\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}-\dfrac{1}{x+3}+\dfrac{1}{x+6}=\dfrac{4}{3}\)
=> \(\dfrac{1}{x+1}-\dfrac{1}{x+3}=\dfrac{4}{3}\)
=> \(\dfrac{2}{\left(x+1\right)\left(x+3\right)}=\dfrac{4}{3}\)
=> 4(x+1)(x+3)=6
=> 4(x2+4x+3)=6
=> 4x2+16x+6=0
=> (4x2+16x+16)-10=0
=> (2x+4)2=10
=> \(\left[{}\begin{matrix}2x+4=\sqrt{10}\\2x+4=-\sqrt{10}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-4}{2}\\x=\dfrac{-\sqrt{10}-4}{2}\end{matrix}\right.\)
29 tháng 10 2018
\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)
ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)
b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)
\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)
6 tháng 11 2018
a/ \(\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=0\)
\(\Leftrightarrow\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^3+1\right)}{x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2+1}=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x=-1\)
Vậy ...
b/ \(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=0\)
\(\Leftrightarrow\dfrac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}=0\)
\(\Leftrightarrow\dfrac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{x^2\left(x^2-1\right)-9\left(x^2-1\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x+2\right)}{x^2-9}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy..
11 tháng 5 2018
\(\text{a) }\dfrac{3}{x^2+5x+4}+\dfrac{2}{x^2+10x+24}=\dfrac{4}{3}+\dfrac{9}{x^2+3x-18}\\ ĐKXĐ:x\ne-1;x\ne-3;x\ne-4;x\ne-6\\ \Rightarrow\dfrac{3}{x^2+4x+x+4}+\dfrac{2}{x^2+6x+4x+24}=\dfrac{4}{3}+\dfrac{9}{x^2+6x-3x-18}\\ \Rightarrow\dfrac{3}{x\left(x+4\right)+\left(x+4\right)}+\dfrac{2}{x\left(x+6\right)+4\left(x+6\right)}=\dfrac{4}{3}+\dfrac{9}{x\left(x+6\right)-3\left(x+6\right)}\\ \Rightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{4}{3}+\dfrac{9}{\left(x-3\right)\left(x+6\right)}\)\(\Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{4}{3}+\dfrac{1}{x-3}-\dfrac{1}{x+6}\\ \Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x+6}-\dfrac{1}{x-3}+\dfrac{1}{x+6}=\dfrac{4}{3}\\ \Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\\ \Rightarrow\dfrac{3\left(x-3\right)}{3\left(x+1\right)\left(x-3\right)}-\dfrac{3\left(x+1\right)}{3\left(x+1\right)\left(x-3\right)}=\dfrac{4\left(x+1\right)\left(x-3\right)}{3\left(x+1\right)\left(x-3\right)}\\ \Rightarrow3x-9-3x-3=4\left(x^2-2x-3\right)\\ \Leftrightarrow4x^2-8x-12=-12\\ \Leftrightarrow4x^2-8x=0\\ \Leftrightarrow4x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)Vậy phương trình có tập nghiệm \(S=\left\{0;2\right\}\)
HL
11 tháng 7 2017
a) \(\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)
= \(\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^3\left(x-1\right)-\left(x-1\right)+2x^2}\)
= \(\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x-1\right)\left(x^3-1\right)+2x^2}\)
= \(\dfrac{\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)+2x^2}\)
= \(\dfrac{\left(x+1\right)^2.\left(x^2-x+1\right)}{\left(x-1\right)^2\left(x^2+x+1\right)+2x^2}\)
Ta thấy mẫu thức của phân thức vốn đã lớn hơn 0 với mọi x, vậy để p/t trên có giá trị bằng 0 thì tử thức phải bằng 0
\(\Rightarrow\left(x+1\right)^2\left(x^2-x+1\right)=0\)
\(\Rightarrow x=-1\)
Vậy x = -1
b) \(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}\)
= \(\dfrac{x^4-x^3+x^3-x^2-4x^2+4}{x^4-x^3+x^3-x^2-9x^2+9}\)
= \(\dfrac{x^3\left(x-1\right)+x^2\left(x-1\right)-4\left(x-1\right)\left(x+1\right)}{x^3\left(x-1\right)+x^2\left(x-1\right)-9\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{\left(x-1\right)\left(x^3+x^2-4x-4\right)}{\left(x-1\right)\left(x^3+x^2-9x-9\right)}\)
= \(\dfrac{x^3+x^2-4x-4}{x^3+x^2-9x-9}\)
= \(\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)-9\left(x+1\right)}\)
= \(\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\) ( ĐKXĐ : \(x\ne\pm3\) )
Để phân thức trên có giá trị bằng 0 thì tử thức phải bằng 0
\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) ( thoả mãn điều kiện xác định )
Vậy x = 2 hoặc x = -2
\(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}\)
\(=\dfrac{x^2.\left(x^2-1\right)-4.\left(x^2-1\right)}{x^2.\left(x^2-1\right)-9.\left(x^2-1\right)}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}\)
\(=\dfrac{x^2-4}{x^2-9}\)
Chúc bạn học tốt!!!
\(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}\)
\(=\dfrac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}\)
\(=\dfrac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{x^2\left(x^2-1\right)-9\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\)