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a, ĐKXĐ: x\(\ne\) 1;-1;2
b, A= \(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)
=\(\left(\frac{2x^2-2x}{2\left(x+1\right)\left(x-1\right)}+\frac{2x+2}{2\left(x+1\right)\left(x-1\right)}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-2}{x+1}\)
=\(\frac{2x^2-2x+2x+2+4x}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{2x^2+4x+2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{2\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{x-2}{x-1}\)
c, Khi x= -1
→A= \(\frac{-1-2}{-1-1}\)
= -3
Vậy khi x= -1 thì A= -3
Câu d thì mình đang suy nghĩ nhé, mình sẽ quay lại trả lời sau ^^
a,ĐKXĐ:x#1; x#-1; x#2
b,Ta có:
A=\(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)
=\(\left(\frac{x\left(x-1\right)2}{\left(x+1\right)\left(x-1\right)2}+\frac{\left(x+1\right)2}{\left(x-1\right)\left(x+1\right)2}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{x-2}\)
=\(\frac{2x^2-2x+2x+2+4x}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{2x^2+4x+2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{2\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{x-2}{x+1}\)
c,Tại x=-1 ,theo ĐKXĐ x#-1 \(\Rightarrow\)A không có kết quả
d,Để A có giá trị nguyên \(\Rightarrow\frac{x-2}{x+1}\)có giá trị nguyên
\(\Leftrightarrow x-2⋮x+1\)
\(\Leftrightarrow x+1-3⋮x+1\)
Mà \(x+1⋮x+1\Rightarrow3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)
Mà theo ĐKXĐ x#2\(\Rightarrow x\in\left\{0;-2;-4\right\}\)
Vậy \(x\in\left\{0;-2;-4\right\}\)thì a là số nguyên
\(ĐKXĐ:x\ne1\)
a) \(A=\left(1+\frac{x^2}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x^2+1-2x}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x-1}{x^2+1}\)
\(\Leftrightarrow A=\frac{\left(2x^2+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x-1}\)
b) Thay \(x=-\frac{1}{2}\)vào A, ta được :
\(A=\frac{2\left(-\frac{1}{2}\right)^2+1}{-\frac{1}{2}-1}\)
\(\Leftrightarrow A=\frac{\frac{3}{2}}{-\frac{3}{2}}\)
\(\Leftrightarrow A=-1\)
c) Để A < 1
\(\Leftrightarrow2x^2+1< x-1\)
\(\Leftrightarrow2x^2-x+2< 0\)
\(\Leftrightarrow2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+\frac{15}{8}< 0\)
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}< 0\)
\(\Leftrightarrow x\in\varnothing\)
Vậy để \(A< 1\Leftrightarrow x\in\varnothing\)
d) Để A có giá trị nguyên
\(\Leftrightarrow2x^2+1⋮x-1\)
\(\Leftrightarrow2x^2-2x+2x-2+3⋮x-1\)
\(\Leftrightarrow2x\left(x-1\right)+2\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow2\left(x+1\right)\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow3⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
a: \(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{2x+1}{x-1}\cdot\dfrac{x+1}{2x+1}=\dfrac{x+1}{x-1}\)
b: Thay x=1/2 vào A, ta được:
\(A=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-1}=\dfrac{3}{2}:\dfrac{-1}{2}=-3\)
c: Để A là số nguyên thì \(x-1+2⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow x\in\left\{2;0;3\right\}\)
a, ĐK : \(x\ne\pm3;\frac{1}{2}\)
\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)
\(=\left(\frac{\left(x-1\right)\left(x-3\right)+2\left(x+3\right)-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{2x-1-2x-1}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}:\left(-\frac{2}{2x+1}\right)\)
\(=\frac{-2x+6}{\left(x+3\right)\left(x-3\right)}.\frac{-\left(2x+1\right)}{2}=\frac{2x+1}{x+3}\)
b, Ta có : \(\left|x+1\right|=\frac{1}{2}\)
TH1 : \(x+1=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2}\)
Thay vào biểu thức A ta được : \(\frac{-1+1}{-\frac{1}{2}+3}=0\)
TH2 : \(x+1=-\frac{1}{2}\Leftrightarrow x=-\frac{3}{2}\)
Thay vào biểu thức A ta được : \(\frac{-3+1}{-\frac{3}{2}+3}=\frac{-2}{\frac{3}{2}}=-\frac{4}{3}\)
c, Ta có : \(P=\frac{x}{2}\Rightarrow\frac{2x+1}{x+3}=\frac{x}{2}\Rightarrow4x+2=x^2+3x\)
\(\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
b, Ta có : \(\frac{2x+1}{x+3}=\frac{2\left(x+3\right)-5}{x+3}=2-\frac{5}{x+3}\)
\(\Rightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
| x + 3 | 1 | -1 | 5 | -5 |
| x | -2 | -4 | 2 | -8 |
a)Ta có : \(4x^2=1\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
mà \(x\ne-\frac{1}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào B , ta được:
\(B=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=\frac{\frac{1}{4}-\frac{1}{2}}{1+1}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}\)
Vậy \(B=-\frac{1}{8}\)khi \(4x^2=1\)
b)Ta có : \(A=\frac{1}{x-1}-\frac{x}{1-x^2}\)
\(=\frac{1}{x-1}+\frac{x}{x^2-1}\)
\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow M=A.B=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x\left(x-1\right)}{2x+1}\)
\(=\frac{x}{x+1}\)
Vậy \(M=\frac{x}{x+1}\)
c)Ta có: \(x< x+1\forall x\)
\(\Rightarrow M=\frac{x}{x+1}< \frac{x+1}{x+1}=1\forall x\ne-1\)
Vậy với mọi \(x\ne-1\)thì \(M< 1\)
Dài quá trôi hết đề khỏi màn hình: nhìn thấy câu nào giải cấu ấy
Bài 4:
\(A=\frac{\left(x-1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2}{\left(x+1\right)\left(x-1\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
a) DK x khác +-1
b) \(dk\left(a\right)\Rightarrow A=\frac{2}{\left(x+1\right)}\)
c) x+1 phải thuộc Ước của 2=> x=(-3,-2,0))
1. a) Biểu thức a có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
Vậy vs \(x\ne2,x\ne-2\) thì bt a có nghĩa
b) \(A=\frac{x}{x+2}+\frac{4-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x+4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{x+2}\)
c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\)
\(\Leftrightarrow x-2=\left(x+2\right).0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)(ko thỏa mãn điều kiện )
=> ko có gía trị nào của x để A=0
a) \(ĐKXĐ:x\ne\pm1\)
\(Q=\frac{1}{2x-2}+\frac{1}{2x+2}+\frac{x^2}{1-x^2}\)
\(\Leftrightarrow Q=\frac{1}{2\left(x-1\right)}+\frac{1}{2\left(x+1\right)}-\frac{x^2}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow Q=\frac{x+1+x-1-2x^2}{2\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow Q=\frac{-2x^2+2x}{2\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow Q=\frac{-1}{x+1}\)
b) Khi \(\left|x+1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=-3\left(tm\right)\end{cases}}\)
Thay \(x=-3\)vào Q ta được :
\(Q=\frac{-1}{-3+1}=\frac{1}{2}\)
c) Để \(Q\)có giá trị nguyên \(\Leftrightarrow-1⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(-1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{-2;0\right\}\)
Vậy để Q có giá trị nguyên \(\Leftrightarrow x\in\left\{-2;0\right\}\)
c) Bạn lấy mỗi giá trị nguyên nhỏ nhất của x = -2 thôi nhé !
Xin lỗi vì đọc nhầm đề