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a, - Để biểu thức trên được xác định thì : \(x^2+x+1\ne0\)
Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Vậy biểu thức luôn được xác định với mọi x .
b, - Để biểu thức trên được xác định thì : \(4x^2+2x+3\ne0\)
Mà \(4x^2+2x+3=\) \(x^2+\frac{x}{2}+\frac{3}{4}=\left(x+\frac{1}{4}\right)^2+\frac{11}{16}>0\)
Vậy biểu thức luôn được xác định với mọi x .
d, - Để biểu thức trên có nghĩa thì : \(3t^2-t+1\ne0\)
Mà \(3t^2-t+1=3\left(t^2-\frac{t}{3}+\frac{1}{3}\right)=3\left(\left(t-\frac{1}{6}\right)^2+\frac{11}{36}\right)>0\)
Vậy biểu thức luôn được xác định với mọi x .
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
a) A= \(\frac{x\left(1-x^2\right)^2}{1+x^2}\): \(\left\{\left[\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right]\left[\frac{\left(1+x\right)\left(1-x+x^2\right)}{1+x}-x\right]\right\}\)
A= \(\frac{x\left(1-x^2\right)^2}{1+x^2}\): (1+x+x2+x)(1-x+x2-x)
A=\(\frac{x\left(1-x^2\right)^2}{1+x^2}\): (1+2x+x2)(1-2x+x2)
A= \(\frac{x\left(1-x^2\right)^2}{1+x^2}\): (1+x)2(1-x)2
A= \(\frac{x\left(1-x^2\right)^2}{1+x^2}\): (1+x)(1+x)(1-x)(1-x)
A= \(\frac{x\left(1-x^2\right)\left(1-x^2\right)}{1+x^2}.\frac{1}{\left(1-x^2\right)\left(1-x^2\right)}\)
A= \(\frac{x}{1+x^2}\)
b)Thay x= \(-\frac{1}{2}\) vào biểu thức A, có:
A= \(\frac{\frac{-1}{2}}{1+\left(\frac{-1}{2}\right)^2}\)
\(\Leftrightarrow\)A= \(\frac{-2}{5}\)
Vậy A= \(\frac{-2}{5}\) khi x=\(-\frac{1}{2}\)
c) Để 2A=1 thì \(\frac{2x}{1+x^2}\)=1
\(\Leftrightarrow\)\(\frac{2x}{1+x^2}\)-1=0
\(\Leftrightarrow\)2x-1-x2=0
\(\Leftrightarrow\)-(2x+1+x2)=0
\(\Leftrightarrow\)x2-2x+1=0
\(\Leftrightarrow\)(x-1)2=0
\(\Leftrightarrow\)x-1=0
\(\Leftrightarrow\)x=1
Vậy x=1 thì 2A=1
a)
\(\begin{array}{l}\frac{2}{{3{\rm{x}}}} + \frac{x}{{x - 1}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{2}{{3{\rm{x}}}} - \frac{x}{{1 - x}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4\left( {1 - x} \right) - 6{{\rm{x}}^2} + 3\left( {6{{\rm{x}}^2} - 4} \right)}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4 - 4{\rm{x}} - 6{{\rm{x}}^2} + 18{{\rm{x}}^2} - 12}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{12{{\rm{x}}^2} - 4{\rm{x}} - 8}}{{6{\rm{x}}\left( {1 - x} \right)}}\end{array}\)
b)
\(\begin{array}{l}\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1}}{{{x^3} - 1}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1 + x\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{ - {x^3} - 1 + {x^3} + {x^2} + x - {x^2} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{x}{{{x^3} - 1}}\end{array}\)
c)
\(\begin{array}{l}\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2\left( {1 - x} \right) - 2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2 - 2{\rm{x}} - 2{\rm{x}} - 4}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{ - 4{\rm{x - 2}}}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{\left( { - 4{\rm{x}} - 2} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 8{\rm{x}} - 2{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 6{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {4{{\rm{x}}^2} - 1} \right)}}\end{array}\)
d)
\(\begin{array}{l}1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}.\frac{{1 + x - 1}}{{1 - {x^2}}}\\ = 1 + \frac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\frac{x}{{1 - {x^2}}}\\ = 1 + \frac{{ - {x^2}\left( {{x^2} - 1} \right)}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}\\ = 1 + \frac{{ - {x^2}}}{{{x^2} + 1}}\\ = \frac{{{x^2} + 1 - {x^2}}}{{{x^2} + 1}}\\ = \frac{1}{{{x^2} + 1}}\end{array}\)
a) Ta có: \(A=\left(\frac{3}{2x+4}+\frac{x}{2-x}+\frac{2x^2+3}{x^2-4}\right):\frac{2x-1}{4x-8}\)
\(=\left(\frac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\frac{2x\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{4\left(x-2\right)}{2x-1}\)
\(=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}\)
\(=\frac{2x^2-x}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}\)
\(=\frac{x\left(2x-1\right)\cdot4\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)\cdot\left(2x-1\right)}\)
\(=\frac{2x}{x+2}\)
b)
ĐKXĐ: \(x\notin\left\{2;-2;\frac{1}{2}\right\}\)
Để A<2 thì A-2<0
hay \(\frac{2x}{x+2}-2< 0\)
\(\Leftrightarrow\frac{2x}{x+2}-\frac{2\left(x+2\right)}{x+2}< 0\)
\(\Leftrightarrow\frac{2x-2x-4}{x+2}< 0\)
\(\Leftrightarrow\frac{-4}{x+2}< 0\)
\(\Leftrightarrow-4;x+2\) khác dấu
mà -4<0
nên x+2>0
hay x>-2
mà \(x\notin\left\{2;-2;\frac{1}{2}\right\}\)
nên \(\left\{{}\begin{matrix}x>-2\\x\notin\left\{\frac{1}{2};2\right\}\end{matrix}\right.\)
Vậy: Để A<2 thì \(\left\{{}\begin{matrix}x>-2\\x\notin\left\{\frac{1}{2};2\right\}\end{matrix}\right.\)
c) Ta có: |x-1|=3
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Thay x=4 vào biểu thức \(A=\frac{2x}{x+2}\), ta được:
\(\frac{2\cdot4}{4+2}=\frac{8}{6}=\frac{4}{3}\)
Vậy: \(\frac{4}{3}\) là giá trị của biểu thức \(A=\frac{2x}{x+2}\) tại x=4
d) Để |A|=1 thì
\(\left[{}\begin{matrix}A=1\\A=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{2x}{x+2}=1\\\frac{2x}{x+2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=x+2\\2x=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=2\\2x+x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\3x=-2\end{matrix}\right.\Leftrightarrow x=\frac{-2}{3}\)
Vậy: Để |A|=1 thì \(x=\frac{-2}{3}\)