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NV
11 tháng 6 2016
ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x^2+x+1\ne0\end{cases}}\)
a/ \(R=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right]\)
\(=1:\left[\frac{x^2+2+\left(x+1\right)\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x}{x^2+x+1}\right)\)
\(=\frac{x^2+x+1}{x}\)
b/ Ta có: \(R=\frac{x^2+x+1}{x}=3+\frac{\left(x-1\right)^2}{x}>3\)
Vậy R > 3
27 tháng 1 2023
a: R-3=(x^2+x-1-3x)/x=(x-1)^2/x
Nếu x>0 thì R-3>0
=>R>3
Nếu x<0 thì R-3<0
=>R<3
c: Để R>4 thì R-4>0
=>\(\dfrac{x^2+x+1-4x}{x}>0\)
=>\(\dfrac{x^2-3x+1}{x}>0\)
TH1: x>0 và x^2-3x+1>0
=>x>0 và \(\left[{}\begin{matrix}x< \dfrac{3-\sqrt{5}}{2}\\x>\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow x>\dfrac{3+\sqrt{5}}{2}\)
mà x nguyên
nên x>3
TH2: x<0 và x^2-3x+1<0
=>x<0 và \(\dfrac{3-\sqrt{5}}{2}< x< \dfrac{3+\sqrt{5}}{2}\)(loại)
30 tháng 1 2022
a: \(P=\dfrac{4x-6-x+1}{2x-3}:\left(\dfrac{6x+1}{2x^2-3x+2x-3}+\dfrac{x}{x+1}\right)\)
\(=\dfrac{3x-5}{2x-3}:\left(\dfrac{6x+1}{\left(x+1\right)\left(2x-3\right)}+\dfrac{x}{x+1}\right)\)
\(=\dfrac{3x-5}{2x-3}:\dfrac{6x+1+2x^2-3x}{\left(x+1\right)\left(2x-3\right)}\)
\(=\dfrac{3x-5}{\left(2x-3\right)}\cdot\dfrac{\left(2x-3\right)\left(x+1\right)}{2x^2+3x+1}\)
\(=\dfrac{3x-5}{2x+1}\)
b: \(P-\dfrac{3}{2}=\dfrac{3x-5}{2x+1}-\dfrac{3}{2}=\dfrac{6x-10-6x-3}{2\left(2x+1\right)}=\dfrac{-7}{2\left(2x+1\right)}\)
28 tháng 6 2017
a.ĐKXĐ \(x\ne0,x\ne1\),\(x\ne-1\)
B=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2-1}{x^3-x}.\frac{x^3+x}{\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x.\left(x^2+1\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2+1}{\left(x-1\right)^2}\)
=\(\frac{3-x^2}{\left(x-1\right)^2}\)
b.TH1 x=3\(\Rightarrow\)B=\(\frac{3-3^2}{2^2}=\frac{-3}{2}\)
TH2 x=-1\(\Rightarrow\)B=\(\frac{3-\left(-1\right)^2}{4}=\frac{1}{2}\)
c.B=-1\(\Leftrightarrow\frac{3-x^2}{\left(x-1\right)^2}=-1\)\(\Leftrightarrow x^2-3=x^2-2x+1\)\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
d.B+2=\(\frac{3-x^2}{\left(x-1\right)^2}+2=\frac{x^2-4x+5}{\left(x-1\right)^2}=\frac{\left(x-2\right)^2+1}{\left(x-1\right)^2}\ge0\)với mọi x\(\Rightarrow B\)>-2
13 tháng 1 2019
\(a,M=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)
\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)
12 tháng 4 2021
a) ĐKXĐ: \(x\ne1\)
Ta có: \(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=7\left(nhận\right)\end{matrix}\right.\)
Thay x=7 vào B, ta được:
\(B=\dfrac{1}{7-1}=\dfrac{1}{6}\)
Vậy: Khi \(x^2-8x+7=0\) thì \(B=\dfrac{1}{6}\)
12 tháng 4 2021
b) Ta có: \(A=\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2+2+x^2-1}{x^3-1}\)
\(=\dfrac{2x^2+1}{x^3-1}\)
27 tháng 2 2020
a) \(P=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\right)\)
\(=\frac{x-1}{2}:\left(\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=\frac{x-1}{2}:\frac{x^2+2+\left(x^2-x\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)\(=\frac{x-1}{2}.\frac{\left(x-1\right)\left(x^2+x+1\right)}{x^2-2x+1}\)
\(=\frac{\left(x-1\right)^2\left(x^2+x+1\right)}{2\left(x-1\right)^2}=\frac{x^2+x+1}{2}\)
b) Ta thấy :
\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\frac{x^2+x+1}{2}>0\Rightarrow P=\left|P\right|\)
c) Lại có :
\(x^2+x+1\ge\frac{3}{4}\Rightarrow\frac{x^2+x+1}{2}\ge\frac{3}{8}\)
Dấu = xảy ra khi :
\(x^2+x+1=\frac{3}{4}\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy MinP = 3/4 ⇔ x = -1/2
27 tháng 2 2020
a, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x^2+x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\end{matrix}\right.\)
=> \(x\ne1\)
- Ta có : \(P=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}-\frac{1}{x-1}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x\left(x-1\right)}{x^3-1}-\frac{x^2+x+1}{x^3-1}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2+2+x\left(x-1\right)-x^2-x-1}{x^3-1}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2+2+x^2-x-x^2-x-1}{x^3-1}\right)\)
=> \(P=\frac{x-1}{2}:\left(\frac{x^2-2x+1}{x^3-1}\right)\)
=> \(P=\frac{\left(x-1\right)\left(x^3-1\right)}{2\left(x-1\right)^2}=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{2\left(x-1\right)^2}\)
=> \(P=\frac{x^2+x+1}{2}\)
b, Ta có : \(\left|P\right|=\left|\frac{x^2+x+1}{2}\right|=\frac{\left|x^2+x+1\right|}{2}\)
=> \(\left|P\right|=\frac{\left|x^2+x+\frac{1}{4}+\frac{3}{4}\right|}{2}=\frac{\left|\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right|}{2}\)
- Ta thấy : \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
=> \(\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}>0\)
=> \(\left|\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\right|=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\)
=> \(\left|P\right|=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\)
Vậy \(\left|P\right|=P=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{2}\)
\(R=\frac{x^2+x+1}{x}=\frac{3x+x^2-2x+1}{x}=\frac{3x+\left(x-1\right)^2}{x}=3+\frac{\left(x-1\right)^2}{x}\ge3\)
\(R=\frac{x^2+x+1}{x}=\frac{3x+x^2-2x+1}{x}=\frac{3x+\left(x-1\right)^2}{x}=3+\frac{\left(x-1\right)^2}{x}\ge3\)