rut gon /x-3/+/4+x/+x-5=Q
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20 tháng 4 2020
\(Q=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
\(\Leftrightarrow\) \(Q=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(2-x\right)}+\frac{5}{\left(x+3\right)\left(2-x\right)}+\frac{-1}{\left(x+3\right)\left(2-x\right)}\)
\(\Rightarrow\) \(Q=\left(x-2\right)\left(x+2\right)+5-1\)
\(\Leftrightarrow\) \(Q=x^2-4+5-1\)
\(\Leftrightarrow\) \(Q=x^2\)
Thay \(Q=\frac{-3}{4}\) ta được:
\(x^2=\frac{-3}{4}\)
Vì \(\frac{-3}{4}>0\forall x\)
\(\Rightarrow\) Pt vô nghiệm
Vậy không có giả trị nào của x thỏa mãn \(Q=\frac{-3}{4}\)
Chúc bn học tốt!!
HD
2
25 tháng 6 2018
\(\frac{x^4-y^4}{y^3-x^3}=\frac{\left(x^2\right)^2-\left(y^2\right)^2}{\left(y-x\right)\left(y^2+xy+x^2\right)}=-\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=-\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=-\frac{\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
JM
1
N
30 tháng 4 2018
\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-1\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{x-4}{x-1}\)
MT
9 tháng 1 2016
Điều kiện : x>=0
\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{1}-x}{\sqrt{1}+\sqrt{x}}=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
4 tháng 7 2015
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
HN
0
Ta có :
TH1 : \(x< -4;\)ta có :
\(Q=\left[-\left(x-3\right)\right]+\left[-\left(x+4\right)\right]+x-5\)
\(=3-x-x-4+x-5\)
\(=-6-x\)
TH2 : \(-4\le x< 3;\)ta có :
\(Q=\left[-\left(x-3\right)\right]+\left(4-x\right)+x-5\)
\(=3-x+4-x+x-5\)
\(=2-x\)
TH3 : \(x\ge3;\)ta có :
\(Q=\left(x-3\right)+\left(4+x\right)+x-5\)
\(=x-3+4+x+x-5\)
\(=3x-4\)