\(\frac{2-x}{x-3}-1=\frac{5}{x}\)
Giai giup minh phuong trinh nay a
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14 tháng 2 2019
\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
\(\Rightarrow\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)\(\Rightarrow\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)
\(\Rightarrow\left(x-2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=\left(x-2004\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
Với \(x-2004\ne0\)
\(\Rightarrow\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(KTM\right)\)
Với \(x-2004=0\)
\(\Rightarrow x=2004\)
NT
1
PC
13 tháng 4 2016
\(A=\frac{\left(x-1\right)-5\sqrt{x-1}+6}{\sqrt{x-1}\cdot\left(\sqrt{x-1}-3\right)}=\frac{\left(\sqrt{x-1}-2\right)\cdot\left(\sqrt{x-1}-3\right)}{\sqrt{x-1}\cdot\left(\sqrt{x-1}-3\right)}\) Đk x\(\ne\) 1;10
\(A=\frac{\sqrt{x-1}-2}{\sqrt{x-1}}=1-\frac{2}{\sqrt{x-1}}\)
BH
21 tháng 2 2019
\(\frac{2}{\left(x+3\right)\left(x+1\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}-\frac{2}{x+3}+\frac{2}{x+3}-\frac{2}{x+5}+\frac{2}{x+5}-\frac{2}{x+7}=\frac{2}{9}\)
\(\frac{2}{x+1}-\frac{2}{x+7}=\frac{2}{9}\\ \Rightarrow\frac{2x+14-2x-2}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}\\ \Rightarrow\frac{12}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}=\frac{12}{54}\)
\(\Rightarrow\left(x+1\right)\left(x+7\right)=54\\ \Rightarrow x^2+8x-54=0\Rightarrow x=-4\pm\sqrt{70}\)
20 tháng 3 2022
a) \(\left(x+\frac{7}{4}\right)\times\frac{3}{2}=6\)
\(\Leftrightarrow\left(x+\frac{7}{4}\right)=6\div\frac{3}{2}\)
\(\Leftrightarrow x+\frac{7}{4}=4\)
\(\Leftrightarrow x=4-\frac{7}{4}\)
\(\Leftrightarrow x=\frac{9}{4}\)
b) \(x\div\frac{3}{5}+\frac{2}{5}=\frac{9}{5}\)
\(\Leftrightarrow x\div\frac{3}{5}=\frac{9}{5}-\frac{2}{5}\)
\(\Leftrightarrow x\div\frac{3}{5}=\frac{7}{5}\)
\(\Leftrightarrow x=\frac{7}{5}\times\frac{3}{5}\)
\(\Leftrightarrow x=\frac{21}{25}\)
c) \(\frac{1}{2}\div3+x=\frac{5}{3}\)
\(\Leftrightarrow\frac{1}{6}+x=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}-\frac{1}{6}\)
\(\Leftrightarrow x=\frac{3}{2}\)
14 tháng 2 2018
a) Ta có: \(\frac{x+a}{x+2}+\frac{x-2}{x-a}=2\left(1\right)\)
Với a = 4
Thay vào phương trình (t) ta được:
\(\frac{x+2}{x+2}+\frac{x-2}{x-2}=2\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2-4+x^2-4=2\left(x^2-4\right)\)
\(\Leftrightarrow2x^2=2x^2-8\)
\(\Leftrightarrow0x=-8\)
Vậy phương trình vô nghiệm
b) Nếu x = -1
\(\Rightarrow\frac{-1+a}{-1+2}+\frac{-1-2}{-1-a}=2\)
\(\Leftrightarrow\frac{-1+a}{1}+\frac{-3}{-1-a}=2\)
\(\Leftrightarrow\frac{\left(-1+a\right)\left(-1-a\right)}{-1-a}+\frac{-3}{-1-a}=\frac{2\left(-1-a\right)}{-1-a}\)
\(\Leftrightarrow1+a-a-a^2-3=-2-2a\)
\(\Leftrightarrow-a^2+2a=-2-1+3\)
\(\Leftrightarrow a\left(2-a\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=0\\2-a=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
Vậy a = {0;2}
NĂM MỚI VUI VẺ
16 tháng 3 2020
1.: Áp dụng BĐT Cauchy-Schwarz cho 3 số dương
\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)
FS
28 tháng 2 2019
\(\frac{x+10}{2000}+\frac{x+20}{1990}+\frac{x+30}{1980}+\frac{x+40}{1970}=-4\)
\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)
\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)
Vì \(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}>0\)
\(\Rightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
BH
28 tháng 2 2019
\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)
\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)
mà\(\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)\ne0\Rightarrow\left(x+2010\right)=0\\ \Rightarrow x=-2010\)