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1 tháng 2 2022
a: \(\Leftrightarrow2\sqrt{3x}+12-4x+5\sqrt{3}=0\)
\(\Leftrightarrow-4x+2\sqrt{3}\cdot\sqrt{x}+12+5\sqrt{3}=0\)
Đặt \(\sqrt{x}=a\left(a>=0\right)\)
Phương trình trở thành \(-4a^2+2\sqrt{3}a+12+5\sqrt{3}=0\)
\(\Delta=\left(2\sqrt{3}\right)^2-4\cdot\left(-4\right)\cdot\left(12+5\sqrt{3}\right)\)
\(=12+16\left(12+5\sqrt{3}\right)\)
\(=12+192+80\sqrt{3}=204+80\sqrt{3}\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}a_1=\dfrac{-2\sqrt{3}-\sqrt{204+80\sqrt{3}}}{-8}=\dfrac{2\sqrt{3}+\sqrt{204+80\sqrt{3}}}{8}\left(nhận\right)\\a_2=\dfrac{-2\sqrt{3}+\sqrt{204+80\sqrt{3}}}{-8}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow a=\dfrac{2\sqrt{3}+2\sqrt{26+20\sqrt{3}}}{8}=\dfrac{\sqrt{3}+\sqrt{26+20\sqrt{3}}}{4}\)
\(\Leftrightarrow x=a^2\simeq5,66\)
c: \(\Leftrightarrow x\sqrt{2}+5\sqrt{2}-4x-5-4\sqrt{2}=0\)
\(\Leftrightarrow x\left(\sqrt{2}-4\right)+\sqrt{2}-5=0\)
\(\Leftrightarrow x=\dfrac{5-\sqrt{2}}{\sqrt{2}-4}=\dfrac{-18-\sqrt{2}}{14}\)
d: \(\Leftrightarrow\dfrac{7x+1-4x-4002}{2001}=\dfrac{3x+2}{2003}-1\)
\(\Leftrightarrow3x-4001=0\)
hay x=4001/3
27 tháng 2 2019
\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)
\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)
\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)
\(\Leftrightarrow2015-x=0\)
\(\Leftrightarrow x=2015\)
KL : PT có nghiệm \(S=\left\{2015\right\}\)
10 tháng 8 2016
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2002}+1\right)+\left(\frac{-x}{2003}+1\right)\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow\) \(x=2003\)
10 tháng 8 2016
↔ \(\frac{2-x}{2001}+1\)\(=\left(\frac{1-x}{2002}+1\right)+\left(\frac{x}{2003}+1\right)\)
↔ \(\frac{2003-x}{2001}\) \(=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
↔ \(\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
↔ x = 2003
BD
3
YN
17 tháng 2 2020
\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}=4\)
\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+3}{2003}-1\right)=4-4=0\)
\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}=0\)
\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x-2000=0\) ( do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\ne0\) )
\(\Leftrightarrow x=2000\)
Vậy x = 2000
Đây là cách của lớp 7 nha
@@ Học tốt
17 tháng 2 2020
\(\frac{x}{2000}\)- 1+\(\frac{x+1}{2001}\)-1+\(\frac{x+2}{2002}\)-1+\(\frac{x+3}{2003}\)-1=0
<=>\(\frac{x-2000}{2000}\)+ \(\frac{x-2000}{2001}\)+ \(\frac{x-2000}{2002}\)+ \(\frac{x-2000}{2003}\)=0
<=>\(\left(x-2000\right)\)\(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)=0
Do \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)khác 0
=> \(x-2000=0\)<=> \(x=2000\)
KC
0
6 tháng 7 2019
a) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
b) Sửa đề :
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=300\)
c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)
\(\Leftrightarrow x=2004\)
Vậy....
FB
1
3 tháng 2 2020
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2001}+1\right)+\left(\frac{-x}{2003}+1\right)\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow\left(2003-x\right)=0\) (vì \(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\))
\(\Leftrightarrow x=2003\).
Vậy tập nghiệm của phương trình là \(S=\left\{2003\right\}\).