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a) x+1/2004 + 1 + x+2/2003 +1 - x+3/2002 +1 - x+4/2001 +1
=> x+2005/2004 + x+2005/2003 - x+2005/2002 - x+2005/2001=0
=> (x + 2005) ( 1/2004+1/2003 - 1/2002 - 1/2001) =0
ta thấy 1/2004+1/2003-1/2002-1/2001 # 0
=> x+2005=0 => x=-2005
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Rightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
\(\Rightarrow x-100=0\).Do \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\ne0\)
\(\Rightarrow x=100\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{x-45}{55}-1-\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
(x-100)(\(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}=0\)
-> x-100 = 0 -> x = 100
mà \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\) khác 0
Vậy x = 100
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Leftrightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\Leftrightarrow\frac{x-45-55}{55}+\frac{x-47-53}{47}-\frac{x-55-45}{45}-\frac{x-53-47}{47}=0\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{47}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
Vậy pt có tập nghiệm S = { 100 }
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
dễ thôi mà
Áp dụng tỉ lệ thức, ta có:
\(\Leftrightarrow\frac{108x-4970}{2915}=\frac{92x-4970}{2115}\Rightarrow\left(108x-4970\right)2115=2915\left(92x-4970\right)\)
=>x=100
mấy câu này dễ mà :V câu a+c lấy mỗi phân số trừ cho 1 ra tử chung rút ra thì tính b+d thì cộng một tử chung rồi lại tính tiếp thôi
a, Mình nghĩ là đề sai .
b, Ta có : \(\frac{x-45}{55}+\frac{x-47}{45}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{45}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-45}{55}-\frac{55}{55}+\frac{x-47}{53}-\frac{53}{53}=\frac{x-55}{45}-\frac{45}{45}+\frac{x-53}{47}-\frac{47}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
=> \(x=100\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{100\right\}\)
c, Ta có : \(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
=> \(\frac{2-x}{2010}-1=\frac{1-x}{2011}+\frac{-x}{2012}\)
=> \(\frac{2-x}{2010}+1=\frac{1-x}{2011}+1+\frac{-x}{2012}+1\)
=> \(\frac{2-x}{2010}+\frac{2010}{2010}=\frac{1-x}{2011}+\frac{2011}{2011}+\frac{-x}{2012}+\frac{2012}{2012}\)
=> \(\frac{2012-x}{2010}=\frac{2012-x}{2011}+\frac{2012-x}{2012}\)
=> \(\frac{2012-x}{2010}-\frac{2012-x}{2011}-\frac{2012-x}{2012}=0\)
=> \(\left(2012-x\right)\left(\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)
=> \(2012-x=0\)
=> \(x=2012\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2012\right\}\)