Tìm x:
a) \(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
b) \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
thực hiện phép tính:
a)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
b)\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
nên x + 1 = 0 => x = -1
Vậy x = -1
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)
\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)
nên 2004 + x = 0 => x = -2004
Vậy x = -2004
=))
Tìm số hữu tỉ x , biết rằng
e,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
f, \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
Tìm x
a) \(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
b) \(\frac{x+1}{3}=\frac{x-2}{4}\)
Tìm số hữu tỉ x, biết:
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}+\frac{x+1}{13}+\frac{x+1}{14}\)
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}+\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(Tìm\) \(x,\) \(biết\) \(:\)
\(a)\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}+=\frac{x+1}{13}+\frac{x+1}{14}\\ b)\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\)
\(\Leftrightarrow x=0-2004\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004.\)
Chúc bạn học tốt!
\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
Tìm x
\(\frac{x+32}{11}=\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
\(\Rightarrow\frac{x+32}{11}+\frac{x+23}{12}-\frac{x+38}{13}-\frac{x+27}{14}=0\)
\(\Rightarrow\left(\frac{x+32}{11}-\frac{x+38}{13}\right)+\left(\frac{x+23}{12}-\frac{x+27}{14}\right)=0\)
\(\Rightarrow\frac{2x-2}{11.13}+\frac{2x-2}{12.14}=0\)
\(\Rightarrow\frac{2x-2}{1}.\left(\frac{1}{11.13}+\frac{1}{12.14}\right)=0\)
vì \(\left(\frac{1}{11.13}+\frac{1}{12.14}\right)\ne0\)
mà \(\frac{2x-2}{1}.\left(\frac{1}{11.13}+\frac{1}{12.14}\right)=0\)
=> \(\frac{2x-2}{1}=0\Rightarrow2x-2=0\Rightarrow2x=2\Rightarrow x=1\)
a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\) b) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
c) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
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
Tìm x biết
a) x+2x+3x+4x+...+100x=-213
b)\(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
c)3(x-2)+2(x-1)=10
d)\(\frac{x+1}{3}=\frac{x-2}{4}\)
e)\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
f)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
a) x + 2x + 3x + ... +100x = -213
=> x . (1 + 2 + 3 +... + 100) = - 213
=> x . 5050 = -213
=> x = - 213 : 5050
=> x = -213/5050
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)
=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)
=> \(x.\frac{1}{4}=\frac{1}{6}\)
=> \(x=\frac{1}{6}:\frac{1}{4}\)
=> \(x=\frac{2}{3}\)
c) 3(x-2) + 2(x-1) = 10
=> 3x - 6 + 2x - 2 = 10
=> 3x + 2x - 6 - 2 = 10
=> 5x - 8 = 10
=> 5x = 10 + 8
=> 5x = 18
=> x = 18:5
=> x = 3,6
d) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> \(4\left(x+1\right)=3\left(x-2\right)\)
=>\(4x+4=3x-6\)
=> \(4x-3x=-4-6\)
=> \(x=-10\)
Cho \(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\). Tìm x.