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(x+2)/11+(x+2)/12+(x+2)/13=(x+2)/14+(x+2)15
<=> (x+2)/11+(x+2)/12+(x+2)/13 - (x+2)/14 - (x+2)/15 = 0
<=> (x+2)(1/11+1/12+1/13 - 1/14 - 1/15 ) = 0
vì: (1/11+1/12+1/13 - 1/14 - 1/15 ) khác 0 nên x-2 = 0 => x=2
Sorry mink mới lớp 5 nên ko thể giúp bn lm bài toán này thành thật xin lỗi
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Dễ thấy \(\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}\)nên biểu thức trong ngoặc thứ hai \(\ne\)0
Do đó \(x+1=0\)\(\Rightarrow x=0-1=-1\)
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+4}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}\)nên biểu thức trong ngoặc thứ hai phải \(\ne\)0
Do đó \(x+2004=0\)\(\Rightarrow x=0-2004=-2004\)
` 1, 5 . ( -2/25 = 3/2 . (-2)/25 = -6/50 = -3/25 `
` 1 3/5 . (-3)/4 = 8/5 . (-3)/4 = -24/20 = -6/5 `
`-15/4 : 21/(-1) = -15/4 . (-1)/21 = 15/84 = 5/28 `
` ( -2 1/7 ) : ( -1 1/14 ) = (-15)/7 : (-15)/14 = -15/7 . (-14)/15 = 210/105 = 2 `
`@animephamdanhv.`
a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)
\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)
=>x+1=0
hay x=-1
c: |x-2|=13
=>x-2=13 hoặc x-2=-13
=>x=15 hoặc x=-11
d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)
=>7|x-2|=5/3
=>|x-2|=5/21
=>x-2=5/21 hoặc x-2=-5/21
=>x=47/21 hoặc x=37/21
a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
a. 2(x-1) + (x+2) - (x+3) = 15 - (x+1)
=>2x-2+x+2-x-3=15-x-1
=>(2x+x-x)-2+2-3=15-1-x
=>2x-3=14-x
=>3x=17
=>x=17/3
b. x+1/15 + x+2/14 = x+4/12 + x+5/11
\(\Rightarrow\frac{x+1}{15}+1+\frac{x+2}{14}+1=\frac{x+4}{12}+1+\frac{x+5}{11}+1\)
\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}=\frac{x+16}{12}+\frac{x+16}{11}\)
\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{12}-\frac{x+16}{11}=0\)
\(\Rightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\right)=0\)
\(\Rightarrow x+16=0\).Do \(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\ne0\)
=>x=-16
\(\approx1,409\)mình không tính được kết quả cụ thể
tính cụ thể một chút!