x+1/10+x+1/11+x+1/12=x+1/13+x+1/14
giup minh vs nha cac bn
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a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)
\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)
\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)
\(\Leftrightarrow x=\frac{-33}{28}\)
b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Ta có:
A = 1 + 3 + 5 + 7 +... + 101
A = \(\frac{102.51}{2}=2601\)
M = 16 - 18 + 20 - 22 + 24 - 26 + .. + 64 - 66 + 68
M = ( 16 - 18 ) + ( 20 - 22 ) + ( 24 - 26 ) + ... + ( 64 - 66 ) + 68
M = (- 2 + - 2 + -2 + ... + - 2 ) + 68
M = 25/2 . ( - 2 ) + 68
M = -25 + 68
M = 43
H = ( 1 + 2 + 3 +...+ 99 ) x ( 13 x 15 - 12 x 15 - 15 )
H = ( 1 + 2 + 3 +...+ 99 ) x { (13 - 12 - 1) x 15 }
H = ( 1 + 2 + 3 +...+ 99 ) x 0
H = 0
G = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + 13 + 14 - ... + 301 + 302
G = ( 1 + 2 ) + ( -3 - 4 ) + ( 5 + 6 ) + ( -7 - 8 ) + ( 9 + 10 ) + ( - 11 - 12 ) + ( 13 + 14 ) -...+ ( 301 + 302 )
G = ( 3 - 7 ) + ( 11 - 15 ) + ( 19 - 23 ) + 27 - ... + 603
G = -4 + - 4 + -4 + 27 - ... + 603
G = 75 x ( -4 ) + 603
G = -300 + 603
G = 303
2.
a) 1 + 2 + 3 + 4 +...+ 99 + 100 + 2 x X = 5052
= > \(\frac{100.101}{2}\)+ 2 x X = 5052
= > 5050 + 2 x X = 5052
= > 2X = 2
= > X = 1
\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
Vậy tập nghiệm của ohuowng trình là \(S=\left\{\dfrac{1}{3};3;4\right\}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
=> x+1 =0 => x =-1
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
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
=> x + 1 = 0 ( vì 1/10 + 1/11 + 1/12 - 1/13 - 1/14 khac 0 )
=> x = -1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
mà 1/10 > 1/13; 1/11>1/14
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
=> x + 1 = 0
x = -1
x+110+x+111+x+112=x+113+x+114x+110+x+111+x+112=x+113+x+114
⇒x+110+x+111+x+112−x+113−x+114=0⇒x+110+x+111+x+112−x+113−x+114=0
⇒(x+1).(110+111+112−113−114)=0⇒(x+1).(110+111+112−113−114)=0
mà 1/10 > 1/13; 1/11>1/14
⇒110+111+112−113−114≠0⇒110+111+112−113−114≠0
=> x + 1 = 0
x = -1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow x+1=0\)( do \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\))
\(\Rightarrow x=-1\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=0-1=-1\)