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U
1
13 tháng 4 2018
a) x(x-8)-x(x+1)=2
x2 -8x -x2-x=2
-9x=2
\(x=-\frac{2}{9}\)
b) (x+3)5 - 7(x+9)=0
5x + 15 -7x -63=0
-2x - 48 =0
-2x=48
x=-24
c)4(x-7)+7(x-2)=11
4x -28 + 7x -14=11
11x -42=11
11x=11+42
11x=53
x=\(\frac{53}{11}\)
B
4
24 tháng 9 2021
\(\dfrac{x+4}{8}+\dfrac{x+3}{9}=\dfrac{x+2}{10}+\dfrac{x+1}{11}\)
\(\Leftrightarrow\left(\dfrac{x+4}{8}+1\right)+\left(\dfrac{x+3}{9}+1\right)=\left(\dfrac{x+2}{10}+1\right)+\left(\dfrac{x+1}{11}+1\right)\)
\(\Leftrightarrow\dfrac{x+12}{8}+\dfrac{x+12}{9}-\dfrac{x+12}{10}-\dfrac{x+12}{11}=0\)
\(\Leftrightarrow\left(x+12\right)\left(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}\right)=0\)
\(\Leftrightarrow x=-12\)( do \(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}\ne0\))
OY
24 tháng 9 2021
\(\dfrac{x+4}{8}+\dfrac{x+3}{9}=\dfrac{x+2}{10}+\dfrac{x+1}{11}\)
\(\dfrac{x+4}{8}+1+\dfrac{x+3}{9}+1=\dfrac{x+2}{10}+1+\dfrac{x+1}{11}+1\)
\(\dfrac{x+12}{8}+\dfrac{x+12}{9}=\dfrac{x+12}{10}+\dfrac{x+12}{11}\)
\(\dfrac{x+12}{8}+\dfrac{x+12}{9}-\dfrac{x+12}{10}-\dfrac{x+12}{11}=0\)
\(\Rightarrow\left(x+12\right).\left(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}\right)=0\)
Vì \(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}\ne0\) nên \(x+12=0\)
\(\Rightarrow x=-12\)
T
5 tháng 6 2019
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
T
5 tháng 6 2019
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
VT
5 tháng 8 2016
a) x=-213:(1+2+3+4+...+100)<=>x=-213/100
b) x-x=-1/3-2/4 <=> 0= -5/6 (vô lý )
c) x=-0,8119408369
d) x= 0.0258907758
25 tháng 1 2017
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
NT
1
MA
7 tháng 9 2016
\(\frac{x+4}{8}+\frac{x+3}{9}=\frac{x+2}{10}+\frac{x+1}{11}\)
\(\Leftrightarrow\frac{x+4}{8}+1+\frac{x+3}{9}+1=\frac{x+2}{10}+1+\frac{x+1}{11}+1\)
\(\Leftrightarrow\frac{x+12}{8}+\frac{x+12}{9}=\frac{x+12}{10}+\frac{x+12}{11}\)
\(\Leftrightarrow\frac{x+12}{8}+\frac{x+12}{9}-\frac{x+12}{10}-\frac{x+12}{11}=0\)
\(\Leftrightarrow\left(x+12\right)\left(\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}\right)=0\)
Mà: \(\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}\ne0\)
\(\Rightarrow x+12=0\)
\(\Rightarrow x=-12\)