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Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2009}{2011}\)
Đặt tổng vế trái là A
Ta có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\)
\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)\div2}\right)\)
\(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\)
\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)
\(\frac{1}{2}A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)
\(A=\left(\frac{1}{2}+\frac{1}{x+1}\right):\frac{1}{2}\)
\(A=1+\frac{1}{\left(x+1\right)\div2}\)
\(\Rightarrow1+\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}-1=\frac{2009}{2011}-\frac{2011}{2011}=-\frac{2}{2011}\)
\(\Rightarrow-\frac{2}{-\left(x+1\right)}=-\frac{2}{2011}\)
\(\Rightarrow-\left(x+1\right)=2011\)
\(\Rightarrow x+1=-2011\)
\(\Rightarrow x=-2011-1=-2012\)
\(a.2^6.\left(x-2\right)=104\)
\(x-2=104:2^6\)
\(x-2=1,652\)
\(x=1,625+2\)
\(x=3,625\)
\(b.2\times4^{x+1}=128\)
\(4^{x+1}=128:2\)
\(4^{x+1}=64\)
\(4^{x+1}=4^3\)
\(\Rightarrow x+1=3\)
\(x=3-1\)
\(\Leftrightarrow x=3\)
\(c.227-5\left(x+8\right)=3^6:3^3\)
\(227-5\left(x+8\right)=3^3\)
\(227-5\left(x+8\right)=27\)
\(5\left(x+8\right)=227-27\)
\(5\left(x+8\right)=200\)
\(x+8=200:5\)
\(x+8=40\)
\(x=40-8\)
\(x=32\)
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Bài 1:
a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)
b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)
a ) 15 : ( x + 2 ) = 3
x + 2 = 15 : 3
x + 2 = 5
x = 5 - 2
x = 3
b ) 541 - ( 218 + x ) = 73
218 + x = 541 - 73
218 + x = 468
x = 468 - 218
x = 250
c ) 20 : ( x + 1 ) = 2
x + 1 = 20 : 2
x + 1 = 10
x = 10 - 1
x = 9
d ) 96 - 3 . ( x + 1 ) = 42
3 . ( x + 1 ) = 96 - 42
3 . ( x + 1 ) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 - 1
x = 17
6.|2-x|= |-15+3|
=> 6.|2 - x| = | - 12 |
=> 6. |2 - x| = 12
=> |2 - x| = 12 : 6
=> |2 - x| = 2
=>\(\orbr{\begin{cases}2-x=2\\2-x=-2\end{cases}}\)=>\(\orbr{\begin{cases}x=2-2\\x=2-\left(-2\right)\end{cases}}\)=>\(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
| 2 - x | = | - 15 + 3 |
=> | 2 - x | = | - 12 |
=> | 2 - x | = 12
=> 2 - x = 12 hoặc 2 - x = - 12
TH1 : 2 - x = 12
=> x = 14
TH2 : 2 - x = - 12
=> x = - 10
Vậy....