ai giải bài này được ko: 2- 7/9- 5/6 và 1-1/3 -1/6 -1/12-1/24 có ai làm được hông
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\(1-\frac{1}{3}-\frac{1}{6}-\frac{1}{12}-\frac{1}{24}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+\frac{1}{12}-\frac{1}{24}\)
\(=1-\frac{1}{24}\)
\(=\frac{23}{24}\)
\(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{2}{4}\)
= \(\dfrac{24}{60}\) + \(\dfrac{40}{60}\) + \(\dfrac{30}{60}\)
= \(\dfrac{64}{60}\) + \(\dfrac{30}{60}\)
= \(\dfrac{47}{30}\)
\(\dfrac{2}{6}+\dfrac{3}{12}\)
= \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\)
= \(\dfrac{7}{12}\)
\(\dfrac{5}{6}\) + \(\dfrac{1}{3}\)
= \(\dfrac{5}{6}\) + \(\dfrac{2}{6}\)
= \(\dfrac{7}{6}\)
\(\dfrac{1}{3}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{6}\)
= \(\dfrac{4}{12}\) + \(\dfrac{5}{12}\) + \(\dfrac{10}{12}\)
= \(\dfrac{9}{12}\) + \(\dfrac{10}{12}\)
= \(\dfrac{19}{12}\)
\(\dfrac{5}{8}\) + \(\dfrac{4}{7}\)
= \(\dfrac{35}{56}\) + \(\dfrac{32}{56}\)
= \(\dfrac{67}{56}\)
\(\dfrac{7}{3}\) + \(\dfrac{8}{7}\)
= \(\dfrac{49}{21}\) + \(\dfrac{24}{21}\)
= \(\dfrac{73}{21}\)
\(\dfrac{1}{5}+\dfrac{5}{35}\)
= \(\dfrac{7}{35}\) + \(\dfrac{5}{35}\)
= \(\dfrac{12}{35}\)
1+2-3-4+5+6-7-8+...........+2001+2002-2003-2004
= (1+2-3-4) + (5+6-7-8) +........+ (2001+2002-2003-2004) + 2005 + 2006 - 2007
= (- 4) + (- 4) + .........+ (- 4) + 2005 + 2006 - 2007
= (- 4) x 501 + 2005 + 2006 - 2007
= - 2004 + 2005 +2006 - 2007
= 1 + 2006-2007
= 2007-2007
= 0
Mình giải rồi nhé nhớ k
a: \(2x+5⋮x+1\)
=>\(2x+2+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
b: \(5x+9⋮x+2\)
=>\(5x+10-1⋮x+2\)
=>\(-1⋮x+2\)
=>\(x+2\in\left\{1;-1\right\}\)
=>\(x\in\left\{-1;-3\right\}\)
c: \(2x+11⋮x+3\)
=>\(2x+6+5⋮x+3\)
=>\(5⋮x+3\)
=>\(x+3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-2;-4;2;-8\right\}\)
d: \(4x+9⋮2x+1\)
=>\(4x+2+7⋮2x+1\)
=>\(7⋮2x+1\)
=>\(2x+1\in\left\{1;-1;7;-7\right\}\)
=>\(2x\in\left\{0;-2;6;-8\right\}\)
=>\(x\in\left\{0;-1;3;-4\right\}\)
e: \(6x+7⋮3x+1\)
=>\(6x+2+5⋮3x+1\)
=>\(5⋮3x+1\)
=>\(3x+1\in\left\{1;-1;5;-5\right\}\)
=>\(3x\in\left\{0;-2;4;-6\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)
g: \(10x+13⋮5x+1\)
=>\(10x+2+11⋮5x+1\)
=>\(11⋮5x+1\)
=>\(5x+1\in\left\{1;-1;11;-11\right\}\)
=>\(5x\in\left\{0;-2;10;-12\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{5};2;-\dfrac{12}{5}\right\}\)
0+1+2+3+4+5+6+7+8+9+10
= ( 0 + 10 ) + ( 1 + 9 ) + ( 2 + 8 ) + ( 3 + 7 ) + ( 4 + 6 ) + 5
= 10 + 10 + 10 + 10 + 10 + 5
= 10 x 5 + 5
= 50 + 5
= 55
a,\(\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}\)
=\(\frac{1}{2}+\frac{3}{4}+\frac{-3}{4}+\frac{4}{5}\)
=\(\left(\frac{3}{4}+\frac{-3}{4}\right)+\frac{1}{2}+\frac{4}{5}\)
= \(0+\frac{1}{2}+\frac{4}{5}=\frac{13}{10}\)
Nhiều quá bạn ơi
\(2-\frac{7}{9}-\frac{5}{6}\)
\(=\frac{x}{y}-\frac{a}{b}\)
(Tự làm đi chứ )
ai còn thức hông giải hộ tớ bài này đi