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Ta có:
\(D=1.2+2.3+3.4+4.5+...+99.100\)
\(\Leftrightarrow3D=1.2.\left(3-0\right)+2.3+\left(4-1\right)+3.4+\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)
\(\Leftrightarrow3D=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)
\(\Leftrightarrow3D=99.100.101\Leftrightarrow D=\frac{99.100.101}{3}=333300\)
\(B=1.3+2.4+3.5+4.6+...+99.101\)
\(\Leftrightarrow B=\left(1.3+3.5+...+99.101\right)+\left(2.4+4.6+...+98.100\right)\)
\(\Leftrightarrow6B=\left(1.3.\left(5-\left(-1\right)\right)+3.5.\left(7-1\right)+...+99.101.\left(103-97\right)\right)+\left(2.4.\left(6-0\right)+4.6.\left(8-2\right)+...+98.100.\left(102-96\right)\right)\)
\(\Leftrightarrow B=\frac{99.101.103+3}{6}+\frac{98.100.102}{6}=338250\)
Vì các bước gần tương tự như bài a) nên mình bỏ bước.
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)
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a) \(215+x=400\)
\(\Rightarrow x=400-215\)
\(\Rightarrow x=185\)
b) \(12,5-2x=\dfrac{1}{5}\)
\(\Rightarrow2x=12,5-\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{123}{10}\)
\(\Rightarrow x=\dfrac{123}{10}:2\)
\(\Rightarrow x=\dfrac{123}{20}\)
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1> a) \(\frac{5}{7}x4:\frac{5}{9}=\frac{5}{7}:\frac{5}{9}x4=\frac{5}{7}x\frac{9}{5}x4=\frac{9}{7}x4=\frac{9x4}{7}=\frac{36}{7}\)
\(b,8x\frac{2}{3}:\frac{1}{2}=8x\frac{2}{3}x\frac{2}{1}=8x2x\frac{2}{3}=16x\frac{2}{3}=\frac{32}{3}\)
\(c,6:\frac{3}{5}-\frac{7}{6}x\frac{6}{7}=6x\frac{5}{3}-1=10-1=9\)
\(\frac{21}{5}x\frac{10}{11}+\frac{57}{11}=\frac{42}{11}+\frac{57}{11}=\frac{99}{11}=9\)
2) a) \(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}\)
\(x=\frac{35}{9}x\frac{6}{35}\)
\(x=\frac{2}{3}\)
b) \(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{5}-\frac{1}{6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)x10-X=10\)
\(\frac{5}{6}x10-X=0\)
\(X=\frac{5}{6}x10=\frac{25}{3}\)
Đúng nha !!!!
1/a/\(\frac{5}{7}\cdot4:\frac{5}{9}=\frac{20}{7}:\frac{5}{9}=\frac{20}{7}\cdot\frac{9}{5}=\frac{36}{7}\)
b/\(8\cdot\frac{2}{3}:\frac{1}{2}=\frac{16}{3}:\frac{1}{2}=\frac{16}{3}\cdot\frac{2}{1}=\frac{32}{3}\)
c/\(6:\frac{3}{5}-\frac{7}{6}\cdot\frac{6}{7}=6\cdot\frac{5}{3}-1=10-1=9\)
2/a/\(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}=\frac{35}{9}\cdot\frac{6}{35}\)
\(x=\frac{2}{3}\)
b/\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\cdot10-x=0\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\cdot10-x=0\)
\(\left(\frac{30}{60}+\frac{10}{60}+\frac{5}{60}+\frac{2}{30}\right)\cdot10-x=0\)
\(\frac{47}{60}\cdot10-x=0\)
\(\frac{47}{6}-x=0\)
\(x=\frac{47}{6}-0\)
\(x=\frac{47}{6}\)
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\(a̸\)
\(\frac{3}{5}-\frac{1}{2}.x=\frac{1}{4}\)
\(\frac{1}{2}.x=\frac{3}{5}-\frac{1}{4}\)
\(\frac{1}{2}.x=\frac{7}{20}\)
\(\Rightarrow\frac{7}{10}\)
\(b̸\)
\(11,3+2\left[x-\frac{1}{3}\right]=\frac{25}{6}\)
\(2\left[x-\frac{1}{3}\right]=\frac{25}{6}-11,3\)
\(2\left[x-\frac{1}{3}\right]=\frac{-107}{15}\)
\(x-\frac{1}{3}=\frac{-107}{15}:2\)
\(x-\frac{1}{3}=\frac{-107}{30}\)
\(x=\frac{-107}{30}+\frac{1}{3}\)
\(x=\frac{-97}{30}\)
\(a)\frac{3}{5}-\frac{1}{2}x=\frac{1}{4}\)
\(\implies\frac{1}{2}x=\frac{3}{5}-\frac{1}{4}\)
\(\implies\frac{1}{2}x=\frac{7}{20}\)
\(\implies x=\frac{7}{20}:\frac{1}{2}\)
\(\implies x=\frac{7}{10}\)
Vậy...
\(b) 11,3+2(x-\frac{1}{3})=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x-2.\frac{1}{3}=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x-\frac{2}{3}=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x=\frac{25}{6}+\frac{2}{3}\)
\(\implies \frac{113}{10}+2x=\frac{29}{6}\)
\(\implies 2x=\frac{29}{6}-\frac{113}{10}\)
\(\implies 2x=\frac{-97}{15}\)
\(\implies x=\frac{-97}{30}\)
Vậy..
\(c)5x-435+2x+140+3x=565\)
\(\implies (5x+2x+3x)+(-435+140)=565\)
\(\implies 10x+(-295)=565\)
\(\implies 10x=565-(-295)\)
\(\implies 10x=860\)
\(\implies x=86\)
Vậy...
~ hok tốt a~
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\(\frac{70}{3}-\left(x+4\frac{1}{5}\right)=16\)
\(\Rightarrow x+4\frac{1}{5}=\frac{22}{3}\)
\(\Rightarrow x+\frac{21}{5}=\frac{22}{3}\)
\(\Rightarrow x=\frac{22}{3}-\frac{21}{5}\)
\(\Rightarrow x=\frac{47}{15}\)
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A = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+..+9\right)}{1\times2+2\times3+3\times4+...+19\times20}\)
\(=\frac{\frac{1\times\left(1+1\right)}{2}+\frac{2\times\left(2+1\right)}{2}+\frac{3\times\left(3+1\right)}{2}...+\frac{9\times\left(9+1\right)}{2}}{1\times2+2\times3+3\times4+...+19\times20}\)
\(=\frac{\frac{1\times2}{2}+\frac{2\times3}{2}+\frac{3\times4}{2}+...+\frac{9\times10}{2}}{1\times2+2\times3+3\times4+...+9\times10}\)
\(=\frac{\frac{1}{2}\times\left(1\times2+2\times3+3\times4+...+9\times10\right)}{1\times2+2\times3+3\times4+...+9\times10}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\)
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4 + x = 7 - 2x
4 + x + 2x = 7
4 + 3x = 7
3x = 7 - 4
3x = 3
x = 3 : 3
x =1
5 + x = 8,3 + 4,7 - x
5 + x + x = 8,3 + 4,7
5 + 2x = 13
2x = 13 - 5
2x = 8
x = 8 : 2
x= 4
ta có :
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+\frac{2}{4\times5\times6}+\frac{2}{5\times6\times7}\)
\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+\frac{1}{4\times5}-\frac{1}{5\times6}+\frac{1}{5\times6}-\frac{1}{6\times7}\)
\(=\frac{1}{2}-\frac{1}{42}=\frac{10}{21}\text{ vậy }A=\frac{5}{21}\)