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ta phân tích thành
\(\frac{1}{1\cdot2}\)x\(\frac{2\cdot2}{2\cdot3}\)x\(\frac{3\cdot3}{3\cdot4}\)x......x\(\frac{5\cdot5}{5\cdot6}\)x\(\frac{6\cdot6}{6\cdot7}\)
Tử nhân tử mẫu nhân mẫu ta có
1x2x2x3x3x.........x5x5x6x6
1x2x2x3x3x4x....x5x6x6x7
Rút gọn ta có
\(\frac{1}{7}\)
Vậy A=\(\frac{1}{7}\)
\(\frac{5}{16}:0,125+1,456:\frac{7}{25}+\frac{9}{2}.0,8\)
\(=\frac{5}{16}:\frac{125}{1000}+\frac{1456}{1000}:\frac{7}{25}+\frac{9}{2}.\frac{8}{10}\)
\(=\frac{5}{16}.\frac{1000}{125}+\frac{1456}{1000}.\frac{25}{7}+\frac{18}{5}\)
\(=\frac{5}{2}+\frac{26}{5}+\frac{18}{5}\)
\(=\frac{25}{10}+\frac{52}{10}+\frac{36}{10}\)
\(=\frac{25+52+36}{10}=\frac{113}{10}\)
a) 3/7 + 4/9 + 4/7 + 5/9
= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )
= 7/7 + 9/9
= 1 + 1
= 2
b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5 ) + 5/5
= 2 + 2 + 2 + 2 + 1
= 2 x 4 + 1
= 8 +1
= 9
c) 1/8 + 1/12 + 3/8 + 5/12
= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)
= 4/8 + 6/12
= 1/2 + 1/2
= 2/4 = 1/2
mỏi tay rồi
(\(\frac{5}{16}\) : 0,125 ) + ( 1,456 : \(\frac{7}{25}\) ) +( \(\frac{9}{2}\) x 0,8 )
= 2,5 + 5,2 + 3,6
= 7,7 + 3,6
= 11,3
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}\)
\(=\frac{5}{6}\)
\(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\)\(\frac{1}{5.6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{5}{6}\)
Hok tốt
\(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}=\frac{1.\left(100-2\right)+2\left(100-3\right)+3\left(100-4\right)+...+98\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{1.100-1.2+2.100-2.3+3.100-3.4+...+98.100-98.99}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{\left(1.100+2.100+3.100+...+98.100\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{100\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}-1\)
Ta có: 1+2+3+...+98=98.99:2=4851
Đặt B=1.2+2.3+3.4+...+98.99 => 3B=1.2.3+2.3.3+3.4.3+...+98.99.3 = 1.2.3+2.3.(4-1)+3.4(5-2)+...+98.99(100-97)
=> 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99 = 98.99.100
=> B=33.98.100. Thay vào A được:
\(A=\frac{100.4851}{33.98.100}-1=\frac{3}{2}-1=\frac{1}{2}\)
1.
a.
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
b.
Tích có 100 thừa số
=> n = 100
\(\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times\left(100-100\right)\)
\(=\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times0\)
\(=0\)
2.
a.
\(135\times789789-789\times135135=1001\times\left(135\times789-789\times135\right)=1001\times0=0\)
b.
\(\left(28\times9696-96\times2828\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=\left[101\times\left(28\times96-96\times28\right)\right]\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=\left(101\times0\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=0\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=0\)
3.
a.
\(\left[\left(x+32\right)-17\right]\times2=42\)
\(\left(x+32\right)-17=\frac{42}{2}\)
\(\left(x+32\right)-17=21\)
\(x+32=21+17\)
\(x+32=38\)
\(x=38-32\)
\(x=6\)
b.
\(125+\left(145-x\right)=175\)
\(145-x=175-125\)
\(145-x=50\)
\(x=145-50\)
\(x=95\)
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