TÍNH GIÁ TRỊ BIỂU THỨC
\(A=\frac{4}{3}-\frac{8}{15}+\frac{12}{35}-...+\frac{188}{8835}-\frac{192}{9215}+\frac{196}{9603}\)
tính :
A=4/3-8/15+12/35-..............+188/8835-192/9215+196/9603
Cho mình cách giải đi
Tính nhanh:
A=4/3-8/15+12/35-...+188/8835-192/9215+196/9603
tinh nhanh
4/3-8/15+12/35-16/63+....+188/8832-192/9215+196/9603
Tính giá trị của biểu thức \(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)\left(1+\frac{1}{24}\right)...\left(1+\frac{1}{9603}\right)\)
(1+1/3)(1+1/8)(1+1/15)...(1+1/9603)=4/3 . 9/8 . 16/15 ... 9604/9603
= (2.2)/(1.3) . (3.3)/(2.4) . (4.4)/(3.5) ... (98.98)/(97.99)
=(2.2.3.3.4.4...98.98)/(1.3.2.4.3.5...97.99)
=(2.3.4...98)/(1.2.3...97) . (2.3.4..98)/(3.4.5...99)
=98/1 .2/99 =169/99 .
1+1/3)(1+1/8)(1+1/15)...(1+1/9603)=4/3 . 9/8 . 16/15 ... 9604/9603
= (2.2)/(1.3) . (3.3)/(2.4) . (4.4)/(3.5) ... (98.98)/(97.99)
=(2.2.3.3.4.4...98.98)/(1.3.2.4.3.5...97.99)
=(2.3.4...98)/(1.2.3...97) . (2.3.4..98)/(3.4.5...99)
=98/1 .2/99 =169/99
\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x\frac{24}{25}x\frac{35}{36}x\frac{48}{49}x\frac{63}{64}\)
Tính giá trị biểu thức
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
Tính giá trị của biểu thức sau (tính hợp lí, nếu có thể):
a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\);
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} = \frac{{ -29}}{{35}}\end{array}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)
\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ = (- 2).45,3 = - 90,6\end{array}\)
a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)
b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)
\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)
c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)
d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)
Tính giá trị của các biểu thức sau:
\(\begin{array}{l}a)\frac{{{3^{12}} + {3^{15}}}}{{1 + {3^3}}}\\b)2:{\left( {\frac{1}{2} - \frac{2}{3}} \right)^2} + 0,{125^3}{.8^3} - {( - 12)^4}:{6^4}\end{array}\)
\(\begin{array}{l}a)\frac{{{3^{12}} + {3^{15}}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}} + {3^{12}}{{.3}^3}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}}.(1 + {3^3})}}{{1 + {3^3}}}\\ = {3^{12}}\\b)2:{\left( {\frac{1}{2} - \frac{2}{3}} \right)^2} + 0,{125^3}{.8^3} - {( - 12)^4}:{6^4}\\ = 2:{\left( {\frac{3}{6} - \frac{4}{6}} \right)^2} + {(0,125.8)^3} - {12^4}:{6^4}\\ = 2:{\left( {\frac{{ - 1}}{6}} \right)^2} + {1^3} - {(\frac{{12}}{6})^4}\\ = 2:\frac{1}{{36}} + 1 - {2^4}\\ = 2.36 + 1 - 16\\ = 72 + 1 - 16=57\end{array}\)
Tính nhanh:
\(A=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9215}\)
Ta có:
\(A=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9215}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{95.97}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
Vậy \(A=\frac{96}{97}\)
\(A=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9215}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{95.97}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}\)
\(=1-\frac{1}{97}=\frac{96}{97}\)
Chúc bạn hok tốt! :))
Tính giá trị biểu thức:
A= \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
Ta có: \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(\Leftrightarrow A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\Rightarrow2A=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow A=\frac{10}{11}:2=\frac{5}{11}\)
Vậy \(A=\frac{5}{11}\)
A = \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
A = \(1-\frac{1}{11}\)
A = \(\frac{10}{11}\)
A = \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
A = \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
A = \(\frac{1}{2}.\frac{10}{11}=\frac{1}{11}\)