Ruat gọn biểu thức:
T=\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).....\left(\dfrac{1}{98}+1\right).\left(\dfrac{1}{99}+1\right)\)
Rút gọn biểu thức sau:
T =\(\left[\dfrac{1}{2}+1\right]\times\left[\dfrac{1}{3}+1\right]\times\left[\dfrac{1}{4}+1\right]\times.......\times\left[\dfrac{1}{98}+1\right]\times\left[\dfrac{1}{99}+1\right]\)
Giúp mik với
\(T=\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).......\left(\dfrac{1}{98}+1\right).\left(\dfrac{1}{99}+1\right) \) \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}....\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
\(T=\left|\dfrac{1}{2}+1\right|\left|\dfrac{1}{3}+1\right|\left|\dfrac{1}{4}+1\right|.....\left|\dfrac{1}{98}+1\right|\left|\dfrac{1}{99}+1\right|\)
\(T=\left|\dfrac{3}{2}\right|.\left|\dfrac{4}{3}\right|.\left|\dfrac{5}{4}\right|......\left|\dfrac{99}{98}\right|.\left|\dfrac{100}{99}\right|\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5.....99.100}{2.3.4.....98.99}=\dfrac{100}{2}=50\)
\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}=\dfrac{3.4.5...99.100}{2.3.4...98.99}=\dfrac{100}{2}=50\)
TÍNH GIÁ TRỊ BIỂU THỨC:\(\left(1+\dfrac{2}{3}\right)\cdot\left(1+\dfrac{2}{4}\right)\cdot\left(1+\dfrac{2}{5}\right)\cdot...\cdot\left(1+\dfrac{2}{97}\right)\cdot\left(1+\dfrac{2}{98}\right)\)
Rút gọn biểu thức D=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(D=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2014}{2015}=\dfrac{1.2.3....2014}{2.3.4....2015}\)
\(D=\dfrac{1}{2015}\)
\(D=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2015}\right)\)
\(D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2014}{2015}=\dfrac{1.2.3...2014}{2.3.4...2015}\)
\(D=\dfrac{1}{2015}\)
Thu gọn các biểu thức sau
A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)
B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{299}\right)\)
C = \(-\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{333333}{424242}\right)\)
A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)
= \(\left(-2\right).\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{215}{214}\right)\)
= \(\dfrac{\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right)...\left(-215\right)}{1.2.3.4...214}\)
= \(\dfrac{2.3.4.5...215}{1.2.3.4...214}\)
= \(\dfrac{215}{1}=215\)
B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)....\left(-1\dfrac{1}{299}\right)\)
= \(\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{300}{299}\right)\)
= \(\dfrac{\left(-3\right).\left(-4\right).\left(-5\right)...\left(-300\right)}{2.3.4...299}\)
= \(\dfrac{3.4.5...300}{2.3.4.5...299}\)
= \(\dfrac{300}{2}=150\)
C = \(-\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{333333}{424242}\right)\)
= \(-\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{11}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
= \(-\dfrac{7}{4}.\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
= \(-\dfrac{7}{4}.33.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)\)
= \(-\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
= \(-\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
= \(-\dfrac{7}{4}.33.\dfrac{4}{21}\)
= \(-\dfrac{7}{4}.\dfrac{4}{21}.33=-\dfrac{1}{3}.33=-11\)
Tính giá trị biểu thức sau:
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)\cdot...\cdot\left(\dfrac{1}{99}-1\right)\cdot\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).........................\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)\left(\dfrac{1}{4}-\dfrac{4}{4}\right)................\left(\dfrac{1}{99}-\dfrac{99}{99}\right)\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(A=\left(\dfrac{-1}{2}\right)\left(\dfrac{-2}{3}\right)\left(\dfrac{-3}{4}\right)...................\left(\dfrac{-98}{99}\right)\left(\dfrac{-99}{100}\right)\)
\(A=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right).........................\left(-98\right)\left(-99\right)}{2.3.4....................98.99.100}\)
\(A=\dfrac{-1}{100}\)
Ta có
A = \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)(99 thừa số)
A = \(\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
A = \(\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right).\left(-100\right)}{2.3.4....98.99.100}\)
A = \(\dfrac{\left(-1\right).\left(-1\right).\left(-1\right)....\left(-1\right)}{1.1.1...1.1.1}\) (100 số -1, 99 số 1)
A = \(\dfrac{-1}{1.1.1.1...1.1.1}\)
A = \(\dfrac{-1}{1}\)
A = -1
Vậy A = -1
Tính giá trị biểu thức :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}\right)-\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{6}{7}\right)+...+\left(100+...+\dfrac{99}{100}\right)\)
rút gọn biểu thức :
N = 1 + \(\left(\dfrac{1}{2}\right)\) + \(\left(\dfrac{1}{2}\right)^2\) + \(\left(\dfrac{1}{2}\right)^3\) + ... + \(\left(\dfrac{1}{2}\right)^{100}\)
\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)
\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)
\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)
\(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)
\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)
\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)
Tính giá trị biểu thức sau
\(A=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right).....\left(\dfrac{1}{99}\right)+1\)
\(A=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}=\dfrac{100}{2}=50\)
Vậy A = 50
Ta có:
A=\(^{\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).....\left(\dfrac{1}{99}\right)+1}\)
A= \(\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{1}{99}+1\)
A=\(\dfrac{1}{2}+1\)
A=\(\dfrac{3}{2}\)
Giải:
\(A=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right).....\left(\dfrac{1}{99}+1\right).\)
\(A=\left(\dfrac{1}{2}+\dfrac{2}{2}\right)\left(\dfrac{1}{3}+\dfrac{3}{3}\right)\left(\dfrac{1}{4}+\dfrac{4}{4}\right).....\left(\dfrac{1}{99}+\dfrac{99}{99}\right).\)
\(A=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{100}{99}.\)
\(A=\dfrac{3.4.5.....100}{2.3.4.....99}.\)
\(A=\dfrac{100}{2}.\) (dùng tính chất rút gọn của phân số).
\(A=50.\)
Vậy \(A=50.\)
~ Học tốt!!! ... ~
Rút gọn biểu thức :
a) \(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{4}\right).\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)
b) \(\left(10+1\right).\left(10^2+1\right)\left(10^3+1\right)...\left(10^{2n}+1\right)\)
a, \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{4}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{16}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
...
\(=\left(1-\dfrac{1}{2^{2n}}\right)\left(1+\dfrac{1}{2^{2n}}\right).2=\left(1-\dfrac{1}{2^{4n}}\right).2=2-\dfrac{1}{2^{4n-1}}\)
Vậy ...
b,Sửa đề: \(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
Ta có:\(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
\(=\left(10-1\right).\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^2-1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^4-1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
...
\(=\left(10^{2n}-1\right)\left(10^{2n}+1\right).\dfrac{1}{9}=\left(10^{4n}-1\right).\dfrac{1}{9}=\dfrac{10^{4n}}{9}-\dfrac{1}{9}\)
Vậy ...