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\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-...-\frac{1}{6}-\frac{1}{2}\)
\(-B=\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\)
\(-B=\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(-B=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+...+\frac{1}{2}-1\)
\(-B=\frac{1}{10}-1\)
\(-B=\frac{9}{10}\)
=> \(B=\frac{-9}{10}\)
\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}\)
\(=-\frac{79}{90}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1}{4}\)
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
Cách 1:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{30}{6}+\frac{10}{6}-\frac{9}{6}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)
\(=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
\(=-\frac{15}{6}\)
\(=-\frac{5}{2}\)
Cách 2:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2+0-\frac{1}{2}\)
\(=-\frac{4}{2}-\frac{1}{2}\)
\(=-\frac{5}{2}\)
= - ( 1/2 +1/6+1/12+1/20+ 1/30+ 1/42+ 1/56+ 1/72+ 1/90)
= - ( 1 - 1/2 + 1/2 -1/3 +1/3 -1/4 +1/4 - 1/5 +1/5 -1/6 +1/6 -1/7 +1/7 -1/8 +1/8 -1/9 +1/9 -1/10)
= - ( 1- 1/10 )
= -9/10
\(-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\left(-\frac{1}{90}\right)+\left(-\frac{1}{72}\right)+\left(-\frac{1}{56}\right)+......+\left(-\frac{1}{2}\right)\)
\(=\left(-1\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{9.10}\right)\)
\(=\left(-1\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\left(-1\right).\left(1-\frac{1}{90}\right)=\frac{\left(-1\right).89}{90}=-\frac{89}{90}\)
nha
C1: dễ nên tự làm nhé
C2: \(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{2}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=6-5-3+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)\)
\(=-2-\frac{1}{2}=\frac{-4}{2}-\frac{1}{2}=\frac{-5}{2}\)
A = \(\frac{4\left(\frac{1}{78}-\frac{1}{60}-\frac{1}{306}\right)}{9\left(\frac{1}{78}-\frac{1}{60}-\frac{1}{306}\right)}:\frac{1+\frac{2}{71}-\frac{5}{121}}{-13\left(1+\frac{2}{71}-\frac{5}{121}\right)}=\frac{4}{9}:-\frac{1}{13}\)
Đoán thôi
Cách 1:
A = \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
A = \(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
A = \(\frac{-15}{6}=\frac{-5}{2}\)
Cách 2:
A = \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
A = \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
A = \(6-5-3-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}+\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\)
A = \(\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}-\frac{3}{2}-\frac{5}{2}\right)\)
A = \(-2-0+\left(2-\frac{5}{2}\right)\)
A = \(-2+\left(2-\frac{5}{2}\right)\)
A = \(-2+2-\frac{5}{2}\)
A = \(0-\frac{5}{2}\)
A = \(\frac{-5}{2}\)
Cách 1:A=(6−2/3+1/2)−(5+5/3−3/2)−(3−7/3+5/2)
= ( 36/6 - 4/6 + 3/6)-(30/6 + 10/6 - 9/6) -(18/6 -14/6 +15/6)
= 35/6 - 31/6 - 19/6
= -15/2
Cách 2: A=(6−2/3+1/2)−(5+5/3−3/2)−(3−7/3+5/2)
= 6- 2/3 +1/2 - 5 - 5/3 + 3/2 - 3 + 7/3 -5/2
= (6-5-3) + ( -2/3-5/3+7/3) + (1/2+3/2-5/2)
= -2 + 0 - 1/2
= -2 - 1/2 = -4/2 - 1/2 = -5/2
Gọi biểu thức trên là : A
\(A=2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}\)
\(A=2+\frac{1}{2+\frac{1}{2+\frac{1}{\frac{5}{4}}}}\)
\(A=2+\frac{1}{2+\frac{1}{2+\frac{4}{5}}}\)
\(A=2+\frac{1}{2+\frac{1}{\frac{14}{5}}}\)
\(A=2+\frac{1}{2+\frac{5}{14}}\)
\(A=2+\frac{1}{\frac{33}{14}}\)
\(A=2+\frac{14}{33}\)
\(A=\frac{80}{33}\)
Vậy : \(2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}=\frac{80}{33}\)
viết k hiểu bn ạ