\(1+\frac{1}{1+\frac{1}{2}}+\frac{1}{1+\frac{1}{2}+\frac{1}{2}}+\left(\frac{1}{2}\right)^2-1\)
GIÚP MK VS Ạ
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\(\Leftrightarrow2.\left(\frac{-1}{2}\right).\left(\frac{2}{3}\right)^2-3\left(-\frac{1}{3}\right)^2.\frac{2}{9}:x=3.\left(-\frac{1}{2}\right)-\frac{2}{3}\)
\(\Leftrightarrow-\frac{4}{9}-\frac{1}{3}.\frac{2}{9}:x=-\frac{3}{2}-\frac{2}{3}\)
\(\Leftrightarrow-\frac{4}{6}-\frac{2}{27}:x=-\frac{13}{6}\)
\(\Leftrightarrow\frac{2}{27}:x=-\frac{4}{9}:\frac{-13}{6}\)
\(\Leftrightarrow\frac{2}{27}:x=\frac{31}{18}\)
\(\Leftrightarrow x=\frac{2}{27}:\frac{31}{18}\)
\(\Rightarrow x=\frac{4}{93}\)
Vậy \(x=\frac{4}{93}\)
a) \(\frac{\left(5.2\right)}{3.2}-\frac{1}{2}x+\frac{1}{3}+\frac{1}{5}=\frac{\left(3.2\right)}{5}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{2}x+\frac{8}{15}=\frac{6}{5}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{2}{3}=\frac{1}{2}x\)
\(\Leftrightarrow\)\(-\frac{1}{6}=\frac{1}{2}x\)
\(\Leftrightarrow\)x=-1/3
b) VT= \(\frac{\left(3.5.4.2\right)}{5.2.3}=4\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right):6+4=4:\frac{2}{3}=6\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right):6=2\)
\(\Leftrightarrow x-\frac{1}{2}=12\)
=> x= 12,5
cho 3 k
\(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{10^2}\right)\)
=> \(\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)\)\(...\left(1-\frac{1}{10}\right)\cdot\left(1+\frac{1}{10}\right)\)
=> \(\left(1-\frac{1}{2}\right)\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\cdot\cdot\frac{9}{10}\cdot\frac{10}{11}\)
=> \(\frac{1}{2}\cdot\frac{3\cdot2\cdot4\cdot\cdot\cdot9\cdot10}{2\cdot3\cdot3\cdot\cdot\cdot10\cdot11}=\frac{1}{2}\cdot\frac{11}{10}=\frac{11}{20}\)
Chúc bn học tốt !
cho mk 3 k nha bn
thanks nhìu
bài này mk ko copy, ko chép mạng, tự nghĩ mất 6 phút .
có công thức rùi nha !
chúc bn học tốt
\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)
\(=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)
\(=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)
\(=-4-\frac{1}{2}\)
\(=-\frac{9}{2}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)
\(A=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)
\(A=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)
\(A=-4+\frac{1}{2}-\frac{4}{3}\)
\(A=-\frac{29}{6}\)
\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(\frac{x^2+1-\left(x+1\right)}{x+1}\right)\left(\frac{4x-2\left(x-1\right)}{x\left(x-1\right)}\right)\)
\(=\left(\frac{x^2+1-x-1}{x+1}\right)\left(\frac{4x-2x+2}{x\left(x-1\right)}\right)=\left(\frac{x^2-x}{x+1}\right)\left(\frac{2\left(x+1\right)}{x\left(x-1\right)}\right)=\frac{2x\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}=2\)
Vậy Q = 2
Hình như đề là rút gọn thì phải.
Giải
\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\left(\frac{x^2}{x}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(x-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\frac{4\left(x-1\right)}{x-1}-\frac{2\left(x-1\right)}{x}=4-\frac{2x-2}{x}\)
=\(1+\frac{1}{\frac{3}{2}}+\frac{1}{2}+\frac{1}{4}-1\)
=\(\frac{2}{3}+\frac{1}{2}+\frac{1}{4}\)
=\(\frac{17}{12}\)