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Tìm x :
a) \(x-5=49:7\)
\(\Leftrightarrow x-5=7\)
\(\Leftrightarrow x=7+5\)
\(\Leftrightarrow x=12\)
Vậy : \(x=12\)
b) \(2x+6=24\)
\(\Leftrightarrow2x=24-6=18\)
\(\Leftrightarrow x=18:2\)
\(\Leftrightarrow x=9\)
Vậy : \(x=9\)
c) \(\frac{1}{3}:x+\frac{1}{2}=5\)
\(\Leftrightarrow\frac{1}{3}:x=5-\frac{1}{2}=\frac{9}{5}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{5}\)
\(\Leftrightarrow x=\frac{5}{27}\)
Vậy : \(x=\frac{5}{27}\)
d) \(\frac{1}{6}.x-\frac{1}{3}=2\)
\(\Leftrightarrow\frac{1}{6}.x=2-\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}:\frac{1}{6}\)
\(\Leftrightarrow x=10\)
Vậy : \(x=10\)
e) \(\frac{x}{27}=\frac{3}{x}\)
\(\Leftrightarrow x.x=27.3\)
\(\Leftrightarrow x^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\) mà \(x\in N\)
\(\Rightarrow x=9\)
Vậy : \(x=9\)
g) \(1200:24-\left(17-x\right)=36\)
\(\Leftrightarrow50-17+x=36\)
\(\Leftrightarrow33+x=36\)
\(\Leftrightarrow x=36-33\)
\(\Leftrightarrow x=3\)
Vậy : \(x=3\)
h) \(674-\left(12+x\right)=427\)
\(\Leftrightarrow12+x=674-427=247\)
\(\Leftrightarrow x=247-12\)
\(\Leftrightarrow x=230\)
Vậy : \(x=230\)
k) \(36.\left(x-9\right)=900\)
\(\Leftrightarrow x-9=900:36\)
\(\Leftrightarrow x-9=25\)
\(\Leftrightarrow x=25+9\)
\(\Leftrightarrow x=34\)
m) \(1,2:x+3,8:x=2,5\)
\(\Leftrightarrow\left(1,2-3,8\right):x=2,5\)
\(\Leftrightarrow-2,6:x=2,5\)
\(\Leftrightarrow x=\frac{-2,6}{2,5}=-\frac{26}{25}\)
Vậy : \(x=-\frac{26}{25}\)
n) \(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\frac{9}{10}.x=\frac{1}{3}\)
\(\Leftrightarrow\frac{9}{20}.x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{20}\)
\(\Leftrightarrow x=\frac{20}{27}\)
Vậy : \(x=\frac{20}{27}\)
\(A=\frac{17}{23}\cdot\frac{8}{16}\cdot\frac{23}{17}\cdot\left(-80\right)\cdot\frac{3}{4}\)\(=\frac{17\cdot4\cdot2\cdot23\cdot16\cdot\left(-5\right)\cdot3}{23\cdot16\cdot17\cdot4}\)
=> \(A=\frac{2\cdot\left(-5\right)\cdot3}{1}=-30\)
\(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)
=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{7}{12}-\frac{7}{12}\right)\)
=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\cdot0=0\)
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)
\(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)
\(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)
\(=100.\frac{2}{101}\)\(=\frac{200}{101}\)
\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)
\(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)
\(=\frac{1}{1994}\) (Giản ước còn lại như này)
Ta có:
\(B=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{1999}\right).\left(1+\frac{1}{2000}\right)\)
\(=\frac{3}{2}.\frac{4}{3}...\frac{2000}{1999}.\frac{2001}{2000}=\frac{3.4....2000.2001}{2.3...1999.2000}=\frac{2001}{2}\)(Tối giản)(Chắc zậy)
Xét vế phải : \(\frac{1}{m+1}+\frac{a.\left(m+1\right)-b}{b.\left(m+1\right)}=\frac{1}{m+1}+\frac{a\left(m+1\right)}{b\left(m+1\right)}-\frac{b}{b\left(m+1\right)}=\frac{1}{m+1}+\frac{a}{b}-\frac{1}{m+1}=\frac{a}{b}\) = vế trái
Vậy suy ra đpcm