$\frac{1}{12}+\frac{1}{22}+\frac{1}{32}$+\frac{1}{42}+\frac{1}{52}$
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\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=1-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1-\left(1-\frac{1}{10}\right)\)
\(=1-\frac{9}{10}\)
\(=\frac{1}{10}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}...\frac{1}{7x8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)\(-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
b,
a, \(\frac{7}{4x}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)
\(\frac{7}{4x}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=22\)
\(\frac{7}{4x}\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]=22\)
\(\frac{7}{4x}\left[33.\left(\frac{35}{420}+\frac{21}{420}+\frac{14}{420}+\frac{10}{420}\right)\right]=22\)
\(\frac{7}{4x}\left[33.\frac{4}{21}\right]=22\)
\(\frac{7}{4x}.\frac{44}{7}\)=22
\(\frac{11}{x}=22\)
x=11:22
x=\(\frac{1}{2}\)
b,\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right).x=1\)
Đặt A\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
Ta có :\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow4A=4.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)
\(\Rightarrow4A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}\)
\(\Rightarrow4A=\frac{32+16+8+4+2+1}{64}=\frac{63}{64}\)
\(\Rightarrow A=\frac{63}{64}:4=\frac{63}{256}\)
\(\Rightarrow\frac{63}{256}.x=1\)
\(\Leftrightarrow x=1:\frac{63}{256}=\frac{256}{63}\)
A= đã cho.
=>-A=1/20+1/30+1/42+1/56+1/72+1/90+1/110.
=>-A=1/4*5+1/5*6+1/6*7+1/7*8+1/8*9+1/9*10+1/10*11.
=>-A=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11.
=>-A=1/4-1/11=7/44.
=>A=-7/44.
thay số 1/52 là 1/56 mới đúng,mk làm rồi.
tk mk nha các bn.
-chúc ai tk mk học giỏi-
1.\(\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Leftrightarrow\left(x+1\right).4=\left(x-2\right).3\)
\(\Leftrightarrow4x+4=3x-6\)
<=>4x-3x=-6-4
<=>x=-10
2.\(\frac{52}{2x-1}=\frac{13}{30}\)
<=>52.30=(2x-1).13
<=>1560=26x-13
<=>-26x=-13-1560
<=>-26x=-1573
<=>x=60,5
3.\(\frac{2x-3}{x+1}=\frac{4}{7}\)
<=>(2x-3).7=(x+1).4
<=>14x-21=4x+4
<=>14x-4x=4+21
<=>10x=25
<=>x=2,5
4.\(\frac{2x+3}{42}=\frac{3x-1}{32}\)
<=>(2x+3).32=42(3x-1)
<=>64x+96=126x-42
<=>64x-126x=-42-96
<=>-62x=-138
<=>x=69/31