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Ta có : \(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
\(\Leftrightarrow2\cdot\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{x}{98}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{x}{98}\)
\(\Leftrightarrow\dfrac{2}{7}=\dfrac{x}{98}\Rightarrow x=28\)
Vậy $x=28$
`A =2/15 +2/35 +2/63 +... +2/339`
`= 2/(3.5) +2/(5.7) + 2/(7.9) + ...+2/(19.21)`
`= 1/3 -1/5 +1/5 -1/7 +1/7 -1/9 +... 1/19 -1/21`
`= 1/3 -1/21 = 7/21 -1/21`
`=6/21 = 2/7`
1) \(x:\dfrac{2}{3}=150\)
\(\Leftrightarrow x=150.\dfrac{2}{3}\)
\(\Leftrightarrow x=100\).
2) \(\dfrac{35}{9}:x=\dfrac{35}{6}\)
\(\Leftrightarrow x=\dfrac{35}{9}:\dfrac{35}{6}\)
\(\Leftrightarrow x=\dfrac{2}{3}\).
3) \(\dfrac{49}{7}:x=\dfrac{49}{5}\)
\(\Leftrightarrow x=\dfrac{49}{7}:\dfrac{49}{5}\)
\(\Leftrightarrow x=\dfrac{5}{7}\).
4) \(1-\left\{5\dfrac{4}{9}+x-7\dfrac{7}{18}\right\}:15\dfrac{3}{5}=0\)
\(\Leftrightarrow1-\left\{\dfrac{49}{9}+x-\dfrac{133}{18}\right\}:\dfrac{78}{5}=0\)
\(\Leftrightarrow\left\{\dfrac{-35}{18}+x\right\}:\dfrac{78}{5}=1-0\)
\(\Rightarrow\dfrac{-35}{18}+x=1.\dfrac{78}{5}\)
\(\Leftrightarrow\dfrac{-35}{18}+x=\dfrac{78}{5}\)
\(\Rightarrow x=\dfrac{1579}{90}\).
Gọi a,b,c.. cho dễ nhé.Thớt vui tính quá, dấu phẩy cũng không viết hộ con dân =)))
a, \(x:\dfrac{2}{3}=150\)
\(\Leftrightarrow x=150.\dfrac{2}{3}\)
\(\Leftrightarrow x=100\)
Vậy...
b, \(\dfrac{35}{9}:x=\dfrac{35}{6}\)
\(\Leftrightarrow x=\dfrac{35}{9}:\dfrac{35}{6}\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy...
c, \(\dfrac{49}{7}:x=\dfrac{49}{5}\)
\(\Leftrightarrow x=\dfrac{49}{7}:\dfrac{49}{5}\)
\(\Leftrightarrow x=\dfrac{5}{7}\) Vậy...
d, \(1-\left\{5\dfrac{4}{9}+x-7\dfrac{7}{18}\right\}:15\dfrac{3}{4}=0\)
\(\Leftrightarrow1-\left\{\dfrac{49}{9}+x-\dfrac{133}{18}\right\}:\dfrac{63}{4}=0\)
\(\Leftrightarrow1-\left\{x-\dfrac{35}{18}\right\}:\dfrac{63}{4}=0\)
\(\Leftrightarrow1-\left(\dfrac{\left(18x-35\right).4}{18.63}\right)=0\)
\(\Leftrightarrow1-\left(\dfrac{72x-140}{1134}\right)=0\)
\(\Leftrightarrow1-\dfrac{72x-140}{1134}=0\)
\(\Leftrightarrow\dfrac{1134-72x+140}{1134}=0\)
\(\Leftrightarrow1274-72x=0\)
\(\Leftrightarrow72x=1274\)
\(\Leftrightarrow x=\dfrac{637}{36}\)
Vậy...
a: =>4/x=y/-21=4/7
=>x=7; y=-12
b: =>xy=63
mà x>y
nên \(\left(x,y\right)\in\left\{\left(9;7\right);\left(21;3\right);\left(63;1\right);\left(-7;-9\right);\left(-3;-21\right);\left(-1;-63\right)\right\}\)
c: =>xy=45
mà x<y<0
nên \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
Giải:
\(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{19.21}=\dfrac{x}{49}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)
\(2.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=\dfrac{x}{49}\)
\(2.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{x}{49}\)
\(2.\dfrac{2}{7}=\dfrac{x}{49}\)
\(\dfrac{4}{7}=\dfrac{x}{49}\)
\(\Rightarrow x=\dfrac{4.49}{7}=28\)
Chúc bạn học tốt!
\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
2 . \(\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{399}=\dfrac{x}{49}\)
2 . \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{19.21}=\dfrac{x}{49}\)
2 . ( \(\dfrac{1}{3}-\dfrac{1}{21}\) ) = \(\dfrac{x}{49}\)
2 . \(\dfrac{2}{7}\) = \(\dfrac{x}{49}\)
=> \(\dfrac{4}{7}=\dfrac{x}{49}\)
=> \(\dfrac{21}{49}=\dfrac{x}{49}\)
=> \(x=21\)
Vậy \(x=21\)