Tìm x :
\(\dfrac{4}{15}\) + \(\dfrac{4}{35}\) + \(\dfrac{4}{63}\) + ... + \(\dfrac{4}{399}\) = \(\dfrac{x}{49}\)
K
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KM
2
15 tháng 5 2021
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!
15 tháng 5 2021
\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
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 \(x=21\)
C
24 tháng 9 2017
\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=4.\left(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{399}\right)=4.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{19.21}\right)=4.\left[\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\right]=4.\left[\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\right]=2.\left(\dfrac{7-1}{21}\right)=\dfrac{12}{21}=\dfrac{4}{7}\)
AH
Akai Haruma
Giáo viên
30 tháng 4 2022
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
AH
Akai Haruma
Giáo viên
30 tháng 4 2022
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
HH
4 tháng 2 2018
a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)
<=> \(25x+10-80x+10=24x+12-30\)
<=> \(25x-80x-24x=12-30-10-10\)
<=> \(-79x=-38\)
<=> \(x=\dfrac{-38}{-79}\)
\(x=\dfrac{38}{79}\)
b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)
<=> \(30x-12x+30+5x+40=210+10x-10\)
<=> \(30x-12x+5x-10x=210-10-30-40\)
<=> \(13x=130\)
<=> \(x=\dfrac{130}{13}\)
\(x=10\)
c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)
<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)
<=> \(28x+28+60x+120+105x+420+2520=0\)
<=> \(28x+60x+105x=-28-120-420-2520\)
<=> \(193x=-3088\)
<=> \(x=\dfrac{-3088}{193}\)
\(x=-16\)
d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)
<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)
<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)
<=> \(22968x=8199576\)
<=> \(x=\dfrac{8199576}{22968}\)
\(x=357\)
TL
3
TC
13 tháng 3 2023
`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`
16 tháng 2 2022
\(a,\dfrac{x}{4}=\dfrac{15}{20}\\ \Rightarrow\dfrac{x}{4}=\dfrac{3}{4}\\ \Rightarrow x=3\\ b,\dfrac{15}{x}=\dfrac{25}{35}\\ \Rightarrow\dfrac{15}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{15}{x}=\dfrac{15}{21}\\ \Rightarrow x=21\\ c,\dfrac{x}{5}=\dfrac{26}{65}\\ \Rightarrow\dfrac{x}{5}=\dfrac{2}{5}\\ \Rightarrow x=2\\ d,\dfrac{3}{x}=\dfrac{51}{85}\\ \Rightarrow\dfrac{3}{x}=\dfrac{3}{5}\\ \Rightarrow x=5\)
19 tháng 3 2022
a,x4=1520⇒x4=34⇒x=3b,15x=2535⇒15x=57⇒15x=1521⇒x=21c,x5=2665⇒x5=25⇒x=2d,3x=5185⇒3x=35⇒x=5
8 tháng 8 2018
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) \(1-\left\{5\dfrac{4}{9}+x-7\dfrac{7}{18}\right\}:15\dfrac{3}{5}=0\)
\(\Leftrightarrow1-\left\{\dfrac{49}{9}+x-\dfrac{133}{18}\right\}:\dfrac{78}{5}=0\)
\(\Leftrightarrow\left\{\dfrac{-35}{18}+x\right\}:\dfrac{78}{5}=1-0\)
\(\Rightarrow\dfrac{-35}{18}+x=1.\dfrac{78}{5}\)
\(\Leftrightarrow\dfrac{-35}{18}+x=\dfrac{78}{5}\)
\(\Rightarrow x=\dfrac{1579}{90}\).
TT
8 tháng 8 2018
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...
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 $x=28$