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 $x=28$

\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=4.\left(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{399}\right)=4.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{19.21}\right)=4.\left[\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\right]=4.\left[\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\right]=2.\left(\dfrac{7-1}{21}\right)=\dfrac{12}{21}=\dfrac{4}{7}\)

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:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$

$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$

$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$

$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$

$\frac{1}{2}(1-\frac{1}{21})=-x$

$\frac{10}{21}=-x$

$\Rightarrow x=\frac{-10}{21}$

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:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$

$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$

$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$

$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$

$\frac{1}{2}(1-\frac{1}{21})=-x$

$\frac{10}{21}=-x$

$\Rightarrow x=\frac{-10}{21}$

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. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)

<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)

<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)

<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)

<=> \(22968x=8199576\)

<=> \(x=\dfrac{8199576}{22968}\)

\(x=357\)

Đề là giải PT nha các bn

`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`

$A=2/15+2/35+2/63+...+2/339$

$=2/(3.5)+2/(5.7)+2/(7.9)+...+2/(19.21)$

$=1/3-1/5+1/5-1/7+1/7-1/9+...1/19-1/21$

$=1/3-1/21=7/21-1/21$

$=6/21=2/7$

sai rồi,2/7 mới đúng,bài này không cần nhân 2

\(a,\dfrac{x}{4}=\dfrac{15}{20}\\ \Rightarrow\dfrac{x}{4}=\dfrac{3}{4}\\ \Rightarrow x=3\\ b,\dfrac{15}{x}=\dfrac{25}{35}\\ \Rightarrow\dfrac{15}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{15}{x}=\dfrac{15}{21}\\ \Rightarrow x=21\\ c,\dfrac{x}{5}=\dfrac{26}{65}\\ \Rightarrow\dfrac{x}{5}=\dfrac{2}{5}\\ \Rightarrow x=2\\ d,\dfrac{3}{x}=\dfrac{51}{85}\\ \Rightarrow\dfrac{3}{x}=\dfrac{3}{5}\\ \Rightarrow x=5\)

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

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 \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)