a x=4

\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)

`a, 8x=64`

`=>x= 64:8`

`=> x=8`

`b, (-5)x=25`

`=>x=25:(-5)`

`=>x=-5`

`c,4x+1=21`

`=>4x=21-1`

`=>4x=20`

`=>x=20:4`

`=>x=5`

`d, (-3)x-1=8`

`=>(-3)x=8+1`

`=>(-3)x=9`

`=>x=9:(-3)`

`=>x=(-3)`

\(a,\dfrac{5}{8}=\dfrac{x}{14}\)

\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)

Vậy \(x=8,75\)

\(b,\dfrac{x}{6}=-\dfrac{1}{3}\)

\(\Rightarrow x=-\dfrac{1.6}{3}=-2\)

Vậy \(x=-2\)

\(c,-\dfrac{3}{5}=\dfrac{x}{10}\)

\(\Rightarrow x=-\dfrac{3.10}{5}=-6\)

Vậy \(x=-6\)

câu d đã có đáp án

mik đang cần gấp ak

\(+\text{)}\dfrac{5}{8}=14x\)⇒\(x=\dfrac{5}{112}\)

\(+\text{)}\dfrac{x}{6}=-\dfrac{1}{3}\)⇒\(x=-2\)

\(+\text{)}-\dfrac{3}{5}=\dfrac{x}{10}\)⇒\(x=-6\)

\(+\text{)}\dfrac{3}{5}=-\dfrac{9}{11}\)⇒\(\dfrac{33}{55}=\dfrac{45}{55}\)

\(+\text{)}\dfrac{x}{2}=\dfrac{2}{x}\)⇒\(x=+-2\)

\(+\text{)}\dfrac{x}{-5}=\dfrac{-5}{x}\)⇒\(x=+-5\)

\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x}{y}=\dfrac{3}{7}\)

\(\dfrac{x}{y}-1=\dfrac{-5}{19}\Rightarrow\dfrac{x}{y}=\dfrac{14}{19}\)

Vô lí => không có x,y thỏa mãn

a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}\)

nên \(\dfrac{x}{y}=\dfrac{3}{7}\)

b) Ta có: \(\dfrac{x}{y-1}=\dfrac{5}{-19}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y-1}{-19}\)

hay \(\dfrac{x}{5}=\dfrac{1-y}{19}\)

Bài 3: 

a chia 36 dư 12 số đó có dạng \(a=36k+12\left(k\in N\right)\)

\(\Rightarrow a=4\left(9k+3\right)\) nên a chia hết cho 4

Mà: \(9k\) ⋮ 3 ⇒ \(9k+3\) không chia hết cho 3

Nên a không chia hết cho 3 

Bài 4:

a) \(x\in B\left(7\right)\) \(\Rightarrow x\in\left\{0;7;14;21;28;35;42;49;...\right\}\)

Mà: \(x\le35\)

\(\Rightarrow x\in\left\{0;7;14;21;28;35\right\}\)

b) \(x\inƯ\left(18\right)\Rightarrow x\in\left\{1;2;3;6;9;18\right\}\)

Mà: \(4< x\le10\)

\(\Rightarrow x\in\left\{6;9\right\}\)

Bài 5:

a) 6 chia hết cho x 

\(\Rightarrow x\inƯ\left(6\right)\)

\(\Rightarrow x\in\left\{1;2;3;6\right\}\)  

b) \(8\) chia hết cho \(x+1\)

\(\Rightarrow x+1\inƯ\left(8\right)\)

\(\Rightarrow x+1\in\left\{1;2;4;8\right\}\)

\(\Rightarrow x\in\left\{0;1;3;7\right\}\)

c) 10 chia hết cho \(x-2\)

\(\Rightarrow x-2\inƯ\left(10\right)\)

\(\Rightarrow x-2\in\left\{1;2;5;10\right\}\)

\(\Rightarrow x\in\left\{3;4;7;12\right\}\)

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

` 8/23 . 46/24 =1/3 .x`

`=>8/23 . 23/12 =1/3 . x`

`=> 1/3 . x=2/3`

`=>x=2/3 : 1/3`

`=>x=2/3 . 3`

`=> x= 6/3`

`=>x=2`

`----`

`1/5 : x= 1/5-1/7`

`=>1/5 : x=  7/35 - 5/35`

`=> 1/5 :x= 2/35`

`=>x= 1/5 : 2/35`

`=>x=1/5 . 35/2`

`=>x=7/2`

`----`

`4/9 - (x-1/2)^2 =1/3`

`=> (x-1/2)^2 =4/9-1/3`

`=> (x-1/2)^2 =4/9- 3/9`

`=> (x-1/2)^2 =1/9`

`=> (x-1/2)^2 = (+- 1/3)^2`

`@ TH1`

`x-1/2=1/3`

`=>x=1/3+1/2`

`=>x= 2/6 + 3/6`

``=>x= 5/6`

`@ TH2`

`x-1/2=-1/3`

`=>x=-1/3 +1/2`

`=>x= -2/6 + 3/6`

`=>x=1/6`

`----`

`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`

`=> 3,2 . x-22/15 : 11/3 = 7/10`

`=>  3,2 . x-22/15 = 7/10 . 11/3`

`=>  3,2 . x-22/15 =77/30`

`=> 3,2 .x= 77/30 + 22/15`

`=> 3,2 .x=121/30`

`=>x= 121/30. 5/16`

`=>x= 121/96`

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)

\(a,3\cdot x-15=x+35\)

\(\Rightarrow3x-x=35+15\)

\(\Rightarrow 2x=50\)

\(\Rightarrow x = 50:2\)

\(\Rightarrow x= 25\)

\(b,(8x-16)(x-5)=0\)

\(+, TH1: 8x-16=0\)

\(\Rightarrow8x=16\)

\(\Rightarrow x = 16:8\)

\(\Rightarrow x=2\)

\(+,TH2: x-5=0\)

\(\Rightarrow x =5\)

\(c,x(x+1)=2+4+6+8+10+...+2500\)  \(^{\left(1\right)}\)

Đặt \(A=2+4+6+8+10+...+2500\)

Số các số hạng của \(A\) là: \(\left(2500-2\right):2+1=1250\left(số\right)\)

Tổng \(A\) bằng: \(\left(2500+2\right)\cdot1250:2=1563750\)

Thay \(A=1563750\) vào \(^{\left(1\right)}\), ta được:

\(x\left(x+1\right)=1563750\)

\(\Rightarrow x\left(x+1\right)=1250\cdot1251\)

\(\Rightarrow x =1250\)

#\(Toru\)