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\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)
\(a,\dfrac{5}{8}=\dfrac{x}{14}\)
\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)
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
\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x}{y}=\dfrac{3}{7}\)
\(\dfrac{x}{y}-1=\dfrac{-5}{19}\Rightarrow\dfrac{x}{y}=\dfrac{14}{19}\)
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: \(4< x\le10\)
\(\Rightarrow x\in\left\{6;9\right\}\)
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`
\(a,3\cdot x-15=x+35\)
\(\Rightarrow3x-x=35+15\)
\(\Rightarrow 2x=50\)
\(\Rightarrow x = 50:2\)
\(\Rightarrow x= 25\)
\(b,(8x-16)(x-5)=0\)
\(+, TH1: 8x-16=0\)
\(\Rightarrow8x=16\)
\(\Rightarrow x = 16:8\)
\(\Rightarrow x=2\)
\(+,TH2: x-5=0\)
\(\Rightarrow x =5\)
\(c,x(x+1)=2+4+6+8+10+...+2500\) \(^{\left(1\right)}\)
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:
\(x\left(x+1\right)=1563750\)
\(\Rightarrow x\left(x+1\right)=1250\cdot1251\)
\(\Rightarrow x =1250\)
#\(Toru\)
2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
a) \(\dfrac{x}{4}\)=\(\dfrac{1}{x}\)→x2=4→x=2 hoặc x=-2
b) \(\dfrac{1}{5}\) = x:4-\(\dfrac{1}{10}\)→x:4=\(\dfrac{1}{5}\)+\(\dfrac{1}{10}\)→x:4=\(\dfrac{2+1}{10}\)→x:4=\(\dfrac{3}{10}\)→x=\(\dfrac{3}{10}\)x4→x=\(\dfrac{12}{10}\)→x=\(\dfrac{6}{5}\)