a)\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)

\(\dfrac{x}{3}=2\Rightarrow x=6\)

\(\dfrac{y}{7}=2\Rightarrow y=14\)

b)\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{5}=2\Rightarrow x=10\)

\(\dfrac{y}{2}=2\Rightarrow y=4\)

Bài 2:

\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)

\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)

\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)

a/ \(\dfrac{x}{7}=\dfrac{18}{14}\)

\(\Leftrightarrow14x=7.18\)

\(\Leftrightarrow14x=126\)

\(\Leftrightarrow x=9\)

Vậy ....

b/ \(6:x=1\dfrac{3}{4}:5\)

\(\Leftrightarrow6:x=\dfrac{7}{20}\)

\(\Leftrightarrow x=\dfrac{120}{7}\)

Vậy ...........

bao giờ bạn cần vậy

nói đi mik giải giùm cho nha!

\(\dfrac{x}{7}=\dfrac{18}{14}\)

\(\Rightarrow x=\dfrac{18.7}{14}=9\)

Vậy x=4

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

a)\(\dfrac{x}{7}=\dfrac{18}{14}\)

\(\Rightarrow x=\dfrac{7.18}{14}=9\)

b)\(6:x=1\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{6}{x}=\dfrac{7}{4}\)

\(\Rightarrow x=\dfrac{6.4}{7}=\dfrac{24}{7}\)

c)5,7:0,35=(-x):0,45

\(\Leftrightarrow\dfrac{114}{7}=\dfrac{-x}{0,45}\)

\(\Rightarrow\left(-x\right)=\dfrac{114.0,45}{7}=\dfrac{-513}{70}\)

\(a,x=\dfrac{18.7}{14}\)

\(x=\dfrac{8}{2}=4\)

\(b,\)\(x=6.5:1\dfrac{3}{4}\)

\(x=30:\dfrac{7}{4}\)

\(x=30.\dfrac{4}{7}=\dfrac{120}{7}\)

\(c,-x=\)\(\dfrac{5,7.0,45}{0,35}\)

\(-x=\)\(\dfrac{2,565}{0,35}\)

\(x=\dfrac{-2,565}{0,35}\)

a)\(\dfrac{x}{7}=\dfrac{18}{14}\) b) 6:x=\(1\dfrac{3}{4}\):5 c) 5,7 : 0,35= -x: 0,45

\(\Rightarrow\) x.14= 18.7 \(\Rightarrow\) 6:x=\(\dfrac{7}{20}\) \(\Rightarrow\)\(\dfrac{114}{7}=\dfrac{-x}{0,45}\)

\(\Rightarrow\)x.14=126 \(\Rightarrow\)x=\(\dfrac{10}{21}\) \(\Rightarrow53,1=\left(-x\right).7\)

\(\Rightarrow\)x=9 \(\Rightarrow x=7,3\)

CHÚC BẠN HỌC TỐT

a, \(\dfrac{x}{7}\)=\(\dfrac{18}{14}\)

\(x.14=7.18\)

\(x.14=126\)

\(\Rightarrow x=9\)

b,\(6:x=1^3_4:5\)

\(6:x=\dfrac{7}{20}\)

\(x=\dfrac{7}{120}\)

C

a) \(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\)và \(x+y-z=69\)

Theo đề bài, ta có:

\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{5}\times\dfrac{1}{8}=\dfrac{y}{6}\times\dfrac{1}{8}\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}\)(1)

\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{8}\times\dfrac{1}{6}=\dfrac{z}{7}\times\dfrac{1}{6}\Rightarrow\dfrac{y}{48}=\dfrac{z}{42}\)(2)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{40}=\dfrac{3}{2}\Rightarrow x=\dfrac{40\times3}{2}=60\\\dfrac{y}{48}=\dfrac{3}{2}\Rightarrow y=\dfrac{48\times3}{2}=72\\\dfrac{z}{42}=\dfrac{3}{2}\Rightarrow z=\dfrac{42\times3}{2}=63\end{matrix}\right.\)

Vậy \(\Rightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)

Ta có:\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)(Nhân 2 vế với \(\dfrac{1}{4}\))

\(\dfrac{y}{8}=\dfrac{x}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(Nhân 2 vế với \(\dfrac{1}{3}\))

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)và x+y-z=6

Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:

\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)

Vì \(\dfrac{x}{20}=3\Rightarrow x=20.3=60\)

\(\dfrac{y}{24}=3\Rightarrow y=24.3=72\)

\(\dfrac{z}{21}=3\Rightarrow z=3.21=63\)

Vậy x=60; y=72; z=63

a) \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)và \(2x-3y+z=6\)

Theo đề bài, ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{3}\times\dfrac{1}{3}=\dfrac{y}{4}\times\dfrac{1}{3}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\)(1)

\(\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{3}\times\dfrac{1}{4}=\dfrac{z}{5}\times\dfrac{1}{4}\Rightarrow\dfrac{y}{12}=\dfrac{z}{20}\)(2)

Từ (1) và (2), ta có: \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

\(\Leftrightarrow\dfrac{x}{9}\Rightarrow\dfrac{2x}{18};\dfrac{y}{12}\Rightarrow\dfrac{3y}{36}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x}{18}=3\Rightarrow x=\dfrac{18\times3}{2}=27\\\dfrac{3y}{36}=3\Rightarrow y=\dfrac{36\times3}{3}=36\\\dfrac{z}{20}=3\Rightarrow z=20\times3=60\end{matrix}\right.\)

Vậy \(\Rightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)

a) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/3 = y/4 = x/3 + y/4 = 28/7 = 4

=> x = 4 × 3 = 12

=> y = 4 × 4 = 16

Vậy x = 12, y = 16

B) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/2 = y/(-5) = x/2 - y/(-5) = (-7)/7 = -1

=> x = -1 × 2 = -2

=> y = -1 × -5 = 5

Vậy x = -2, y = 5

C) làm tương tự như bài a, b

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: