a:=>3x=15

=>x=5

b: =>x+3=0,96

=>x=-2,04

c: =>x^2=36

=>x=6 hoặc x=-6

`a, 3/4=(3x)/20`

`3x*4=3*20`

`3x*4=60`

`3x=60 \div 4`

`3x=15`

`x=15 \div 3`

`x=5`

`b, (1,2)/(x+3)=5/4`

`1,2*4=(x+3)*5`

`4,8=(x+3)*5`

`x+3= 4,8 \div 5`

`x+3=0,96`

`x=0,96-3`

`x=-2,04`

`c, (x^2)/32=9/8`

`x^2*8=32*9`

`x^2*8=288`

`x^2=288 \div 8`

`x^2=36`

`x^2=(+-6)^2`

`-> \text {x= 6 hoặc -6}`

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

a) _{^{\(\dfrac{3}{5}+\dfrac{11}{20}=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)}}

b) _{^{\(\dfrac{5}{8}-\dfrac{4}{9}=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)}}

c) _{^{\(\dfrac{9}{16}\times\dfrac{4}{3}=\dfrac{3}{4}\)}}

d) _{^{\(\dfrac{4}{7}:\dfrac{8}{11}=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)}}

e) _{^{\(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{4}{5}\times\dfrac{5}{2}=\dfrac{3}{5}+2=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)}}

a, 23/20

b, 13/72

c, 3/4

d,11/14

e, 13/5

a)\(=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)

b)\(=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)

c)\(=\dfrac{9\times4}{16\times3}=\dfrac{3}{4}\)

d)\(=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)

e)\(=\dfrac{3}{5}+\dfrac{4}{2}=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)

liên quan

a. 7/8 x 2/5 =7/20

b.9/4-5/6=17/12

c.4/7x2/5=8/35

d.3/5+2=13/5

e. 4-3/5=17/5

g.3x4/9=4/3

h.9/5:2=9/10

a) \(\dfrac{14}{40}=\dfrac{7}{20}\)

b) \(\dfrac{27}{12}-\dfrac{10}{12}=\dfrac{17}{12}\)

c) \(\dfrac{8}{35}\)

d) \(\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)

e) \(\dfrac{20}{5}-\dfrac{3}{5}=\dfrac{17}{5}\)

g) \(\dfrac{12}{9}=\dfrac{4}{3}\)

h) \(\dfrac{9}{5}x\dfrac{1}{2}=\dfrac{9}{10}\)

`@LQuyen`

\(\dfrac{7}{8}\times\dfrac{2}{5}=\dfrac{14}{40}\)

\(\dfrac{9}{4}-\dfrac{5}{6}=\dfrac{54}{24}-\dfrac{20}{24}=\dfrac{34}{24}\)

\(\dfrac{4}{7}\times\dfrac{2}{5}=\dfrac{8}{35}\)

\(\dfrac{3}{5}+\dfrac{2}{1}=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)

\(\dfrac{4}{1}-\dfrac{3}{5}=\dfrac{20}{5}-\dfrac{3}{5}=\dfrac{17}{5}\)

\(3\times\dfrac{4}{9}=\dfrac{12}{9}\)

\(\dfrac{9}{5}:2=\dfrac{9}{10}\)

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

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)

Do đó: x=5; y=5; z=17

\(a,\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Rightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm10\\y=\pm15\\z=\pm20\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)\) có giá trị là hoán vị của \(\left(\pm10;\pm15;\pm20\right)\)

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)

=>\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)

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

\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}=\dfrac{-3x-4y+5z+3-12-25}{-6-16+30}=2\)

=>x-1=4; y+3=8; z-5=12

=>x=5; y=5; z=17

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

=>\(\dfrac{4\left(3x-2y\right)}{4.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)

=>\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

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

\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)

=>\(\dfrac{12x-8y}{16}=0\)

=>12x-8y=0

=>12x=8y

=>\(\dfrac{12x}{24}=\dfrac{8y}{24}\)

=>\(\dfrac{x}{2}=\dfrac{y}{3}\)(1)

Lại có \(\dfrac{8y-6z}{4}=0\)

=>8y-6z=0

=>8y=6z

=>\(\dfrac{8y}{24}=\dfrac{6z}{24}\)

=>\(\dfrac{y}{3}=\dfrac{z}{4}\)(2)

từ (1) và (2)=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)