a: =>6/x=x/24

=>x^2=144

=>x=12 hoặc x=-12

b: =>x(1-7/12+3/8)=5/24

=>x*19/24=5/24

=>x=5/24:19/24=5/19

c: =>(x-1/3)^2=1+3/4+1/2=9/4

=>x-1/3=3/2 hoặc x-1/3=-3/2

=>x=11/6 hoặc x=-7/6

d: =>(x-3)^2=16

=>x-3=4 hoặc x-3=-4

=>x=-1 hoặc x=7

e: =>9/x=-1/3

=>x=-27

f: =>x-1/2=0 hoặc -x/2-3=0

=>x=1/2 hoặc x=-6

a, đk x khác 0

<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)

b, <=> 36x +252 = -360 <=> x = -17 

c. đk x khác -1 

<=> (x+1)^2 = 16 

TH1 : x + 1 = 4 <=> x = 3 (tm)

TH2 : x + 1 = -4 <=> x = -5 (tm) 

d, đk x khác 1/2 

<=> (2x-1)^2 = 81 

TH1 : 2x - 1 = 9 <=> x = 5 (tm) 

TH2 : 2x - 1 = -9 <=> x = -4 (tm) 

a: \(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

b: =>x+7/15=-2/3

=>x+7=-10

hay x=-17

c: \(\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)

hay \(x\in\left\{3;-5\right\}\)

a) \(\dfrac{x}{4}=\dfrac{4}{x}\)=>x²=4.4=16 =>x²=4²

=>x=2 hay x=-2.

b) \(\dfrac{x+7}{15}=-\dfrac{24}{36}\)=>\(\dfrac{x+7}{15}=-\dfrac{2}{3}\)=>x+7=-\(\dfrac{2}{3}.15\)=-10 =>x=-17

c)\(\dfrac{x+1}{8}=\dfrac{2}{x+1}\)=>(x+1)²=2.8=16=4²

=>x+1=4 hay x+1=-4

=>x=3 hay x=-5.

d) \(\dfrac{2x-1}{\left(-3\right)^2}=\dfrac{\left(-3\right)^2}{2x-1}\)=>\(\dfrac{2x-1}{9}=\dfrac{9}{2x-1}\)=>(2x-1)²=9²

=>2x-1=9 hay 2x-1=-9

=>x=5 hay x=-4.

x=\(\dfrac{4}{15}\) : \(\dfrac{-2}{3}\)

x=\(\dfrac{-2}{5}\)

a: Ta có: \(x\cdot\dfrac{-2}{3}=\dfrac{4}{15}\)

\(\Leftrightarrow x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-2}{5}\)

b: Ta có: \(x\cdot\dfrac{-7}{19}=\dfrac{-13}{24}\)

\(\Leftrightarrow x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{247}{168}\)

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

x + 3/2 = 37/24

 x = 37/24 - 3/2

x = 1/24

x+3/2=37/24

x=37/24-3/2

x=1/24

x + 3/2 = 37/24

x = 37/24 - 3/2

x = 1/24

vậy x = ...

a) x=24/35 -2/7

x=14/35

b) x=7/8+5/6

x=41/24

c) x-11/5=3/5

x=11/5+3/5

x=14/5

tick cho mình nhé

giúp mình với ạ, mình sẽ tick

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{3y}{6}=\dfrac{3z}{9}=\dfrac{x-3y+3z}{5+6-9}=\dfrac{24}{2}=12\\ \Rightarrow\left\{{}\begin{matrix}x=60\\y=24\\z=36\end{matrix}\right.\)

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{3y}{6}=\dfrac{3z}{9}\)

Áp dụng t/c DTSBN, ta có:

\(\dfrac{x}{5}=\dfrac{3y}{6}=\dfrac{3z}{9}=\dfrac{x-3y+3z}{5-6+9}=\dfrac{24}{8}=3\)

\(\dfrac{x}{5}=3\Rightarrow x=15\)

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

\(\dfrac{z}{3}=3\Rightarrow=9\)

Vậy...

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}\text{ và }x-3y+3z=24\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau:}\)

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{x-3y+3z}{5-3.2+3.3}=\dfrac{24}{8}=3\)

\(\Rightarrow x=3.5=15\)

\(y=3.2=6\)

\(z=3.3=9\)

\(\Leftrightarrow\dfrac{x}{-2}=\dfrac{1}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}\)

=>x=1; y=36; z=12

`1-(5 3/8+x-7 5/24):(-16 2/3)=0`
`=>1-(5+3/8+x-7-5/24):(-50/3)=0`
`=>1=(x-2-11/6):(-50/3)`
`=>1=(x-11/6):(-50/3)`
`=>x-11/6=-50/3`
`=>x=-89/6`
Vậy `x=-89/6`

\(1-\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow1-\left(\dfrac{43}{8}+x-\dfrac{173}{24}\right):\dfrac{50}{3}=0\)

\(\Leftrightarrow1-\left(\dfrac{129}{24}-\dfrac{173}{24}+x\right).\dfrac{3}{50}=0\)

\(\Leftrightarrow1+\dfrac{11}{6}-x.\dfrac{3}{50}=0\)

\(\Leftrightarrow\dfrac{17}{6}-x.\dfrac{3}{50}=0\)

\(\Leftrightarrow x.\dfrac{3}{50}=\dfrac{17}{6}\)

\(\Leftrightarrow x=\dfrac{425}{9}\)

-Chúc bạn học tốt-

\(1-\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)=0\) 

\(1-\left(\dfrac{43}{8}:\dfrac{-50}{3}+x:\dfrac{-50}{3}-\dfrac{173}{24}:\dfrac{-50}{3}\right)=0\) 

\(1-\dfrac{-129}{400}-x:\dfrac{-50}{3}+\dfrac{-173}{400}=0\) 

\(\left(1+\dfrac{129}{400}+\dfrac{-173}{400}\right)-x:\dfrac{-50}{3}=0\) 

                        \(\dfrac{89}{100}-x:\dfrac{-50}{3}=0\) 

                                    \(x:\dfrac{-50}{3}=\dfrac{89}{100}-0\) 

                                    \(x:\dfrac{-50}{3}=\dfrac{89}{100}\) 

                                               \(x=\dfrac{89}{100}.\dfrac{-50}{3}\) 

                                               \(x=\dfrac{-89}{6}\)

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