3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)

=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)

=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)

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

4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)

=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)

=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)

=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)

=> \(3x=\dfrac{1}{9}\)

=> \(x=\dfrac{1}{9}:3\)

=> \(x=\dfrac{1}{27}\)

\(\dfrac{6}{x}=\dfrac{24}{x-27}\)

Theo định nghĩa phân số bằng nhau, ta có:

\(6\cdot\left(x-27\right)=24x\\ 6x-162=24x\\ -162=24x-6x\\ -162=18x\\ x=\left(-162\right):18\\ x=-9\)

Theo bài ra ta có:

6.(x-27)=x.24

6x-162=x.24

-162=24x-6x

-162=18x

x =(-162):18

x = -9

tick mk nha

b) \(\dfrac{x}{27}=\dfrac{3}{x}\)

\(\Rightarrow x.x=27.3\)

\(\Rightarrow x^2=81\)

\(\Rightarrow x^2=9^2\)

\(\Rightarrow x=9\)

Vậy x=9

bài này áp dụng quy tắc nhân chéo nha :vv

a) \(\dfrac{x}{5}=\dfrac{2}{5}\Leftrightarrow5x=2.5=10\Leftrightarrow x=\dfrac{10}{5}=2\)

b) \(\dfrac{3}{8}=\dfrac{6}{x}\Leftrightarrow3x=6.8=48\Leftrightarrow x=\dfrac{48}{3}=16\)

c)\(\dfrac{1}{9}=\dfrac{x}{27}\Leftrightarrow9x=27\Leftrightarrow x=\dfrac{27}{9}=3\)

d)\(\dfrac{4}{x}=\dfrac{8}{6}\Leftrightarrow8x=4.6=24\Leftrightarrow x=\dfrac{24}{8}=3\)

e) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+4x=-6+20\\ \Leftrightarrow7x=14\Leftrightarrow x=\dfrac{14}{7}=2\)

g) \(\dfrac{x}{-2}=\dfrac{-8}{x}\Leftrightarrow x^2=\left(-2\right)\left(-8\right)=16\\ \Rightarrow x=\pm4\)

a)x=2

b)x=16

c)x=3

d)x=3

e)x=-2

g)x=4

giúp zới

a,Vì \(\dfrac{5}{6}\)=\(\dfrac{x}{24}\) nên ta có: 5.24:6= 20 \(\Rightarrow\)x =20

Mấy câu sau làm tương tự như vậy.

b,x =21

c,x =9

d,x = -5

a) \(\left(3\dfrac{1}{3}-x\right)1\dfrac{1}{6}=\dfrac{7}{24}\)

\(\Leftrightarrow\left(\dfrac{10}{3}-x\right)\dfrac{7}{6}=\dfrac{7}{24}\)

\(\Leftrightarrow\dfrac{10}{3}-x=\dfrac{7}{24}:\dfrac{7}{6}\)

\(\Leftrightarrow\dfrac{10}{3}-x=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{10}{3}-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{37}{12}\).

b) \(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)

\(\Leftrightarrow\left(4,5-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)

\(\Leftrightarrow4,5-2x=\dfrac{4}{3}.\dfrac{3}{4}\)

\(\Leftrightarrow4,5-2x=1\)

\(\Leftrightarrow2x=4,5-1\)

\(\Leftrightarrow2x=3,5\)

\(\Leftrightarrow x=\dfrac{35}{2}\).

\(B=\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\left(1+\dfrac{1}{24}\right).....\left(1+\dfrac{1}{440}\right)\left(1+\dfrac{1}{483}\right)\)

\(B=\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.....\dfrac{441}{440}.\dfrac{484}{483}\)

\(B=\dfrac{9.16.25.....441.484}{8.15.24.....440.483}\)

\(B=\dfrac{3.3.4.4.5.5.....21.21.22.22}{2.4.3.5.4.6.....20.22.21.23}\)

\(B=\dfrac{3.4.5.....21.22}{2.3.4.....20.21}.\dfrac{3.4.5.....21.22}{4.5.6.....22.23}\)

\(B=11.\dfrac{3}{23}=\dfrac{33}{23}\)

B = \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{121}{120}.\dfrac{144}{143}\)

B = \(\dfrac{4.9.16.25...121.144}{3.8.15.24....120.143}\)

B = \(\dfrac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)

B = \(\dfrac{2.3.4.5...11.12}{1.2.3.4.5...10.11}.\dfrac{2.3.4.5...11.12}{3.4.5.6.7...12.13}\)

B = 12 . \(\dfrac{2}{13}\)

B = \(\dfrac{24}{13}\)

a: \(\Leftrightarrow-\dfrac{720}{150}=-4.8< x< \dfrac{-63}{210}=-0.3\)

mà x là số nguyên

nen \(x\in\left\{-4;-3;-2;-1\right\}\)

b: \(\Leftrightarrow-\dfrac{125}{27}< x< \dfrac{120}{210}=\dfrac{4}{7}\)

mà x là số nguyên

nên \(x\in\left\{-4;-3;-2;-1;0\right\}\)

Ta có :

\(-\dfrac{24}{-6}=\dfrac{x}{3}\)

\(\Rightarrow x=\dfrac{-24\cdot3}{-6}=12\)

=> TA CÓ :

\(\dfrac{12}{3}=\dfrac{4}{y^2}\)

\(\Rightarrow y^2=\dfrac{4\cdot3}{12}=1\)

\(\Rightarrow y=\pm1\)

=> Ta có :

\(\dfrac{4}{1}=\dfrac{z^3}{-2}\)

\(\Rightarrow z^3=\dfrac{4\cdot\left(-2\right)}{1}=-8\)

\(\Rightarrow z=-2\)

Vậy x= 12 ; y = \(\pm1\) ;z=-2