a) \(\frac{x}{6}^2=\frac{24}{25}\)

\(\Rightarrow x^2.25=6.24\)

\(\Rightarrow x^2.25=144\)

\(\Rightarrow x^2=144\div25\)

\(\Rightarrow x^2=5,76=2,4^2=\left(-2,4^2\right)\)

\(\Rightarrow x\in\left\{2,4;-2,4\right\}\)

Vậy \(x\in\left\{2,4;-2,4\right\}\)

b) \(\frac{72-x}{7}=\frac{x-40}{9}\)

\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)

\(\Rightarrow648-9x=7x-280\)

\(\Rightarrow\left(-280\right)-648\) \(=-9x-7x\)

\(\Rightarrow-928=-16x\)

\(\Rightarrow x=58\)

Vậy \(x=58\)

ớ ko có vế phải seo tính đc

à quên =10 chờ tí nhé

\(\frac{x-2}{3.4}+\frac{x-2}{4.5}+...+\frac{x-2}{8.9}=10\)

tới đây tự nghĩ là ra

a) (72-x).9=7.(x-40)

   648-9x=7x-280

648+280=7x+9x (chuyển vế)

928=16x

x=928:16=58

b) (37-x).7=3.(x+13)

 259-7x=3x+39

259-39=3x+7x

220=10x

x=220:10=22

c)(x-1).(x+3)=(x-2).(x+2)

x²+3x-x-3=x²+2x-2x-4

(x²-x²)+(3x-x-2x+2x)=-4+3

0+2x=-1

x=-1/2

\(\frac{x-40}{9}=\frac{72-x}{7}\)

=> 7 ( x - 40 ) = 9 ( 72 - x )

     7 x - 280   = 648 - 9 x

     7 x + 9 x   = 648 + 280

       16 x        = 928

            x        = 58

\(\frac{x-40}{9}=\frac{72-x}{7}\)

\(\Rightarrow\left(x-40\right).7=9.\left(72-x\right)\)

\(\Rightarrow648-9x=7x-280\)

\(\Rightarrow\left(-280\right)-648=\left(-9x\right)-7x\)

\(\Rightarrow-928=-16x\)

\(\Rightarrow x=58\)

Tính giá trị biểu thức:

d) D=\(\frac{3x-2y}{x-3y}\)với \(\frac{x}{y}=\frac{10}{3}\)

=> (72 - x) . 9 = (x - 40) . 7

=> 648 - 9x = 7x - 280

=> 648 + 280 = 7x + 9x

=> 928 = 16x

=> x = 58

=> (72-x).9=(x-40).7

=> 648-9x=7x-280

=> 648+280-9x-7x=0

=> 928-16x=0

=> 928=16x

=> x=928:16

=> x=58

x =58 . Tớ hơi bị đẳng cấp đấy

\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\left(\frac{3}{9}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\cdot\frac{2}{9}=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\frac{2}{9}\)

\(x-2=\frac{16}{9}\cdot\frac{9}{2}\)

\(x-2=8\)

\(x=8+2\)

\(x=10\)

Vậy \(x=10\)

\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\)\(=\frac{16}{9}\)

\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right)\left(\frac{2}{9}\right)=\frac{16}{9}\)

2(x-2)=16

x-2=8

x=10
 

=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9

Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72

=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9

=> Sáng = 91.(1/3-1/9)

=> Sáng = 2/9

Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72

Ta được :

x-2.2/9=16/9

giờ thì tự làm nha

=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9

Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72

=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9

=> Sáng = 91.(1/3-1/9)

=> Sáng = 2/9

Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72

Ta được :

x-2.2/9=16/9

giờ thì tự làm nha

Ai k mk mk k lại

=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9

Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72

=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9

=> Sáng = 91.(1/3-1/9)

=> Sáng = 2/9

Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72

Ta được :

x-2.2/9=16/9

a.

\(\frac{x}{4}=\frac{3}{2}\)

\(x=\frac{3}{2}\times4\)

\(x=6\)

b.

\(\frac{x}{16}=\frac{9}{x}\)

\(x\times x=16\times9\)

\(x^2=144\)

\(x^2=\left(\pm12\right)^2\)

\(x=\pm12\)

Vậy \(x=12\) hoặc \(x=-12\)

c.

\(\frac{x^2}{6}=\frac{24}{25}\)

\(x^2=\frac{24}{25}\times6\)

\(x^2=\frac{144}{25}\)

\(x^2=\left(\pm\frac{12}{5}\right)^2\)

\(x=\pm\frac{12}{5}\)

Vậy \(x=\frac{12}{5}\) hoặc \(x=-\frac{12}{5}\)

d.

\(\frac{72-9}{7}=\frac{x-40}{9}\)

\(\frac{x-40}{9}=\frac{63}{7}\)

\(x-40=\frac{63}{7}\times9\)

\(x-40=81\)

\(x=81+40\)

\(x=121\)