a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)

\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)

nên x-17=0

hay x=17

Vậy: x=17

b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)

\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)

\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)

nên x+20=0

hay x=-20

Vậy: x=-20

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

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

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

Giải:

a)-3/10-(-1/5)+x)=-3/2

             -1/5+x   =-3/10-(-3/2)

              -1/5+x   =6/5

                       x   =6/5-(-1/5)

                       x   =7/5

b)-(-x+3/4)-12/8.(-32/15)=-(-1/2)

           x-3/4+16/5          =1/2

           x-3/4                =1/2-16/5

           x                           =-27/10

           x                           =-27/10+3/4

           x                           =-39/20

c)x-3/x+5=4/3

=>(x-3).3=4.(x+5)

     3x-9   =4x+20

     3x-4x =20+9

     -1x     =29

        x     =-29

Câu b cậu nên tính lại cho kĩ nhé, ấn máy tính dễ nhầm lắm đấy!

Mk phải ấn: -(39/20+3/4)-12/8.-32/15=1/2

Vì x là số âm mà đằng trước x là dấu ''-'' nên -(-39/20)=39/20 ; -(-1/2)=1/2

Chúc bạn học tốt!

Bạn ơi mik ra \(\dfrac{x^3+45x-54}{12\left(x-3\right)\left(x+3\right)}\) có đúng không bạn?

\(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}\Rightarrow x=\dfrac{1}{10}\)

\(\dfrac{1}{2}.x+\dfrac{1}{2}=\dfrac{5}{2}\Leftrightarrow\dfrac{1}{2}.x=\dfrac{5}{2}-\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}.x=2\Leftrightarrow x=4\)

\(\dfrac{1}{2}-\dfrac{2}{3}.x=\dfrac{7}{12}\Leftrightarrow\dfrac{2}{3}.x=\dfrac{1}{2}+\dfrac{7}{12}\Leftrightarrow\dfrac{2}{3}.x=\dfrac{13}{12}\Leftrightarrow x=\dfrac{13}{12}:\dfrac{2}{3}\Leftrightarrow x=\dfrac{13}{8}\)

h) x/y = 9/10 ⇒  y/10 = x/9

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

y/10 = x/9 = (y - x)/(10 - 9) = 120/1 = 120

*) x/9 = 120 ⇒ x = 120.9 = 1080

*) y/10 = 120 ⇒ y = 120.10 = 1200

Vậy x = 1080; y = 1200

k) x/y = 3/4

⇒ x/3 = y/4

⇒ 5y/20 = 3x/9

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

5y/20 = 3x/9 = (5y - 3x)/(20 - 9) = 33/11 = 3

*) 3x/9 = 3 ⇒ x = 3.9:3 = 9

*) 5y/20 = 3 ⇒ y = 3.20:5 = 12

Vậy x = 9; y = 12

a: ĐKXĐ: x<>0; x<>-1

PT =>x+1-2x=3

=>1-x=3

=>x=-2(nhận)

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

=>x-3=5(2x-3)

=>10x-15=x-3

=>9x=12

=>x=4/3(nhận)

c: ĐKXĐ: x<>0; x<>2

PT =>x(x+2)-x+2=2

=>x^2+2x-x=0

=>x(x+1)=0

=>x=-1

a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)

b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)

c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)

d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)

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