\(\frac{90}{x}-\frac{36}{x-6}=2\)               MTC = x (x-6)         ĐK\(\hept{\begin{cases}x\ne0\\x\ne6\end{cases}}\)

\(\frac{90\left(x-6\right)}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}=\frac{2x\left(x-6\right)}{x\left(x-6\right)}\)

\(\frac{90x-540}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}-\frac{2x^2-12x}{x\left(x-6\right)}=0\)

\(90x-540-36x-2x^2+12x=0\)

\(-2x^2+66x-540=0\)

\(-2x^2+36x+30x-540=0\)

\(-2x\left(x-18\right)+30\left(x-18\right)=0\)

\(\left(x-18\right)\left(-2x+30\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-18=0\\-2x+30=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=18\\x=15\end{cases}}\)

vậy.....

ĐKXĐ:  \(x\ne0;\)  \(x\ne6\)

         \(\frac{90}{x}-\frac{36}{x-6}=2\)

\(\Leftrightarrow\)\(\frac{90\left(x-6\right)}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}=2\)

\(\Leftrightarrow\)\(\frac{90x-540-36x}{x\left(x-6\right)}=2\)

\(\Leftrightarrow\)\(\frac{54x-540}{x\left(x-6\right)}=2\)

\(\Leftrightarrow\)\(54x-540=2x\left(x-6\right)\)

\(\Leftrightarrow\)\(27x-270=x\left(x-6\right)\)

mk lm đc có vậy thôi.  tham khảo nha

cảm ơn câu trả lời của song ngư đúng r mk cx lm ra vậy 

ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)

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

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow x=3\)

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

\(a)\frac{-3}{x}=\frac{2^{10}}{2^{13}}\)

\(\Rightarrow\frac{-3}{x}=\frac{1}{2^3}=\frac{1}{8}\)

\(\Rightarrow-3\cdot8=x\)       \(\Rightarrow x=-24\)

\(b)\frac{7}{x}=\frac{y}{36}=\frac{-14}{28}\)

\(\Rightarrow\frac{7}{x}=\frac{y}{36}=-0,5\)

\(\Rightarrow x=7:\left(-0,5\right)=-14\)

     \(y=-0,5\cdot36=-18\)

Vậy x = -14; y = -18

\(c)\frac{10}{x}=\frac{15}{39}\)

\(\Rightarrow10\cdot39=15x\)             \(\Rightarrow390=15x\)

=> x = 390 : 15 = 26

Ah mk nhầm 1 chút!
x,y thuộc N nên phần a và b x và y không có giá trị thỏa mãn yêu cầu đề bài

Sorry!!!!

Ta có: \(\dfrac{\left(x+3\right)\left(x-3\right)}{3}+2=x\left(1-x\right)\)

\(\Leftrightarrow\dfrac{x^2-9}{3}+\dfrac{6}{3}=\dfrac{3x\left(1-x\right)}{3}\)

\(\Leftrightarrow x^2-9+6=3x-3x^2\)

\(\Leftrightarrow x^2-3-3x+3x^2=0\)

\(\Leftrightarrow4x^2-3x-3=0\)

\(\Delta=9-4\cdot4\cdot\left(-3\right)=9+48=57\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{57}}{8}\\x_2=\dfrac{3+\sqrt{57}}{8}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3-\sqrt{57}}{8};\dfrac{3+\sqrt{57}}{8}\right\}\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

Không biết nãy bị lỗi ở đâu, mình gửi lại:<