ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\)

\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)

\(\Leftrightarrow\left(x-3\right).\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right).\left(x-4\right)\)

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow3x^2-9x-8x+24=0\)

\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\left(\text{thỏa}\right)\)

\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne\left\{2;4\right\}\right)\\ =>\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\\ =>x^2-3x-4x+12+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\\ =>2x^2-11x+16=-x^2+6x-8\\ =>3x^2-17x+24=0\\ =>\left(x-3\right)\left(3x-8\right)=0\\ =>\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\left(TMDK\right)\)

\(\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\dfrac{-\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)

\(\dfrac{x^2-7x+12}{x^2-6x+8}+\dfrac{x^2-4x+2}{x^2-6x+8}=\dfrac{-x^2+6x-8}{x^2-6x+8}\)

\(\dfrac{x^2-7x+12+x^2-4x+2+x^2-6x+8}{x^2-6x+8}=0\)

\(\dfrac{3x^2-17x+22}{x^2-6x+8}=0\)

\(\dfrac{\left(3x-11\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)

\(\dfrac{3x-11}{x-4}=0\)

Với x khác 4 

=> 3x -11 = 0 

=> \(x=\dfrac{11}{3}\)

ĐKXĐ: \(x\ge3\)

\(pt\Leftrightarrow5\sqrt{x-3}+3\sqrt{x-3}-\sqrt{x-3}=7\)

\(\Leftrightarrow7\sqrt{x-3}=7\Leftrightarrow\sqrt{x-3}=1\)

\(\Leftrightarrow x-3=1\Leftrightarrow x=4\left(tm\right)\)

Đề cái phải là : 231-(x-6)=1339:13 ko

231 - ( x - 6 ) = 1339 : 13

231 - ( x - 6 ) = 103

                   x = 231 - 103 + 6

                   x = 128 + 6 

                   x = 134

\(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}-\frac{7}{3}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{9}\)

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

\(\Rightarrow x+\frac{1}{2}=\frac{1}{3}\)

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

\(x=-\frac{1}{6}\)

\(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}=\frac{7}{3}\)\

\(\left(x+\frac{1}{2}\right)^2=\frac{22}{9}-\frac{21}{9}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{9}\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{3}\)

TH1:\(x+\frac{1}{2}=\frac{1}{3}\)

       \(x=\frac{1}{3}-\frac{1}{2}\)

        \(x=-\frac{1}{6}\)

TH2:\(x+\frac{1}{2}=-\frac{1}{3}\)

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

       \(x=-\frac{5}{6}\)

Vậy \(x\in\left\{-\frac{1}{6};-\frac{5}{6}\right\}\)

x  +(22/9 -1/2) =7/3

x   +(44/18-9/18)=7/3 + 35/18

x                         =42/18+35/18

x                          =77/18

k nha bạn                    

a, ĐKXĐ:\(x\ne-3\)

\(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+3}{x+3}\\ \Leftrightarrow x+1=1\\ \Leftrightarrow x=0\left(tm\right)\)

b, ĐKXĐ:\(x>2\)

\(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\\ \Leftrightarrow x^2-4x-2=x-2\\ \Leftrightarrow x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc đề của bạn dễ hiểu hơn nhé.

=8

tíck nha

bn chỉ cần nhan vào rồi tính thôi

\(\Rightarrow2\cdot x+16=0\)

\(2\cdot x=-16\)

\(x=-8\)

=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84

=> (72 - 84 )  - (20x + 36x ) = (30x - 6x ) - 240 + 84

=> -12 - 56x = 24x - 156

=> -12 + 156 = 24x + 56x 

=> 144 = 80x

=> x = 144  : 80

=> x = 9/5

=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84

=> (72 - 84 )  - (20x + 36x ) = (30x - 6x ) - 240 + 84

=> -12 - 56x = 24x - 156

=> -12 + 156 = 24x + 56x 

=> 144 = 80x

=> x = 144  : 80

=> x = 9/5

mn nhanh lên ạaa!!!

Đề bài cụ thể là gì vậy ạ

Tìm x, y ạ