Ta có :

\(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)

\(\Leftrightarrow\left(\frac{392-x}{32}+1\right)+\left(\frac{390-x}{34}+1\right)+\left(\frac{388-x}{36}+1\right)+\left(\frac{386-x}{38}+1\right)+\left(\frac{384-x}{40}\right)=0\)

\(\Leftrightarrow\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)

\(\Leftrightarrow\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)

Mà : \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\ne0\)

\(\Leftrightarrow424-x=0\)

\(\Leftrightarrow x=424\)

Vậy x = 424

a) ĐKXĐ: x - 3 \(\ne\)0                                         x \(\ne\)3

             9 - x² \(\ne\)0                       <=>          x \(\ne\)\(\pm\)3

            x + 3 \(\ne\)0                                       x \(\ne\)-3

      \(\frac{6x-12}{2x^2-18}\) \(\ne\)0                         \(6x-12\ne0\) và \(2x^2-18\ne0\)

               x \(\ne\)\(\pm\)3

<=>     \(x\ne2\) và x \(\ne\)\(\pm\)3

<=> x \(\ne\)\(\pm\)3 và x \(\ne\)2

Ta có: B = \(\left(\frac{x+3}{x-3}+\frac{2x^2-6}{9-x^2}+\frac{x}{x+3}\right):\frac{6x-12}{2x^2-18}\)

 B = \(\left(\frac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{6\left(x-2\right)}{2\left(x^2-9\right)}\)

B = \(\left(\frac{x^2+6x+9-2x^2+6+x^2-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+3\right)}\)

B = \(\frac{3x+15}{\left(x+3\right)\left(x-3\right)}\cdot\frac{\left(x-3\right)\left(x+3\right)}{3\left(x-2\right)}\)

B = \(\frac{3\left(x+5\right)}{3\left(x-2\right)}\)

B = \(\frac{x+5}{x-2}\)

b) (sai đề)

c) Ta có: B = \(\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)

Để B \(\in\)Z <=> 7 \(⋮\)x - 2 <=> x - 2 \(\in\)Ư(7) = {1; -1; 7; -7}

Lập bảng: 

Vậy ...

a) \(\text{ĐKXĐ:}\hept{\begin{cases}x\ne\pm3\\x\ne2\end{cases}}\)

\(B=\left(\frac{x+3}{x-3}+\frac{2x^2-6}{9-x^2}+\frac{x}{x+3}\right):\frac{6x-12}{2x^2-18}\)

\(B=\left[\frac{x+3}{x-3}+\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right].\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)

\(B=\left[\frac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]\)

\(B=\left[\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}-\frac{2x^2-6}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\right].\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)

\(B=\frac{x^2+6x+9-\left(2x^2-6\right)+x^2-3}{\left(x-3\right)\left(x+3\right)}.\frac{2\left(x^2-9\right)}{6\left(x-2\right)}\)

\(B=\frac{3\left(x+5\right)}{\left(x-3\right)\left(x+3\right)}.\frac{2\left(x-3\right)\left(x+3\right)}{6\left(x-2\right)}\)

\(B=\frac{x+5}{x-2}\)

b) Ta có: \(\frac{x+5}{x-2}=1+\frac{7}{x-2}\)

Để B nguyên thì: \(7⋮x-2\)

\(\Rightarrow x-2\inƯ\left(7\right)\)

\(\RightarrowƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta có bảng: 

Vậy: \(x\in\left\{1;-5;9\right\}\)

Mình nghĩ phải sửa lại x+1 thành x-1 nha bạn ơi.

\(\frac{2}{x-1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2}{x-1}+\frac{18}{x^2-x+3x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2}{x-1}+\frac{18}{x\left(x-1\right)+3\left(x-1\right)}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2}{x-1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2\left(x+3\right)+18}{\left(x-1\right)\left(x+3\right)}=\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{2x+6+18}{\left(x-1\right)\left(x+3\right)}=\frac{2x^2-2x-5x+5}{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{2x+24}{\left(x-1\right)\left(x+3\right)}=\frac{2x^2-7x+5}{\left(x-1\right)\left(x+3\right)}\)

\(\Rightarrow2x+24=2x^2-7x+5\)

\(\Leftrightarrow2x+24-2x^2+7x-5=0\)

\(\Leftrightarrow-2x^2+9x+19=0\)

Từ đây giải nốt nha bạn

\(ĐKXĐ:x\ne\pm1;x\ne-3\)

\(\frac{2}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Rightarrow\frac{x^2+2x-3}{\left(x^2-1\right)\left(x+3\right)}+\frac{18\left(x+1\right)}{\left(x^2-1\right)\left(x+3\right)}=\frac{\left(x^2-1\right)\left(2x-5\right)}{\left(x^2-1\right)\left(x+3\right)}\)

\(\Rightarrow\frac{x^2+2x-3}{\left(x^2-1\right)\left(x+3\right)}+\frac{18x+18}{\left(x^2-1\right)\left(x+3\right)}=\frac{\left(x^2-1\right)\left(2x-5\right)}{\left(x^2-1\right)\left(x+3\right)}\)

\(\Rightarrow\frac{x^2+20x+15}{\left(x^2-1\right)\left(x+3\right)}=\frac{\left(x^2-1\right)\left(2x-5\right)}{\left(x^2-1\right)\left(x+3\right)}\)

\(\Rightarrow x^2+20x+15=\left(x^2-1\right)\left(2x-5\right)\)

\(\Rightarrow x^2+20x+15=x^3-5x^2-2x+5\)

\(\Rightarrow x^3-6x^2-22x-10=0\)

Giải nghiệm ta được ba nghiệm:

\(\left(\frac{-2103}{988};\frac{-5056}{9331};8,67\right)\)

nghe quen quen nhớ sương sương nhưng không biết làm

hay cho và kết bạn nha

Áp dụng BĐT Cô-si,ta có : 

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2b\)

Tương tự :....

Cộng lại , ta được :

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)

Dấu "=" xảy ra khi a = b = c