1C

2A

1C        2A

Ta có:\(\left(\frac{6}{x^2-6x}+\frac{1}{x+6}\right):\frac{x^2+36}{x^2-36}\)

   \(=\left(\frac{6\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)}+\frac{x\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x^2-6^2}{x^2+36}\)

     \(=\left(\frac{6x+36+x^2-6x}{x\left(x-6\right)\left(x+6\right)}\right).\frac{\left(x-6\right)\left(x+6\right)}{x^2+36}\)

        \(=\frac{x^2+36}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+36}\)

      \(=\frac{1}{x}\)

Kiểm tra đi bạn phải là \(\frac{1}{x}\)

ĐKXĐ : \(x\ne\pm6\)

\(\frac{36}{x+6}+\frac{36}{x-6}=\frac{9}{2}\)

\(\frac{72\left(x-6\right)}{\left(x+6\right)\left(x-6\right)2}+\frac{72\left(x+6\right)}{\left(x-6\right)\left(x+6\right)2}=\frac{9\left(x+6\right)\left(x-6\right)}{2\left(x+6\right)\left(x-6\right)}\)

\(72\left(x-6\right)+72\left(x+6\right)=9\left(x+6\right)\left(x-6\right)\)

\(72x-432+72x+432=9x^2-324\)

\(144x=9x^2-324\)

\(144x-9x^2+324=0\)

\(-9x^2+144x+324=0\)

\(\Delta=144^2-4.\left(-9\right).324=32400>0\)

Nên phương trình có 2 nghiệm phân biệt 

\(x_1=\frac{-144-\sqrt{32400}}{2.\left(-9\right)}=\frac{-144-180}{-18}=18\)

\(x_2=\frac{-144+\sqrt{32400}}{2.\left(-9\right)}=\frac{-144+180}{-18}=-2\)

Đk : x khác 6 và -6

\(\frac{36}{x+6}+\frac{36}{x-6}=\frac{9}{2}\)

\(< =>\frac{36\left(x-6\right)+36\left(x+6\right)}{\left(x+6\right)\left(x-6\right)}=\frac{9}{2}\)

\(< =>\frac{36x-216+36x+216}{x^2-6x+6x-36}=\frac{9}{2}\)

\(< =>\frac{72x}{x^2-6^2}=\frac{9}{2}\)

\(< =>144x=9x^2-324\)

\(< =>9x^2-144x-324=0\)

Ta có : \(\Delta=\left(-144\right)^2-4.9.\left(-324\right)=32400\)

\(< =>\sqrt{\Delta}=180\)

Vì delta > 0 nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{144+180}{18}=18\)

\(x_2=\frac{144-180}{18}=-2\)

Vậy ...

a, \(\frac{2}{x}=\frac{y}{4}\Leftrightarrow xy=8\)

Suy ra : \(x;y\inƯ\left(8\right)=\left\{1;2;4;8\right\}\)tự lập bảng 

b, Xét : \(\frac{1}{x}=\frac{5}{15}\Leftrightarrow\frac{5}{5x}=\frac{5}{15}\Leftrightarrow x=3\)

\(\frac{y}{21}=\frac{5}{15}\Leftrightarrow15y=105\Leftrightarrow y=3\)

\(\frac{10}{z}=\frac{5}{15}\Leftrightarrow5z=150\Leftrightarrow z=30\)

c, tương tự b 

NÓI LUÔN LÀ RÚT GỌN C/M nghe nặng nề quá

ĐK tồn tại \(x\ne\) {-6,0,6}

\(A=\left(\frac{36+6x+x^2-6}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x^2+36}{\left(x-6\right)\left(x+6\right)}\) Bẫy rồi

đây là dấu nhân hả bạn???

a, \(\dfrac{x^2-49}{x-7}\) + x - 2 = \(\dfrac{\left(x-7\right)\left(x+7\right)}{x-7}\) + x - 2 = x + 7 + x - 2 = 2x + 5

b, \(\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right)\) . \(\dfrac{x^2+6x}{2x-6}\) 

= \(\left(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)

= \(\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)

= \(\left(\dfrac{6\left(2x-6\right)}{x\left(x-6\right)\left(x+6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)

= \(\dfrac{6}{x-6}\)

1. = \(\dfrac{\left(x-7\right)\left(x+7\right)}{x-7}\) + x-2

    = x+7 +x-2

    = 2x-5

2.  = (\(\dfrac{x}{\left(x-6\right)\left(x+6\right)}\) - \(\dfrac{x-6}{x\left(x+6\right)}\) ) \(^{\dfrac{x^2+6x}{2x-6}}\)

     = ( \(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}\) - \(\dfrac{\left(x-6\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\) ) \(\dfrac{x^2+6x}{2x-6}\) 

     = \(\dfrac{x^2-\left(x^2-12x+36\right)}{x\left(x-6\right)\left(x+6\right)}\)  . \(\dfrac{x^2+6x}{2x-6}\)

     = \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\) .  \(\dfrac{x^2+6x}{2x-6}\)

     = \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\) . \(\dfrac{x^2+6x}{2x-6}\)

     = \(\dfrac{12\left(x-3\right)x\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)2\left(x-3\right)}\)

     = \(\dfrac{6}{x-6}\)

Chúc bạn học tốt!