\(R=\frac{43^2-11^2}{\left(36,5\right)^2-\left(27,5\right)^2}\)

    \(=\frac{\left(43-11\right)\left(43+11\right)}{\left(36,5-27,5\right)\left(36,5+27,5\right)}\)

    \(=\frac{32.54}{9.64}\)

    \(=\frac{6}{2}=3\)

Bạn viết sai đề bài rồi

\(S=\frac{97^3+83^3}{180}-97.83\)

  \(=\frac{\left(97+83\right)\left(97^2-97.83+83^2\right)}{180}-97.83\)

   \(=97^2-97.83+83-97.83\)

    \(=\left(97-83\right)^2=14^2=196\)

Trả lời:

\(R=\frac{43^2-11^2}{36,5^2-27,5^2}\)

\(R=\frac{\left(43-11\right).\left(43+11\right)}{\left(36,5-27,5\right).\left(36,5+27,5\right)}\)

\(R=\frac{32.54}{9.64}\)

\(R=3\)

Đề bài sai bạn nhé 

\(S=\frac{97^3+83^3}{180}-97.83\)

\(S=\frac{\left(97+83\right).\left(97^2-97.23+83^2\right)}{180}-97.83\)

\(S=97^2-97.83+83^2-97.83\)

\(S=97^2-2.97.83+83^2\)

\(S=\left(97-83\right)^2\)

\(S=14^2\)

\(S=196\)

\(\frac{\left(43-11\right)\left(43+11\right)}{\left(36,5-27,5\right)\left(36,5+27,5\right)}=\frac{32.54}{9.64}\)

\(=3\)

dùng hằng đẳng thức thứ 3 ta có

(43-11)(43+11) và (36,5-27,5)(36,5+27,5)

Nên =1728 : 576=3

\(R=\frac{43^2-11^2}{36,5^2-27,5^2}\)

\(R=\frac{\left(43-11\right)\left(43+11\right)}{\left(36,5+27,5\right)\left(36,5-27,5\right)}\)

\(R=\frac{32.54}{64.9}\)

\(R=3\)

\(B=\left(\frac{x-4}{x\left(x-2\right)}+\frac{2}{x-2}\right):\left(\frac{x+2}{x}-\frac{x}{x-2}\right)\)

\(< =>B=\left(\frac{x-4}{x\left(x-2\right)}+\frac{2x}{x\left(x-2\right)}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}+\frac{x.x}{x\left(x-2\right)}\right)\)

\(< =>B=\left(\frac{x-4+2x}{x\left(x-2\right)}\right):\left(\frac{x^2-4}{x\left(x-2\right)}+\frac{x^2}{x\left(x-2\right)}\right)\)

\(< =>B=\frac{3x-4}{x\left(x-2\right)}:\frac{x^2-4+x^2}{x\left(x-2\right)}\)

\(< =>B=\frac{3x-4}{x\left(x-2\right)}.\frac{x\left(x-2\right)}{2x^2-4}\)

\(< =>B=\frac{3x-4}{2x^2-4}\)

\(b,\)Với \(x=-2\)thì

 \(B=\frac{3\left(-2\right)-4}{2\left(-2\right)^2-4}=\frac{-6-4}{8-4}=-\frac{10}{4}=-\frac{5}{2}\)

\(ĐKXĐ:x\ne2;x\ne0\)

a

\(B=\left[\frac{x-4}{x\left(x-2\right)}+\frac{2}{x-2}\right]:\left(\frac{x+2}{x}-\frac{x}{x-2}\right)\)

\(=\frac{x-4+2x}{x\left(x-2\right)}:\frac{\left(x+2\right)\left(x-2\right)-x^2}{x\left(x-2\right)}\)

\(=\frac{3x-4}{x^2-4-x^2}=-\frac{3x-4}{4}\)

b

\(B=-\frac{3x-4}{4}=-\frac{3\cdot\left(-2\right)-4}{4}=\frac{5}{2}\)

c

\(\left|B\right|-2x=5\Leftrightarrow\left|B\right|=5+2x\)

\(B=-\frac{3x-4}{4}\Leftrightarrow-\frac{3x-4}{4}\ge0\Leftrightarrow x\le\frac{4}{3}\)

\(B=\frac{3x-4}{4}\Leftrightarrow x>\frac{4}{3}\)

Xét các trường hợp của x thì ra nghiệm bạn nhé

d

\(\left(2-x\right)B=-\frac{\left(2-x\right)\left(3x-4\right)}{4}\)

Để ( 2 - x ).B đạt giá trị nhỏ nhất thì ( 2 - x ) ( 3x - 4 ) đạt giá trị lớn nhất

Casio sẽ giúp chúng ta phần này

e

Để B là số nguyên âm lớn nhất hay \(B=-1\Leftrightarrow-\frac{3x-4}{4}=-1\Leftrightarrow x=\frac{8}{3}\)

g

\(\left|B\right|+3< 2x-1\)

Làm hệt như câu c nhé :D 

\(=\left(\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x}{x+2}\right)\)

\(=\left(\frac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+10-x}{x+2}\right)\)

Đổi 10-x lại thành\(10-x^2\) nha, mk thiếu! sorry!

\(=\left(\frac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)

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

\(=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=-\frac{1}{x-2}\)