\(A=\dfrac{97^3+83^3}{180}-97.83\)

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

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

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

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

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

\(A=14^2\)

\(A=196\)

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

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

    \(=\frac{32.54}{9.64}\)

    \(=\frac{6}{2}=3\)

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 

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

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

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

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

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

\(S=14^2\)

\(S=196\)

Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)

\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)

\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)

\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)

\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)

Bài 2:

\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)

\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

Vì \(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)

\(a,\frac{97^3+83^3}{180}-97.83\)

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

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

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

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

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

\(=14^2=196\)

Giáo án hả :v Nhìn quen quenn :v

5) 3x - 1 < 8

⇔ 3x < 9

⇔ x < 3

4) -8x > 24

<=> x > 32

Vì \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|>0\forall x\)

mà \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)

nên x>0

Với x>0, ta được:

\(x+\frac{1}{101}+x+\frac{2}{101}+x+\frac{3}{101}+...+x+\frac{100}{101}=101x\)

\(\Leftrightarrow100x-101x+\frac{5050}{101}=0\)

\(\Leftrightarrow-x+50=0\)

hay x=50

Vậy: S={50}