a) \(A=\frac{97^3+83^3}{180}-97\cdot83\)

\(A=\frac{\left(97+83\right)\left(97^2-97\cdot83+83^2\right)}{180}-97\cdot83\)

\(A=\frac{180\cdot\left(97^2-97\cdot83+83^2\right)}{180}-97\cdot83\)

\(A=97^2-97\cdot83+83^2-97\cdot83\)

\(A=9409-2\cdot8051+6889\)

\(A=196\)

b) \(B=\left(50^2+48^2+...+2^2\right)-\left(49^2+47^2+...+1^2\right)\)

\(B=50^2+48^2+...+2^2-49^2-47^2-...-1^2\)

\(B=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)

\(B=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\)

\(B=50+49+48+47+...+2+1\)

Số số hạng là : \(\left(50-1\right):1+1=50\)( số )

Tổng B là : \(\left(50+1\right)\cdot50:2=1275\)

Vậy....

a,\(6x-8y=9\)

\(\Rightarrow x=\frac{9+8y}{6}\)

\(y=\frac{6x-9}{8}\)

Vậy....

\(b,11x+18y=120\)

\(\Rightarrow x=\frac{120-18y}{11}\)

\(y=\frac{120-11x}{18}\)

100 nha

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\(A=a^4-2a^3+3a^2-4a+5\)

    \(=\left(a^4-2a^3+a^2\right)+2\left(a^2-2a+1\right)+3\)

      \(=\left(a^2-a\right)^2+2\left(a-1\right)^2+3\ge3\)

Dấu "='' khi a = 1

Vậy ..........

=7 nhá

học tốt

3+4=7

ngày  18/2 là sinh nhật J-hope

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4ac-4bc\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow VT\ge0}\)

Dấu "=" xảy ra <=> a = b = c (đpcm)