\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=1-\dfrac{1}{99}=\dfrac{98}{99}\)

1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x=t\)

\(\Rightarrow BT=\left(t+10\right)\left(t+12\right)-24\)

\(=t^2+22x+96=\left(t+11\right)^2-25\ge-25\)

Vậy GTNN của bt là - 25\(\Leftrightarrow x^2+7x+11=0\)

\(\Delta=7^2-4.11=5\)

\(\orbr{\begin{cases}x_1=\frac{-22+\sqrt{5}}{2}\\x_2=\frac{-22-\sqrt{5}}{2}\end{cases}}\)

2) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x=t\)

\(\RightarrowĐT=\left(t+7\right)\left(t+15\right)-20\)

\(=t^2+22t+85=\left(t+11\right)^2-36\ge-36\)

Vậy GTNN của bt là - 36\(\Leftrightarrow x^2-8x+11=0\)

\(\Delta=\left(-8\right)^2-4.11=20\)

\(\orbr{\begin{cases}x_1=\frac{-22-\sqrt{20}}{2}\\x_2=\frac{-22+\sqrt{20}}{2}\end{cases}}\)

Nhầm, lộn tìm GTNN

a) \(\left(t+11\right)^2-25=\left(x^2+7x-16\right)\left(x^2+7x+36\right)\)

b) \(\left(x^2-8x+5\right)\left(x^2-8x+16\right)\)

\(=\left(x^2-8x+5\right)\left(x-4\right)^2\)

\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)

\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)

\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)

\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)

làm giống như trên

\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)

\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)

P/S: . là nhân nha

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Cái này dễ thế mà ko bít lm