\(5-8x-x^2\)

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

\(=-\left(\left(x+4\right)^2-21\right)\)

\(=21-\left(x+4\right)^2\le21\)

Min bằng 21 \(\Leftrightarrow x=-4\)

Xét\(\Delta ABC\)vuông tại A có \(AC=\frac{1}{2}BC\)

Trên tia đối của tia AC lấy điểm D sao cho AD = AC

Xét \(\Delta ABD\) và \(\Delta ABC\)có :

AB cạnh chung

\(\widehat{BDA}=\widehat{BAC}\)(gt)

AD = AC(gt)

=> \(\Delta ABD=\Delta ABC\)(c.g.c)

=> BD = BC(hai cạnh tương ứng)

Do \(AC=\frac{1}{2}BC,AC=\frac{1}{2}DC\Rightarrow BC=DC\)

\(\Delta BDC\)có BD = BC = DC nên là tam giác đều,do đó \(\widehat{C}=60^0\)

=> \(\widehat{ABC}=30^0\)

E = \(\frac{36}{1\cdot7}+\frac{36}{7\cdot13}+...+\frac{36}{94\cdot100}=\frac{36}{6}\left[\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+...+\frac{1}{94\cdot100}\right]\)

\(=6\left[1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{94}-\frac{1}{100}\right]=6\left[1-\frac{1}{100}\right]\)

\(=6\cdot\frac{99}{100}=\frac{297}{50}\)

F = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)

\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3a+2}-\frac{1}{3a+5}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3a+5}\right]=\frac{1}{6}-\frac{1}{9a+15}\)

G = \(\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{4}{8\cdot12}+\frac{5}{12\cdot17}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{12}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

E=36/1-36/7+36/7-36/13+...+36/94-36/100

  =36-36/100=891/25

thay x = -1 => 0.f(-1) = -1.f(2)

=> 0 = -1.f(2) 

=> f(2) = 0

tương tự, ta thay x = -3

=> -2.f(-3) = -3.0=0

=> -2.f(-3) =0

=> f(-3) = 0

=> f(x) có 2 nghiệm là -3 và 2

k đúng cho tớ nhé ihaha =))))

bn hk hằng đẳng thức chưa ?