=> x^2 = 8.2

x^2 = 16

=> \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

=> (8.2):x=x

16=x*2

=> x=4 hoặc x= -4

\(P=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)

\(P=1+\frac{1}{2}.3+\frac{1}{3}.6+\frac{1}{4}.10+....+\frac{1}{2016}.2033136\)

\(P=1+\frac{3}{2}+4+\frac{5}{2}+....+\frac{2017}{2}\)

\(P=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2017}{2}\)

\(P=\frac{2+3+4+5+....+2017}{2}=\frac{2035152}{2}=1017576\)

Để S lớn nhất thì 1/a, 1/b, 1/c phải  lớn nhất

=> S lớn nhất khi a = b = c = 1

=> S = 1/1 + 1/1 + 1/1 = 3 

tk nha

\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{100^2}-1\right)\)

\(A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right).\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right).....\left(\frac{1}{100^2}-\frac{100^2}{100^2}\right)\)

\(A=\left(-\frac{3}{4}\right).\left(-\frac{8}{9}\right)........\left(-\frac{9999}{10000}\right)\)

\(A=\frac{\left(-3\right).\left(-8\right).....\left(-9999\right)}{4.9...10000}=\frac{1.\left(-3\right).2.\left(-4\right)......99.\left(-101\right)}{2.2.3.3.....100.100}\)

\(A=\frac{\left(1.2.3....99\right).\left[\left(-3\right).\left(-4\right)......\left(-101\right)\right]}{\left(2.3.4....100\right).\left(2.3.4...100\right)}=\frac{1.\left(-101\right)}{100.\left(-1.\right).\left(-1\right)....\left(-1\right).2}=\frac{-101}{100.2}=\frac{-101}{200}\)

Ta thấy \(\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\Rightarrow A< -\frac{1}{2}\)

\(\frac{35\left(27^8+2.9^{11}\right)}{15\left(81^6-12.3^{19}\right)}\)= \(\frac{35\left(\left(3^3\right)^8+2.\left(3^2\right)^{11}\right)}{15\left(\left(3^4\right)^6-4.3.3^{19}\right)}\)= \(\frac{35\left(3^{24}+2.3^{22}\right)}{15\left(3^{24}-4.3^{20}\right)}\)= \(\frac{35\left(3^{22}.3^2+2.3^{22}\right)}{15\left(3^{20}.3^4-4.3^{20}\right)}\)= \(\frac{35\left(3^{22}.\left(9+2\right)\right)}{15\left(3^{20}.\left(81-4\right)\right)}\)= \(\frac{35\left(3^{22}.11\right)}{15\left(3^{20}.77\right)}\)= \(\frac{5.7.3^{22}.11}{5.3.3^{20}.7.11}\)= \(\frac{3^{22}}{3.3^{20}}\)= \(\frac{3^{20}.3.3}{3.3^{20}}\)= \(\frac{3}{1}\)= 3

\(Cm:\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

Gọi biểu thức trên là A, ta có:

3A = 1-2/3+3/3^2-...-100/3^99

3A + A = [1-2/3+3/3^2-...-100/3^99] + [1/3-2/3^2+3/3^3-...-100/3^100]

4A = 1 - 1/3 + 1/3^2 - ... - 1/3^99 - 100/3^99 [1]

Gọi B = 1-1/3 + 1/3^2 - ... - 1/3^99

3B = 3 - 1 + 1/3 - 1/3^2 -...-1/3^2012

3B + B = [3-1+1/3-1/3^2-...-1/3^2012] + [1-1/3 + 1/3^2 - ... - 1/3^99]

4B = 3 - 1/3^99 

=> 4B < 3 => B < 1/4 [2]

Từ [1], [2] => 4A < B < 3/4 => A < 3/16 [đpcm]

MỎI TAY QUỚ

tk nha

Lúc đặt câu hỏi, bạn bấm vào góc trên cùng bên trái để gõ phép tính đẹp. Ý của bạn có phải là:

\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

May minh KO bam dau dc. Thong Cam^-^

_Co 5 duong thang dong quy tai diem O tao thanh 10 tia chung goc O.

_Chon 1 tia noi voi 9 tia chung goc O con lai, ta duoc 9 goc dinh O. Cu lam nhu vay voi 10 tia chung goc O, ta se duoc 9.10 goc dinh O. Nhung neu nhu vay thi moi goc dinh O da duoc tinh 2 lan nen so goc thuc te co trong hinh la: 9.10/2=45( goc chung dinh O)

-------------------------------------------Vay co tat ca 45 goc chung dinh O-----------------------------------------------

                                                               ~Dier Nathan~  

co 4 goc