Ta có:

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)

\(\frac{x-1}{15}=\frac{-60}{1-x}\)

\(\Rightarrow\frac{x-1}{15}=\frac{-60}{-x+1}\)

\(\Rightarrow\frac{x-1}{15}=\frac{-60}{-\left(x-1\right)}\)

\(\Rightarrow\frac{x-1}{15}=\frac{60}{x-1}\)

\(\Rightarrow\left(x-1\right)^2=15.60\)

\(\Rightarrow\left(x-1\right)^2=900\)

\(\Rightarrow x-1=\pm30\)

\(\Rightarrow\left\{\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=31\\x=-29\end{matrix}\right.\)

Vậy...

\(\frac{x}{7}=\frac{x+16}{35}\)

\(\Rightarrow35x=7\left(x+16\right)\)

\(\Rightarrow35x=7x+112\)

\(\Rightarrow35x-7x=112\)

\(\Rightarrow28x=112\)

\(\Rightarrow x=112:28\)

\(\Rightarrow x=4\)

nếu 0<a<180 thì góc O= góc I = \(\frac{a^0}{2}\)

nếu a =180 thì không xác định được số đo của các góc O và I

Bài 2:

a) \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)

Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A< 1+1-\frac{1}{50}\)

\(\Rightarrow A< 2-\frac{1}{50}< 2\)

b) \(B=2^1+2^2+2^3+...+2^{30}\) (Có 30 số hạng)

\(\Rightarrow B=\left(2^1+2^2+...+2^5+2^6\right)+\left(2^7+2^8+2^9+...+2^{12}\right)+...+\left(2^{25}+2^{26}+...+2^{29}+2^{30}\right)\)

(có \(30:6=5\) nhóm)

\(\Rightarrow B=1\left(2^1+2^2+...+2^6\right)+2^6\left(2^1+2^2+...+2^6\right)+.....+2^{24}\left(2^1+2^2+...+2^6\right)\)

\(\Rightarrow B=1.126+2^6.126+2^{12}.126+...+2^{24}.126\)

\(\Rightarrow B=126.\left(1+2^6+2^{12}+...+2^{24}\right)\)

\(\Rightarrow B=21.6.\left(1+2^6+2^{12}+...+2^{24}\right)⋮21\)

\(\Rightarrow B⋮21\)

Bài 1:

a) Ta có:

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)

b) Ta có:

\(71^{50}=\left(71^2\right)^{25}=5041^{25}\)

\(37^{75}=\left(37^3\right)^{25}=50653^{25}\)

Vì \(5041^{25}< 50653^{25}\Rightarrow71^{50}< 37^{75}\)

c) Ta có:

\(\frac{201201}{202202}=\frac{201.1001}{202.1001}=\frac{201}{202}\)

\(\frac{201201201}{202202202}=\frac{201.1001001}{202.1001001}=\frac{201}{202}\)

\(\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)

Để \(3^{2014}+3^a⋮10\)

\(\Rightarrow3^{2014}+3^a\) có tận cùng là 0

\(\Rightarrow3^{2014}+3^a=\left(\overline{.........0}\right)⋮10\)

Ta có: \(3^{2014}=\left(3^4\right)^{503}.3^2=81^{503}.9=\left(\overline{.......1}\right).9=\left(\overline{.......9}\right)\)

\(\Rightarrow3^{2014}+3^a=\left(\overline{......9}\right)+3^a=\left(\overline{.......0}\right)\)

\(\Rightarrow3^a\) có tận cùng là 1

Mà a nhỏ nhất

\(\Rightarrow a=0\)

Ta có: \(x+3z+x+2y=8+9\)

\(\Rightarrow2x+2y+3z=17\)

\(\Rightarrow2x+2y+2z+z=17\)

\(\Rightarrow2\left(x+y+z\right)=17-z\)

Mà \(x+y+z\) có GTLN

\(\Rightarrow17-z\) cũng có GTLN

Mà \(z\ge0\Rightarrow-z\le0\)

\(\Rightarrow17-z\le17\)

\(\Rightarrow17-z\) đạt GTLN là 17 tại \(z=0\)

+) \(x+3z=8\)

Thay \(z=0\)

\(\Rightarrow x+0=8\)

\(\Rightarrow x=8\)

+) \(x+2y=9\)

Thay \(x=8\)

\(\Rightarrow8+2y=9\)

\(\Rightarrow2y=1\)

\(\Rightarrow y=\frac{1}{2}\)

Vậy \(x=8;y=\frac{1}{2};z=0\)

Ta có: \(\frac{x-4}{y-3}=\frac{4}{3}\)

\(\Rightarrow3\left(x-4\right)=4\left(y-3\right)\)

\(\Rightarrow3x-12=4y-12\)

\(\Rightarrow3x-4y=0\)

\(\Rightarrow3x-3y-y=0\)

\(\Rightarrow3\left(x-y\right)-y=0\)

Thay \(x-y=5\)

\(\Rightarrow3.5-y=0\)

\(\Rightarrow15-y=0\)

\(\Rightarrow y=15\)

\(\Rightarrow x-15=5\)

\(\Rightarrow x=15+5\)

\(\Rightarrow x=20\)

Vậy .....