Ta có: \(\left(2x-1\right)^{2014}+\left(y-\dfrac{2}{5}\right)^{2014}+\left|x+y+z\right|=0\)

\(\Rightarrow\left(2x-1\right)^{2014}=0\) (1)

\(\Rightarrow\left(y-\dfrac{2}{5}\right)^{2014}=0\) (2)

\(\Rightarrow\left|x+y+z\right|=0\) (3)

(1) Ta tìm được x:

\(\left(2x-1\right)^{2014}=0\)

\(\Rightarrow2x-1=0\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\dfrac{1}{2}\)

(2) Ta tìm được y:

\(\left(y-\dfrac{2}{5}\right)^{2014}=0\)

\(\Rightarrow y-\dfrac{2}{5}=0\)

\(\Rightarrow y=\dfrac{2}{5}\)

Từ (1) và (2) ta kết hợp với (3) ta sẽ tìm được z:

\(x+y+z=0\) hay \(\dfrac{1}{2}+\dfrac{2}{5}+z=0\)

\(\Rightarrow\dfrac{9}{10}+z=0\)

\(\Rightarrow z=-\dfrac{9}{10}\)

Vậy: \(x=\dfrac{1}{2};y=\dfrac{2}{5};z=-\dfrac{9}{10}\)

\(\left(2x-1\right)^{2014}+\left(y-\dfrac{2}{5}\right)^{2014}+|x+y+z|=0\left(1\right)\)

mà \(\left(2x-1\right)^{2014}\ge0;\left(y-\dfrac{2}{5}\right)^{2014}\ge0\) (với mọi x;y)

\(\left(1\right)\Rightarrow2x-1=0;y-\dfrac{2}{5}=0;|x+y+z|=0\)

\(\Rightarrow x=\dfrac{1}{2};y=\dfrac{2}{5};z=-\dfrac{1}{2}-\dfrac{2}{5}=-\dfrac{9}{10}\)

\(n\left(2n-3\right)-2n\left(n+1\right)=2n^2-3n-2n^2-2n=-5n\)

mà \(-5n⋮5\left(n\in Z\right)\)

⇒đpcm

\(n\left(2n-3\right)-2n\left(n+1\right)=\)

\(=2n^2-3n-2n^2-2n=-5n⋮5\)

\(\Leftrightarrow2^{x+1}.3^y=2^{2x}.3^x\)

\(\Leftrightarrow\dfrac{2^{2x}.3^x}{2^{x+1}.3^y}=1\Leftrightarrow2^{x-1}.3^{x-y}=1\)

\(\Leftrightarrow\dfrac{2^x3^{x-y}}{2}=1\Leftrightarrow2^x.3^{x-y}=2\)

\(\Leftrightarrow2^x.3^{x-y}=2^1.3^0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Ta có 2^{x + 1} . 3^y = 12^x

2^{x + 1} . 3^y = 2^2x . 3^x

⇒ x + 1 = 2x

    x = y 

Vậy x = y = 1

\(C=\dfrac{5122512}{2^2}-512\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\right)\)

Đặt BT trong ngoặc đơn là B

\(\Rightarrow2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)

\(B=2B-B=\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\)

\(\Rightarrow C=\dfrac{5120512+2000}{2^2}-512\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)

\(=\dfrac{512.10001+2^2.500}{2^2}-512\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)

\(=\dfrac{2^9.10001+2^2.500}{2^2}-2^9\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)

\(=2^7.10001+500-2^7+\dfrac{1}{2}=\)

\(=2^7.10000+500+0,5=1280000+500+0,5=1280500,5\)

\(A=x^2-10x+32=x^2-10x+25+9=\left(x-5\right)^2+9\)

mà \(\left(x-5\right)^2\ge0\)

\(\Rightarrow\left(x-5\right)^2+9\ge9\)

\(\Rightarrow Min\left(A\right)=9\)

$A=x^2-10x+32=x^2-10x+25+9={\left(x-5\right)}^2+9$

mà ${\left(x-5\right)}^2\ge 0$

$\Rightarrow {\left(x-5\right)}^2+9\ge 9$

$\Rightarrow Min\left(A\right)=9$

\(\widehat{xAB}+\widehat{BAC}+\widehat{yBC}=180^o\) (1)

xy//BC nên

\(\widehat{xAB}=\widehat{B}\) (góc sole trong) (2)

\(\widehat{yBC}=\widehat{C}\) (góc so le trong) (3)

Từ (1) (2) (3)

\(\Rightarrow\widehat{BAC}+\widehat{B}+\widehat{C}=180^o\)

Ta có: \(35^{25}-35^{24}\)

\(=35^{24}\cdot\left(35-1\right)\)

\(=35^{24}\cdot34\)

Mà: 34 ⋮ 17

\(\Rightarrow35^{24}\cdot34\) ⋮ 17 

Hay \(35^{25}-35^{24}\) ⋮ 17 (đpcm)

\(\left(35^{25}-35^{24}\right)=\left(35.35^{24}-35^{24}\right)=34.35^{24}\)

mà \(34⋮17\)

\(\Rightarrow34.35^{24}⋮17\)

⇒đpcm

\(VT=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{100}\)

\(VP=\dfrac{1}{26}+\dfrac{1}{27}+...\dfrac{1}{50}=\dfrac{2}{26}+\dfrac{2}{28}+...\dfrac{2}{50}...\)

(VP lần lượt triển khai \(\dfrac{1}{26}=\dfrac{2}{26}-\dfrac{1}{26};\dfrac{1}{28}=\dfrac{2}{28}-\dfrac{1}{28}...\))

Tiếp tục \(\dfrac{2}{26}+\dfrac{2}{28}+...\dfrac{2}{50}...=\dfrac{1}{13}+\dfrac{1}{14}+...\dfrac{1}{25}...\)

(VP lần lượt triển khai \(\dfrac{1}{14}=\dfrac{2}{14}-\dfrac{1}{14};\dfrac{1}{16}=\dfrac{2}{16}-\dfrac{1}{16}...\))

Chuyển sang VT để đơn giản phần số đối \(-\dfrac{1}{2};\dfrac{1}{2}...\)

Cuối cùng ta sẽ được \(VT=1;VP=\dfrac{2}{2}\Rightarrow VT=VP\)

⇒Đpcm