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

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

\(-\dfrac{3}{16}+-\dfrac{3}{8}-\dfrac{5}{4}\)

\(=-\dfrac{3}{16}-\dfrac{3}{8}-\dfrac{5}{4}\)

\(=-\dfrac{3}{16}-\dfrac{6}{16}-\dfrac{20}{16}\)

\(=\dfrac{-3-6-20}{16}\)

\(=-\dfrac{29}{16}\)

\(\left(-\dfrac{25}{13}\right)+\dfrac{-9}{17}+\dfrac{12}{13}+\dfrac{25}{17}\)

\(=\left[\left(-\dfrac{25}{13}\right)+\dfrac{12}{13}\right]+\left[\left(-\dfrac{9}{17}+\dfrac{25}{17}\right)\right]\)

\(=-\dfrac{13}{13}+\dfrac{17}{17}\)

\(=-1+1\)

\(=0\)

giúp với

giúp với

\(-\dfrac{1}{12}+\left(-\dfrac{5}{6}\right)-\dfrac{4}{3}=-\dfrac{1}{12}-\dfrac{5}{6}-\dfrac{4}{3}=-\dfrac{1}{12}-\dfrac{10}{12}-\dfrac{16}{12}=-\dfrac{27}{12}=-\dfrac{9}{4}\)

\(6,5-\left(-\dfrac{1}{5}\right)=\dfrac{65}{10}+\dfrac{1}{5}=\dfrac{65}{10}+\dfrac{2}{10}=\dfrac{67}{10}\)

giúp với

\(-\dfrac{18}{10}+0,4=-\dfrac{18}{10}+\dfrac{4}{10}=-\dfrac{14}{10}=-\dfrac{7}{5}\)

giúp với

`@` `\text {Ans}`

`\downarrow`

`-1/8 - 3/20`

`= -5/40 - 6/40`

`= -11/40`

\(-\dfrac{1}{8}-\dfrac{3}{20}=-\dfrac{5}{40}-\dfrac{6}{40}=-\dfrac{11}{40}\)