\(250\times12-\left(242+302\times2,5\right)\)

\(=3000-997\)

\(=2003\)

a)

\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)

b)

\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).

ĐK: \(x-9\ne0\Rightarrow x\ne9\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)

\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)

2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)

\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)

\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)

\(P=3\cdot\dfrac{2}{3}-5\cdot\sqrt{\dfrac{2}{5}}+25\cdot\dfrac{6}{25}=2+6-\sqrt{10}=8-\sqrt{10}\)

Thay \(x=\sqrt{\frac{2}{3}};y=\sqrt{\frac{6}{25}}\) vào biểu thức P ta được:

\(P=3\left(\sqrt{\frac{2}{3}}\right)^2-5\sqrt{\sqrt{\frac{2}{3}}.\sqrt{\frac{6}{25}}}+25\left(\sqrt{\frac{6}{25}}\right)^2\)

\(P=3.\frac{2}{3}-\sqrt{25.\sqrt{\frac{2}{3}}.\sqrt{\frac{6}{25}}}+25.\frac{6}{25}\)

\(P=2-\sqrt{\sqrt{25^2}.\sqrt{\frac{2}{3}}.\sqrt{\frac{6}{25}}}+6\)

\(P=8-\sqrt{\sqrt{25^2.\frac{2}{3}.\frac{6}{25}}}\)

\(P=8-\sqrt{\sqrt{100}}\)

\(P=8-\sqrt{10}\)

Bài này cũng dễ

Chỉ cần thay vào là dc mừ

Sao lại vào câu hỏi hay

hi mình là kudo skibidi

\(\frac{19.5^{22}-5^{13}-25^{18}}{\left(7.5^{17}\right)^2}\)  \(=\frac{19.5^{22}-5^{13}-5^{36}}{7^2.5^{34}}\)   \(=\frac{5^{13}\left(19.5^9-1-5^{23}\right)}{7^2.5^{34}}\)

\(=\frac{19.5^9-1-5^{23}}{7^2.5^{21}}\)

Hình như đề bài có vấn đề bn ak

Xin lỗi bạn mình chép đề bài đúng  mà