\(\left(\sqrt[3]{3}+\sqrt[3]{2}\right)^3=\left(\sqrt[3]{3}\right)^3+3\left(\sqrt[3]{3}\right)^2\sqrt[3]{2}+3\sqrt[3]{3}\left(\sqrt[3]{2}\right)^2+\left(\sqrt[3]{2}\right)^3\)

\(=3+3\sqrt[3]{18}+3\sqrt[3]{12}+2=5+3\sqrt[3]{18}+3\sqrt[3]{12}\)

10.7: Nối a-3, b-2, c-4, d-1.

10.8: Đáp án E.

Để \(\sqrt{-5x}\) có nghĩa \(\Leftrightarrow-5x\ge0\Leftrightarrow x\le0\).

Thanks.

1. There are a lot of vegetables and fruits in his farm.

2. The living room has everything.

a) Chiều cao là: \(1500:120=12,5\left(cm\right)\)

b) Chiều rộng là: \(12,5:2=6,25\left(cm\right)\)

Chiều dài là: \(120:6,25=19,2\left(cm\right)\)

Diện tích xung quanh là: \(S_{xq}=\left(19,2+6,25\right)\cdot2\cdot12,5=636,25\left(cm^2\right)\)

Diện tích toàn phần là: \(S_{tp}=636,25+120\cdot2=876,25\left(cm^2\right)\)

\(\dfrac{-1}{24}-\left[\dfrac{1}{4}-\left(\dfrac{1}{2}-\dfrac{7}{8}\right)\right]\)

\(=\dfrac{-1}{24}-\left[\dfrac{6}{24}-\left(\dfrac{12}{24}-\dfrac{21}{24}\right)\right]\)

\(=\dfrac{-1}{24}-\left[\dfrac{6}{24}-\left(-\dfrac{9}{24}\right)\right]\)

\(=\dfrac{-1}{24}-\dfrac{15}{24}=\dfrac{-16}{24}=\dfrac{-2}{3}\)

a) Với \(0< a\ne1\), ta có:

\(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(\dfrac{1}{\sqrt{a}}+1\right)\)

\(=\left[\dfrac{1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]\cdot\left(\dfrac{1}{\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}}\right)\)

\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1+\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{1+\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2}{1-\sqrt{a}}\)

Vậy \(P=\dfrac{2}{1-\sqrt{a}}\).

b) Ta có: \(9+4\sqrt{2}=8+4\sqrt{2}+1=\left(\sqrt{8}+1\right)^2\).

Thay \(a=\left(\sqrt{8}+1\right)^2\) (TMĐK) vào P, ta có:

\(P=\dfrac{2}{1-\sqrt{\left(\sqrt{8}-1\right)^2}}=\dfrac{2}{1-\sqrt{8}-1}=\dfrac{2}{\sqrt{8}}=\dfrac{\sqrt{2}}{2}\)

Vậy \(P=\dfrac{\sqrt{2}}{2}\) khi \(a=9+4\sqrt{2}\).

c) Để \(P>\dfrac{1}{2}\) thì:

\(\dfrac{2}{1-\sqrt{a}}>\dfrac{1}{2}\Leftrightarrow\dfrac{4}{2\left(1-\sqrt{a}\right)}>\dfrac{1-\sqrt{a}}{2\left(1-\sqrt{a}\right)}\)

\(\Leftrightarrow\dfrac{4}{2\left(1-\sqrt{a}\right)}-\dfrac{1-\sqrt{a}}{2\left(1-\sqrt{a}\right)}>0\)

\(\Leftrightarrow\dfrac{3+\sqrt{a}}{2\left(1-\sqrt{a}\right)}>0\) 

Vì \(a>0\Rightarrow\sqrt{a}>0\Rightarrow3+\sqrt{a}>0\)

Mà \(\dfrac{3+\sqrt{a}}{2\left(1-\sqrt{a}\right)}>0\Rightarrow1-\sqrt{a}>0\) 

\(\Leftrightarrow1>\sqrt{a}\Leftrightarrow a< 1\)

Kết hợp điều kiện: \(0< a< 1\).

Vậy để \(P>\dfrac{1}{2}\) thì \(0< a< 1\).

\(1\dfrac{4}{5}=-0,15-x\)

\(x=-0,15-1\dfrac{4}{5}\)

\(x=-0,15-1,8\)

\(x=-1,95\)

Phân tử khối của khí A là: \(23\times2=46\) (đvC) 

Khí A là khí NO₂.