mình chỉ biết mỗi kq rút gọn thôi còn chi tiết thì mình ko rõ lắm

\(a,\sqrt{75}+2\sqrt{3}-2\sqrt{7}\\ =\sqrt{25\cdot3}+2\sqrt{3}-2\sqrt{7}\\ =5\sqrt{3}+2\sqrt{3}-2\sqrt{7}\\ =7\sqrt{3}-2\sqrt{7}\)

\(b,\sqrt{\left(4-\sqrt{7}\right)^2}-\sqrt{63}\\ =\left|4-\sqrt{7}\right|-\sqrt{9\cdot7}\\ =4-\sqrt{7}-3\sqrt{7}\\ =4-4\sqrt{7}\)

\(c,\dfrac{3}{\sqrt{5}+3}-\dfrac{\sqrt{5}}{\sqrt{5}-3}\\ =\dfrac{3\left(\sqrt{5}-3\right)}{5-3}-\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{5-3}\\ =\dfrac{3\sqrt{5}-9-5-3\sqrt{5}}{2}\\ =\dfrac{-14}{2}\\ =-7\)

\(\Rightarrow A^3=5\sqrt{2}-7-3\sqrt[3]{5\sqrt{2}-7}^2.\sqrt[3]{5\sqrt{2}+7}+3\sqrt[3]{5\sqrt{2}-7}.\sqrt[3]{5\sqrt{2}+7}^2-5\sqrt{2}-7=-14-3.\sqrt[3]{\left(5\sqrt{2}-7\right)\left(5\sqrt{2}+7\right)}\left[\sqrt[3]{5\sqrt{2}-7}-\sqrt[3]{5\sqrt{2}+7}\right]=-14-3\sqrt[3]{1}.A=-14-3A\)

\(\Rightarrow A^3=-14-3A\Leftrightarrow A^3+3A+14=0\Leftrightarrow\left(A+2\right)\left(A^2-2A+7\right)=0\Leftrightarrow\left[{}\begin{matrix}A=-2\\A^2-2A+7>0\left(loại\right)\end{matrix}\right.\)

a: \(\dfrac{x^2-7}{x+\sqrt{7}}=x-\sqrt{7}\)

b: \(\dfrac{x^2-5}{x-\sqrt{5}}=x+\sqrt{5}\)

a) Ta có:

\(A=\left(a-4\right)\left(a+5\right)-\left(a-5\right)\left(a+4\right)\)

\(=\left[\left(a-4\right)a+5\left(a-4\right)\right]-\left[\left(a-5\right)a+4\left(a-5\right)\right]\)

\(=\left[a^2-4a+5a-20\right]-\left[a^2-5a+4a-20\right]\)

\(=a^2-4a+5a-20-a^2+5a-4a+20\)

\(=\left(a^2-a^2\right)+\left(-4a+5a+5a-4a\right)+\left(-20+20\right)\)

\(=0+2a+0\)

\(=2a\)

b) Ta có:

\(B=\left(2-a\right)\left(a+7\right)-\left(a-1\right)\left(a+2\right)\)

\(=\left[\left(2-a\right)a+7\left(2-a\right)\right]-\left[\left(a-1\right)a+2\left(a-1\right)\right]\)

\(=\left[2a-a^2+14-7a\right]-\left[a^2-a+2a-2\right]\)

\(=2a-a^2+14-7a-a^2+a-2a+2\)

\(=\left(2a-7a+a-2a\right)-\left(a^2+a^2\right)+\left(14+2\right)\)

\(=-6a-2a^2+16\)

đi mk giải cho 100 % luôn ko lừa đâu vì mk học lớp 7 mà đối với mk  bài này dễ lắm