Question

[Nguyễn Lê Sao Mai]

Bài 1: Thực hiện phép tính:

a) \(\left(\sqrt{6}-\sqrt{10}\right)\) \(\sqrt{4+\sqrt{ }15}\)

b) \(\left(\sqrt{3}+\sqrt{30}\right)\) \(\sqrt{10-\sqrt{41-4\sqrt{10}}}\)

Bài 2:

Cho A = \(\left(\sqrt{a}-\frac{a+2}{\sqrt{a}+1}\right)\) : \(\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}-4}{1-a}\right)\) với a\(\ge\) 0 , a\(\ne\) 1

a) Rút gọn A

b) Tìm a để A = \(\frac{1}{2}\)

Ai giúp tớ giải đề này với :< cảm ơn nhiều lắm luôn <3

Đề này cần làm gấp ạ //-\\

Answer 1

[Thành Đạt]

a,\(\left(\sqrt{6}-\sqrt{10}\right)\sqrt{4+\sqrt{15}}=\sqrt{6}.\sqrt{4-\sqrt{15}}-\sqrt{10}.\sqrt{4+\sqrt{15}}\)

=\(\sqrt{24+6\sqrt{15}}-\sqrt{40+10\sqrt{15}}=\sqrt{\left(\sqrt{15}+3\right)^2}-\sqrt{\left(\sqrt{15}+5\right)^2}\)

=\(\sqrt{15}+3-\sqrt{15}-5=-2\)

b,\(\left(\sqrt{3}+\sqrt{30}\right)\sqrt{10-\sqrt{41-4\sqrt{10}}}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{40-2\sqrt{40}+1}}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{\left(\sqrt{40}-1\right)^2}}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{40}+1}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{11-2\sqrt{10}}=\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{\left(\sqrt{10}-1\right)^2}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\left(\sqrt{10}-1\right)=9\sqrt{3}\)

2,\(A=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)-a-2}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}\left(1-\sqrt{a}\right)-\sqrt{a}+4}{1-a}\right)\)

\(A=\left(\frac{a+\sqrt{a}-a-2}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}-a-\sqrt{a}+4}{1-a}\right)=\left(\frac{\sqrt{a}+2}{\sqrt{a}+1}\right).\left(\frac{1-a}{4-a}\right)\)

\(A=\frac{\sqrt{a}-2}{\sqrt{a}+1}.\frac{a-1}{a-4}=\frac{\sqrt{a}-1}{\sqrt{a}+2}\)

b, ̣để \(A=\frac{1}{2}\Rightarrow\frac{\sqrt{a}-1}{\sqrt{a}+2}=\frac{1}{2}\Leftrightarrow2\sqrt{a}-2=\sqrt{a}+2\Leftrightarrow\sqrt{a}=4\Leftrightarrow a=16\left(t.m\right)\)

Answer 2

[Thành Đạt]

Bạn oi bài 2 hàng A thú 2 phải là \(\frac{\sqrt{a}-2}{\sqrt{a}+1}\) mình nhầm