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

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

\(S=\frac{\left[\frac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^3+2a\sqrt{a}+b\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)

\(S=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\left(\sqrt{a}\right)^2\sqrt{a}+\left(\sqrt{b}\right)^2\sqrt{b}}{3a^2+3b\sqrt{ab}}+\frac{\sqrt{b}-\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(S=\frac{\left(\sqrt{a}\right)^3-3\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)-\left(\sqrt{b}\right)^3+2\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{3a^2+3b\sqrt{ab}}-\frac{1}{\sqrt{a}+\sqrt{b}}\)

\(S=\frac{3\left(\sqrt{a}\right)^3-3a\sqrt{b}+3\sqrt{a}b}{3a^2+3b\sqrt{ab}}-\frac{1}{\sqrt{a}+\sqrt{b}}\)

\(S=\frac{\sqrt{a}\left(a-\sqrt{ab}+b\right)}{\sqrt{a}\left[\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3\right]}-\frac{1}{\sqrt{a}+\sqrt{b}}\)

\(S=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}\)

\(S=\frac{1}{\sqrt{a}+\sqrt{b}}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)

em ko bieets hu hu

#)Giải :

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

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

\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)

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

#)Giải :

b) Để \(A>-6\Leftrightarrow-2\sqrt{x}>-6\)

\(\Leftrightarrow\sqrt{x}< 3\)

\(\Leftrightarrow x< 9\)

Kết hợp với đkxđ => 0 < x < 9

ĐKXĐ a>0 và a≠1

Rút gọn được A=2+2(a+1)/√a

A=7 → 2+2(a+1)/√a=7→2a-5√a+2=0, giải ra a=4 hoặc a=1/4.

Do a≠1 nên (√a-1)²>0 → a+1>2√a, do đó A>2+2.2√a/√a=6. Vậy A>6 với mọi a>0 và a≠1

Bản trả lời câu a ra hộ mình đi

Viết bằng điện thoại nên hơi khó trình bày, c dựa vào a√a+1=(√a+1)(a-√a+1), a√a-1=(√a-1)(a+√a+1), 

\(VT=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\frac{a+b}{\sqrt{ab}}\right)\)

\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(a-b\right)}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a^2-a\sqrt{ab}-b^2-b\sqrt{ab}-a^2+b^2}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{a+b}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{-\left(a+b\right)\sqrt{ab}}=\sqrt{b}-\sqrt{a}=VP\)

Vậy đẳng thức được chứng minh

Cảm ơn cậu nhiều nha ^^