Lời giải:

ĐK: \(x\geq 1\)

Áp dụng BĐT Cauchy-Schwarz:

\((16x)^2=(13\sqrt{x-1}+5\sqrt{x+1})^2\leq (13+5)(13x-13+5x+5)\)

\(\Leftrightarrow 256x^2\leq 18(18x-8)\)

\(\Leftrightarrow 64x^2-81x+36\leq 0\)

Điều này hiển nhiên vô lý vì \(64x^2-81x+36=(8x-\frac{81}{16})^2+\frac{2655}{256}>0\)

Do đó PT vô nghiệm.

ĐKXĐ: \(-1\le x\le1\).

Đặt \(x^2=a\left(0\le a\le1\right)\).

PT đã cho được viết lại thành:

\(13\sqrt{a-a^2}+9\sqrt{a+a^2}=16\).

Áp dụng bất đẳng thức AM - GM cho hai số thực không âm ta có:

\(a+4\left(1-a\right)\ge2\sqrt{a.4\left(1-a\right)}\)

\(\Rightarrow\sqrt{a-a^2}\le1-\dfrac{3}{4}a\)

\(\Rightarrow13\sqrt{a-a^2}\le13-\dfrac{39}{4}a\); (1)

\(a+\dfrac{4}{9}\left(a+1\right)\ge2\sqrt{a.\dfrac{4}{9}\left(a+1\right)}\)

\(\Rightarrow\sqrt{a\left(a+1\right)}\le\dfrac{13}{12}a+\dfrac{1}{3}\)

\(\Rightarrow9\sqrt{a+a^2}\le\dfrac{39a}{4}+3\). (2)

Cộng vế với vế của (1), (2) ta có \(13\sqrt{a-a^2}+9\sqrt{a+a^2}\le16\).

Mặt khác từ pt đã cho ta có đẳng thức phải xảy ra.

Do đó đẳng thức ở (1) và (2) cũng xảy ra

\(\Leftrightarrow\left\{{}\begin{matrix}a=4\left(1-a\right)\\a=\dfrac{2}{3}\left(1+a\right)\end{matrix}\right.\Leftrightarrow a=\dfrac{4}{5}\Leftrightarrow x=\pm\sqrt{\dfrac{4}{5}}\) (TMĐK).

Vậy...

\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}+\frac{\sqrt{x}-10}{x-4}\)

\(A=\frac{x+2\sqrt{x}+x-3\sqrt{x}+2+\sqrt{x}-10}{x-4}\)

\(A=\frac{2x-8}{x-4}=\frac{2\left(x-4\right)}{x-4}=2\)

\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)

\(B=43+24\sqrt{3}-8\sqrt{20+6\sqrt{3}+8}\)

\(B=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)

\(B=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)

\(B=43+24\sqrt{3}-24\sqrt{3}-8\)

\(B=35\)

Nguyễn Việt Lâm giúp mk nhá, tks bn nhìu :>>

b,\(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}-16\sqrt{x+1}=0\) (dk \(x\ge-1\)

\(\Leftrightarrow\sqrt{x+1}\left(4-3+2-16\right)=0\)

\(\Leftrightarrow\sqrt{x+1}.-13=0\)

\(\Leftrightarrow x=-1\)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)