a, ĐK: \(x\ge1\)

Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=2\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a-b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=b+2\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)

TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)

\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x-1}-1=1\\2\sqrt{x-1}-1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-1}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

f) \(\left(\sqrt{6x+1}-\sqrt{6x-1}\right)^2=\left(\sqrt{6x+1}\right)^2-2\sqrt{\left(6x+1\right)\left(6x-1\right)}+\left(\sqrt{6x-1}\right)^2\)

\(=6x+1+6x-1-2\sqrt{36x^2-1}=12x-2\sqrt{36x^2-1}\)

tương tự các câu khác mình làm tắt chút nha:

c) \(\left(\sqrt{2x+3}+\sqrt{2x-3}\right)^2=2x+3+2x-3-2\sqrt{\left(2x+3\right)\left(2x-3\right)}=4x+2\sqrt{4x^2-9}\)

d) \(\left(\sqrt{2x+y}+\sqrt{2x-y}\right)^2=2x+y+2x-y-2\sqrt{\left(2x+y\right)\left(2x-y\right)}=4x-2\sqrt{4x^2-y^2}\)

\(\left(\sqrt{5x-2}-\sqrt{5x+2}\right)^2=5x-2+5x+2-2\sqrt{\left(5x-2\right)\left(5x+2\right)}=10x-2\sqrt{25x^2-4}\)

b: Đặt \(x^2+5x+4=a\)

\(\Leftrightarrow a=5\sqrt{a+24}\)

\(\Leftrightarrow a^2=25a+600\)

\(\Leftrightarrow a^2-25a-600=0\)

\(\Leftrightarrow\left(a-40\right)\left(a+15\right)=0\)

\(\Leftrightarrow a=-15\)

hay S=∅

Đặt \(\sqrt{5x^2+6x+5}=a,4x=b\left(a\ge0\right)\)

Khi đó Pt

<=> \(a\left(a^2+1\right)=b\left(b^2+1\right)\)

<=>\(\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

MÀ \(a^2+ab+b^2+1>0\)

=> \(a=b\)

=> \(\sqrt{5x^2+6x+5}=4x\)

=> \(\hept{\begin{cases}x\ge0\\11x^2-6x-5=0\end{cases}}\)

=>\(x=1\)

Vậy x=1