a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

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

\(\Leftrightarrow x-1=1\)

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

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=12\\ \Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=9\\ \Leftrightarrow2x=81\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)

a) ĐK: \(x\ge0\)

PT \(\Leftrightarrow\sqrt{4x}\left(\dfrac{3}{4}-1-\dfrac{1}{4}\right)+5=0\)

\(\Leftrightarrow2\sqrt{x}.\left(-\dfrac{1}{2}\right)+5=0\)

\(\Leftrightarrow x=25\) (thỏa)

Vậy \(x=25\)

b) Đk: \(x\le3\)

PT \(\Leftrightarrow\sqrt{3-x}-\sqrt{9\left(3-x\right)}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}=6\)

\(\Leftrightarrow\sqrt{3-x}\left(1-\sqrt{9}+\dfrac{5}{4}.\sqrt{16}\right)=6\)

\(\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow x=-1\) (thỏa)

Vậy \(x=-1\)

2:

a: 

Sửa đề: \(P=\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{\sqrt{1-a^2}}+1\right)\)

\(P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)

\(=\dfrac{2+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{1-a^2}}{2+\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}\)

\(=\sqrt{1-a}\)

b: Khi a=24/49 thì \(P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\)

c: P=2

=>1-a=4

=>a=-3

1a (đkxđ:\(x\ge0\)) \(\Leftrightarrow\dfrac{-1}{2}.\sqrt{4x}+5=0\) \(\Leftrightarrow\sqrt{4x}=10\) \(\Leftrightarrow x=25\) (t/m)

b (đkxđ:\(x\le3\) ) \(\Leftrightarrow\sqrt{3-x}\left(1-3+1,25.4\right)=6\) \(\Leftrightarrow\sqrt{3-x}=2\) \(\Leftrightarrow x=-1\) (t/m)

a:

ĐKXĐ: \(x>=-2\)

\(1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)

=>\(1+\sqrt{\left(x+2\right)\left(x+5\right)}=\sqrt{x+5}+\sqrt{x+2}\)

Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\)(ĐK: a>0 và b>0)

Phương trình sẽ trở thành:

1+ab=a+b

=>ab-a-b+1=0

=>a(b-1)-(b-1)=0

=>(b-1)(a-1)=0

=>\(\left\{{}\begin{matrix}a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)

=>\(\left\{{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\)

=>\(x\in\varnothing\)

b: \(\sqrt{4x^2-2x+\dfrac{1}{4}}=4x^3-x^2+8x-2\)

=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)

=>\(\sqrt{\left(2x-\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)

=>\(\left|2x-\dfrac{1}{2}\right|=4x^3-x^2+8x-2\)(1)

TH1: x>=1/4

\(\left(1\right)\Leftrightarrow4x^3-x^2+8x-2=2x-\dfrac{1}{2}\)

=>\(4x^3-x^2+6x-\dfrac{3}{2}=0\)

=>\(x^2\left(4x-1\right)+1,5\left(4x-1\right)=0\)

=>\(\left(4x-1\right)\left(x^2+1,5\right)=0\)

=>4x-1=0

=>x=1/4(nhận)

TH2: x<1/4

Phương trình (1) sẽ trở thành:

\(4x^3-x^2+8x-2=-2x+\dfrac{1}{2}\)

=>\(x^2\left(4x-1\right)+2\left(4x-1\right)+0,5\left(4x-1\right)=0\)

=>\(\left(4x-1\right)\cdot\left(x^2+2,5\right)=0\)

=>4x-1=0

=>x=1/4(loại)

a, \(\sqrt[3]{\dfrac{2x}{x+1}}.\sqrt[3]{\dfrac{x+1}{2x}}=2\)

⇔ \(\left\{{}\begin{matrix}1=2\\x\ne0\&x\ne-1\end{matrix}\right.\)

Phương trình vô nghiệm

b, x = \(\dfrac{8}{125}\)