\(a,Đk:x\ge0\\ PT\Leftrightarrow4x-8\sqrt{x}-7\sqrt{x}+14=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(4\sqrt{x}-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{49}{4}\end{matrix}\right.\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\sqrt{x+1}-\sqrt{3x}+1-4x^2=0\\ \Leftrightarrow\dfrac{1-2x}{\sqrt{x+1}+\sqrt{3x}}+\left(1-2x\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(\dfrac{1}{\sqrt{x+1}+\sqrt{3x}}+2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\\dfrac{1}{\sqrt{x+1}+\sqrt{3x}}+2x+1=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge0\Leftrightarrow\left(1\right)>0\)

Vậy PT có nghiệm \(x=\dfrac{1}{2}\)

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

=>\(\dfrac{3}{2}\sqrt{x-3}=3\)

=>\(\sqrt{x-3}=2\)

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

=>\(\dfrac{3}{2}\sqrt{x-5}=3\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

a) \(x^2-3=0\Leftrightarrow x^2=3\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x^2}=\sqrt{3}\\\sqrt{x^2}=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{x}\end{matrix}\right.\) vậy \(x=\sqrt{3};x=-\sqrt{3}\)

b) \(x-9\sqrt{x+14}=0\) \(\Leftrightarrow\) \(x=9\sqrt{x+14}\) \(\Leftrightarrow\) \(x^2=81\sqrt{\left(x+14\right)^2}\)

th1 : \(x\ge-14\) thì \(x^2=81\sqrt{\left(x+14\right)^2}\) \(\Leftrightarrow\) \(x^2=81\left(x+14\right)\)

\(\Leftrightarrow\) \(x^2=81x+1134\) \(\Leftrightarrow\) \(x^2-81x-1134=0\)

giải phương trình ra ta có : \(\left\{{}\begin{matrix}x_1=\dfrac{81+9\sqrt{137}}{2}\left(tmđk\right)\\x_2=\dfrac{81-9\sqrt{137}}{2}\left(loại\right)\end{matrix}\right.\)

th2 : \(x< -14\) thì \(x^2=81\sqrt{\left(x+14\right)^2}\) \(\Leftrightarrow\) \(x^2=81\left(-x-14\right)\)

\(\Leftrightarrow\) \(x^2=-81x-1134\) \(\Leftrightarrow\) \(x^2+81x+1134=0\)

giải phương trình ta được : \(\left\{{}\begin{matrix}x_1=-18\left(tmđk\right)\\x_2=-63\left(tmđk\right)\end{matrix}\right.\)

vậy \(x=\dfrac{81+9\sqrt{137}}{2};x=-18;x=-63\)

c) \(4x-\sqrt{x^2}-4x+4=0\)

th1 : \(x\ge0\) thì \(4x-\sqrt{x^2}-4x+4=0\) \(\Leftrightarrow\) \(4x-x-4x+4=0\)

\(\Leftrightarrow\) \(-x+4=0\Leftrightarrow x=4\)

th2 : \(x< 0\) thì \(4x-\sqrt{x^2}-4x+4=0\) \(\Leftrightarrow\) \(4x+x-4x+4=0\)

\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

vậy \(x=-4;x=4\)

h: \(\sqrt{18x}+\sqrt{32x}-14=0\)

\(\Leftrightarrow7\sqrt{2x}=14\)

hay x=2

a) \(\sqrt{4x^2-4x+1}=x-1\)(ĐKXĐ : \(x\ge1\))

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

Vậy phương trình vô nghiệm.

b) \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)

\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}+3=0\)

\(\Leftrightarrow3=0\)(vô lí)

Vậy phương trình vô nghiệm.

c) \(\sqrt{x-5}+\frac{14-x}{3+\sqrt{x-5}}=3\)(ĐKXĐ : \(x\ge5\))

\(\Leftrightarrow3\sqrt{x-5}+x-5+14-x=9+3\sqrt{x-5}\)

\(\Leftrightarrow9=9\)(luôn đúng)

Vậy phương trình luôn luôn có nghiệm với mọi \(\hept{\begin{cases}x\in R\\x\ge5\end{cases}}\)

b:

ĐKXĐ: x>0

 \(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)

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

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

=>x=1

a) ta có \(\sqrt{12x^2+12x+19}+\sqrt{20x^2+20x+14}=-4x^2-4x+6\)

ta có : \(VT\ge\sqrt{16}+\sqrt{9}=7\) và \(VT\le7\)

\(\Rightarrow VT=VP\) \(\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)

b) điều kiện \(x>0\)

ta có : \(\left(x+\dfrac{1}{x}\right)-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)

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

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

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

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

vậy \(x=1\)

Mysterious Person giup mk nha

b:

ĐKXĐ: x>0

 \(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)

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

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

=>x=1