1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

b. \(\sqrt{x-4}+\sqrt{x^2-3x+4}=x\)

(ĐKXĐ: \(x\ge4\))

\(\Leftrightarrow\sqrt{x^2-3x+4}=x-\sqrt{x-4}\)

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

\(\Leftrightarrow x^2-3x+4-x^2-x+4+2\sqrt{x^2-4x}=0\Leftrightarrow-4x+8+2\sqrt{x^2-4x}=0\Leftrightarrow-2\left(2x-4-\sqrt{x^2-4x}\right)=0\Leftrightarrow2x-4-\sqrt{x^2-4x}=0\Leftrightarrow\sqrt{x^2-4x}=2x-4\Leftrightarrow x^2-4x=4x^2+16-16x\Leftrightarrow x^2-4x^2-4x+16x-16=0\Leftrightarrow-3x^2+12x-16=0\Leftrightarrow3x^2-12x+16=0\)

Ta có: \(\Delta=b^2-4ac=\left(-12\right)^2-4.3.16=-48< 0\)

=> pt vô nghiệm.

Vậy pt đã cho vô nghiệm.

ĐK : x > 3/2

Đặt \(\sqrt{3x-2}=a\left(a>0\right)\) . Khi đó pt thành :

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

hay \(\sqrt{3x-2}\left(x-1\right)+x^2-3x+2=0\Leftrightarrow\left(\sqrt{3x-2}-1\right)\left(x-1\right)+x^2-2x+1=0\Leftrightarrow\dfrac{3x-3}{\sqrt{3x-2}+1}\left(x-1\right)+\left(x-1\right)^2=0\Leftrightarrow\dfrac{3\left(x-1\right)^2}{\sqrt{3x-2}+1}+\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2\left(\dfrac{3}{\sqrt{3x-2}+1}+1\right)=0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)

Vì \(\dfrac{3}{\sqrt{3x-2}+1}+1>0\)

Vậy nghiệm của pt là x = 1

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

ĐKXĐ: \(x>1\)

\(3x^2+1=4\)

\(3x^2=3\)

\(x^2=1\)

\(x=\pm1\)

=> Pt vô nghiệm

b) ĐKXĐ: x>-4

\(x^2+3x+4=x+4\)

\(x^2+2x=0\)

\(x\left(x+2\right)=0\)

\(\left[{}\begin{matrix}x=0\\x+2=0\Leftrightarrow x=-2\end{matrix}\right.\)

c) Đkxđ: \(3x-2>0\Leftrightarrow x>\dfrac{2}{3}\)
Pt \(\Leftrightarrow3x^2-x-2=3x-2\)
\(\Leftrightarrow3x^2-4x=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(l\right)\\x=\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{4}{3}\) là nghiệm của phương trình.
d) Đkxđ: \(x\ne1\)
\(2x+3+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\) \(\Leftrightarrow\dfrac{\left(2x+3\right)\left(x-1\right)}{x-1}+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)+4=x^2+3\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy \(x=-2\) là nghiệm của phương trình.

a)\(\sqrt{2x^2+x+6}+\sqrt{x^2+x+2}=x+\dfrac{4}{x}\)

\(pt\Leftrightarrow\sqrt{2x^2+x+6}-3+\sqrt{x^2+x+2}-2=x+\dfrac{4}{x}-5\)

Liên hợp quy đồng nốt