\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

Theo đề: \(\sqrt[3]{x^3+5x^2}-1=\sqrt{\frac{5x^2-2}{6}}\)

\(\Rightarrow\sqrt[3]{x^3+5x^2}=1+\sqrt{\frac{5x^2-2}{6}}\)

\(Đkxđ:x^2\ge\frac{2}{5}\)

Đặt: \(\hept{\begin{cases}\sqrt[3]{x^3+5x^2}=u\\\sqrt{\frac{5x^2-2}{6}}=v\ge0\end{cases}}\)

Ta được: \(\hept{\begin{cases}x^3+5x^2=u^3\\5x^2-2=6v^2\Rightarrow x^3+2=\left(v-1\right)^3+2\Leftrightarrow x=v-1\\u=1+v\end{cases}}\)

Từ trên ta giải được nghiệm: \(x=-6+2\sqrt{7}\)

Ta có:

\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge\sqrt{9}+\sqrt{4}=5\)

\(3-4x-2x^2=5-2\left(x+1\right)^2\le5\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}3\left(x+1\right)^2=0\\5\left(x^2-1\right)^2=0\\2\left(x+1\right)^2=0\end{matrix}\right.\) \(\Rightarrow x=-1\)

Vậy pt có nghiệm duy nhất \(x=-1\)

ĐKXĐ: \(x\ge-\dfrac{1}{2}\)

\(4x^3+4x^2-5x+9=4\sqrt[4]{\left(2x+1\right).2.2.2}\le2x+1+2+2+2\)

\(\Leftrightarrow4x^3+4x^2-7x+2\le0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)^2\le0\)

\(\Leftrightarrow\left(2x-1\right)^2\le0\) (do \(x+2>0\) ; \(\forall x\ge-\dfrac{1}{2}\))

\(\Rightarrow x=\dfrac{1}{2}\)

Vậy pt có nghiệm duy nhất \(x=\dfrac{1}{2}\)

a: Ta có: \(\sqrt{4x^2+4x+3}=8\)

\(\Leftrightarrow4x^2+4x+1+2-64=0\)

\(\Leftrightarrow4x^2+4x-61=0\)

\(\Delta=4^2-4\cdot4\cdot\left(-61\right)=992\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-4-4\sqrt{62}}{8}=\dfrac{-1-\sqrt{62}}{2}\\x_2=\dfrac{-4+4\sqrt{62}}{8}=\dfrac{-1+\sqrt{62}}{2}\end{matrix}\right.\)

PT<=>\(\sqrt{5x+7}=4\sqrt{x+3}\)

<=> \(\begin{cases}x\ge-\frac{7}{4}\\5x+7=16x+48\end{cases}\)

<=> \(\begin{cases}x\ge-\frac{7}{4}\\x=-\frac{41}{11}\end{cases}\)

=> PTVN

\(ĐK:\begin{cases}5x+7\ge0\\x+3>0\end{cases}\) \(\Leftrightarrow\begin{cases}x\ge-\frac{7}{5}\\x>-3\end{cases}\) \(\Leftrightarrow x\ge-\frac{7}{5}\)

\(\frac{\sqrt{5x+7}}{\sqrt{x+3}}=4\)

\(\Leftrightarrow\)\(\frac{5x+7}{x+3}=16\)

\(\Leftrightarrow16\left(x+3\right)=5x+7\)

\(\Leftrightarrow16x+48=5x+7\)

\(\Leftrightarrow16x-5x=7-48\)

\(\Leftrightarrow11x=-41\)

\(\Leftrightarrow x=\frac{-41}{11}\left(KTM\right)\)

Vậy pt vô nghiệm

\(\frac{\sqrt{5x+7}}{\sqrt{x+3}}\) = 4 (1)

đkxđ: \(\begin{cases}5x+7\ge0\\x+3\ge0\end{cases}\Leftrightarrow\begin{cases}x\ge\frac{7}{5}\\x\ge-3\end{cases}}\)

(1) \(\Leftrightarrow\) \(\frac{5x+7}{x+3}\)= 16

      \(\Leftrightarrow\) 5x +7 = 16x+48

      \(\Leftrightarrow\) x = \(\frac{-41}{11}\) (L)

Vậy pt vô nghiệm