c:

ĐKXĐ: 6-5x>=0

=>5x<=6

=>x<=1,2

\(2\sqrt[3]{3x-2}-3\cdot\sqrt{6-5x}+16=0\)

=>\(2\cdot\sqrt[3]{3x-2}+4+12-3\cdot\sqrt{6-5x}=0\)

=>\(2\cdot\left(\sqrt[3]{3x-2}+2\right)+3\left(4-\sqrt{6-5x}\right)=0\)

=>\(2\cdot\frac{3x-2+8}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{16-6+5x}{4+\sqrt{6-5x}}=0\)

=>\(2\cdot\frac{3x+6}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5x+10}{4+\sqrt{6-5x}}=0\)

=>\(\left(2\cdot\frac{3}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5}{4+\sqrt{6-5x}}\right)\left(x+2\right)=0\)

=>x+2=0

=>x=-2(nhận)

d: ĐKXĐ: x>=1

\(\sqrt[3]{x+6}-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2-2\cdot\sqrt{x-1}+2=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}-\left(2-x\right)\left(2+x\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}+\left(x-2\right)\left(2+x\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{2}{1+\sqrt{x-1}}+\left(2+x\right)\right)=0\)

=>x-2=0

=>x=2(nhận)

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

Lời giải:

Đặt $\sqrt[3]{x^2+3x-5}=a; \sqrt[3]{x+2}=b$. Khi đó pt đã cho tương đương với:

$a+b=\sqrt[3]{a^3+b^3-1}+1$

$\Leftrightarrow a+b-1=\sqrt[3]{a^3+b^3-1}$

$\Leftrightarrow (a+b-1)^3=a^3+b^3-1$

$\Leftrightarrow (a+b)^3-3(a+b)^2+3(a+b)-1=a^3+b^3-1$

$\Leftrightarrow 3ab(a+b)-3(a+b)^2+3(a+b)=0$

$\Leftrightarrow ab(a+b)-(a+b)^2+(a+b)=0$

$\Leftrightarrow (a+b)(ab-a-b+1)=0$

$\Leftrightarrow (a+b)(a-1)(b-1)=0$

Nếu $a+b=0\Leftrightarrow \sqrt[3]{x^2+3x-5}=-\sqrt[3]{x+2}$

$\Leftrightarrow x^2+3x-5=-(x+2)$

$\Leftrightarrow x^2+4x-3=0$

$\Leftrightarrow x=-2\pm \sqrt{7}$

Nếu $a-1=0\Leftrightarrow \sqrt[3]{x^2+3x-5}=1$

$\Leftrightarrow x^2+3x-6=0$

$\Leftrightarrow x=\frac{-3\pm \sqrt{33}}{2}$

Nếu $b-1=0\Leftrightarrow \sqrt[3]{x+2}=1$

$\Leftrightarrow x=-1$

1.

ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)

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

\(\Leftrightarrow x^2-4x-3-2\sqrt{\left(x^2-x\right)\left(x+1\right)}=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x}=a>0\\\sqrt{x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow a^2-3b^2-2ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)

\(\Leftrightarrow a=3b\)

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

\(\Leftrightarrow x^2-x=9\left(x+1\right)\)

\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)

2.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:

\(x^3+3\left(x^2-4a^2\right)a=0\)

\(\Leftrightarrow x^3+3ax^2-4a^3=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=x+1\\x^2=4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-4x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=2-2\sqrt{2}\end{matrix}\right.\)

a:ĐKXĐ: 3<=x<=5

Ta có: \(3x^2-17x+24=\sqrt{x-3}+3\sqrt{5-x}\)

=>\(3x^2-12x-5x+20=\sqrt{x-3}-1+3\sqrt{5-x}-3\)

=>\(\left(x-4\right)\left(3x-5\right)=\frac{x-3-1}{\sqrt{x-3}+1}+3\left(\sqrt{5-x}-1\right)\)

=>\(\left(x-4\right)\left(3x-5\right)=\frac{x-4}{\sqrt{x-3}+1}+3\cdot\frac{5-x-1}{\sqrt{5-x}+1}\)

=>\(\left(x-4\right)\left(3x-5-\frac{1}{\sqrt{x-3}+1}+\frac{3}{\sqrt{5-x}+1}\right)=0\)

=>x-4=0

=>x=4(nhận)

b: ĐKXĐ: x>=1

Ta có: \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2+2-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)+\left(x-2\right)\left(x+2\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}+\left(x-2\right)\left(x+2\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+\frac{-2}{1+\sqrt{x-1}}+\left(x+2\right)\right)=0\)

=>x-2=0

=>x=2(nhận)