\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

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

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

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

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

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(a)ĐK:x\ge0\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x=0\left(tm\right)\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=0

\(b)ĐK:x\ge2+\sqrt{7}\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-4x-3=(x-2)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x-3=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\-3=4\end{matrix}\right.\)(vô lý)

Vậy pt vô nghiệm

bn thiếu điều kiện kìa

\(\sqrt[3]{x}+\sqrt[3]{2x-3}=\sqrt[3]{12\left(x-1\right)}\left(1\right)\)

\(\left(1\right)\Leftrightarrow x+2x-3+3.\sqrt[3]{x\left(2x-3\right)}.\left(\sqrt[3]{x}+\sqrt[3]{2x-3}\right)=12x-12\)

\(\Rightarrow\sqrt[3]{12x\left(x-1\right)\left(2x-3\right)}=3x-3\)

\(\Leftrightarrow12x\left(x-1\right)\left(2x-3\right)=[3\left(x-1\right)]^3\)

\(\Leftrightarrow12x\left(2x^2-5x+3\right)=27\left(x^3-3x^2+3x-1\right)\)

\(\Leftrightarrow24x^3-60x^2+36x=27x^3-81x^2+81x-27\)

\(\Leftrightarrow3x^3-21x^2+45x-27=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\). Thử lại ta thấy cả x=1;x=3 đều t/m bài toán

Vậy, pt có tập nghiệm là S={1;3}