Question

[Minh Hoàng]

Giải các phương trình sau :

a) \(\sqrt{x-1}+\sqrt{x-4}=\sqrt{x+4}\)

b) \(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=1\)

c) \(5\sqrt{x^2+5x+8}=x^2+5x+4\)

Answer 1

[Trần Trung Nguyên]

a) ĐK:\(x\ge4\)

\(\sqrt{x-1}+\sqrt{x-4}=\sqrt{x+4}\Leftrightarrow x-1+x-4+2\sqrt{\left(x-1\right)\left(x-4\right)}=x+4\Leftrightarrow9-x=2\sqrt{x^2-5x+4}\left(ĐK:x\le9\right)\Leftrightarrow\left(9-x\right)^2=4\left(x^2-5x+4\right)\Leftrightarrow81-18x+x^2=4x^2-20x+16\Leftrightarrow3x^2-2x-65=0\Leftrightarrow3x^2-15x+13x-65=0\Leftrightarrow3x\left(x-5\right)+13\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(3x+13\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\3x+13=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\left(tm\right)\\x=-\dfrac{13}{3}\left(ktm\right)\end{matrix}\right.\)

Vậy S={5}

b)\(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=1\Leftrightarrow\sqrt[3]{2x-1}-1+\sqrt[3]{x-1}=0\Leftrightarrow\dfrac{2x-1-1}{\left(\sqrt[3]{2x-1}\right)^2+2.\sqrt[3]{2x-1}+1}+\dfrac{x-1}{\left(\sqrt[3]{x-1}\right)^2}=0\Leftrightarrow\left(x-1\right)\left[\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}\right]=0\)(1)

Dễ thấy \(\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}>0\)

Vậy (1)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy S={1}

c) ĐK:\(\left[{}\begin{matrix}x\le-4\\x\ge-1\end{matrix}\right.\)

\(5\sqrt{x^2+5x+8}=x^2+5x+4\left(2\right)\)

Đặt a=x²+5x+4(a\(\ge0\))

(2)\(\Leftrightarrow5\sqrt{a+4}=a\Leftrightarrow25\left(a+4\right)=a^2\Leftrightarrow a^2-25a-100=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=\dfrac{25+5\sqrt{41}}{2}\left(tm\right)\\a=\dfrac{25-5\sqrt{41}}{2}\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow a=\dfrac{25+5\sqrt{41}}{2}\Leftrightarrow\dfrac{25+5\sqrt{41}}{2}=x^2+5x+4\Leftrightarrow25+5\sqrt{41}=2x^2+10x+8\Leftrightarrow2x^2+10x-17-5\sqrt{41}=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=3,045972466\left(tm\right)\\x=-8,045972466\left(tm\right)\end{matrix}\right.\)

Vậy S={-8,045972466;3,045972466}

Answer 2

[Trần Trung Nguyên]

c) ĐK:\(\left[{}\begin{matrix}x\le-4\\x\ge-1\end{matrix}\right.\)

\(5\sqrt{x^2+5x+28}=x^2+5x+4\left(1\right)\)

Đặt a=x²+5x+4(a\(\ge0\))

Vậy \(\left(1\right)\Leftrightarrow5\sqrt{a+24}=a\Leftrightarrow25\left(a+24\right)=a^2\Leftrightarrow a^2-25a-600=0\Leftrightarrow a^2-40a+15a-600=0\Leftrightarrow a\left(a-40\right)+15\left(a-40\right)=0\Leftrightarrow\left(a-40\right)\left(a+15\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a-40=0\\a+15=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=40\left(tm\right)\\a=-15\left(ktm\right)\end{matrix}\right.\)

Vậy ta có a=40\(\Leftrightarrow x^2+5x+4=40\Leftrightarrow x^2+5x-36=0\Leftrightarrow x^2-4x+9x-36=0\Leftrightarrow x\left(x-4\right)+9\left(x-4\right)=0\Leftrightarrow\left(x-4\right)\left(x+9\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-4=0\\x+9=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\left(tm\right)\\x=-9\left(tm\right)\end{matrix}\right.\)

Vậy S={-9;4}