Question

[Đoàn Thị Thanh Loan]

Giải phương trình

1.\(\sqrt{x+4-4\sqrt{x}}+\sqrt{x+9-6\sqrt{x}}=1\)

2. \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

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

4. \(\left(x^2+3x-4\right).\left(x^2+x-6\right)=24\)

Answer 1

[Y]

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

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

+ Ta có : \(\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|\ge\left|\sqrt{x}-2+3-\sqrt{x}\right|=1\)

Dấu "=" \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)

\(\Leftrightarrow2\le\sqrt{x}\le3\Leftrightarrow4\le x\le9\)

2. + \(ĐK:4-2x-x^2\ge0\)

+ VT = \(\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}\)

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

Dấu "=" \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

+ VP \(=-\left(x^2+2x+1\right)+5=-\left(x+1\right)^2+5\le5\forall x\) (2)

Dấu "=" \(\Leftrightarrow x=-1\)

+ Từ (1) và (2) suy ra : pt \(\Leftrightarrow VT=VP=5\Leftrightarrow x=-1\) (TM)

3. + TH1: \(x< 0\) ta có :

\(VT< \sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )

+ TH2 : x = 0 ta có :

\(VT=\sqrt[3]{1}+\sqrt[3]{0}=1\) ( TM )

+ TH3 : x > 0 ta có :

\(VT>\sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )

Vậy x = 0 là nghiệm duy nhất của pt

4. \(\Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+3\right)-24=0\)

\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x-8\right)-24=0\)

\(\Leftrightarrow t\left(t-5\right)-24=0\) ( với \(t=x^2+2x-3\) )

\(\Leftrightarrow t^2-5t-24=0\Leftrightarrow\left(t+3\right)\left(t-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-3\\t=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=-3\\x^2+2x-3=8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\\left(x+1\right)^2=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=2\sqrt{3}-1\\x=-2\sqrt{3}-1\end{matrix}\right.\) ( TM )