Question

Giải pt dạng |A|=B và |A|=|B|

1. |x²-1|=1-4x

2. |4x+1|=x²+2x-4

3. |3x-5|=2x²+x-3

4. x²-2x+8=|x²-1|

5. x²+5x-|3x-2|-5=0

6. x²-5|x-1|-1=0

7. |3x²-2|=|6-x²|

Answer 1

a/ - Với \(x>\frac{1}{4}\) PT vô nghiêm

- Với \(x\le\frac{1}{4}\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(1-4x\right)^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x^2+4x-2=0\\x^2-4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{6}\left(l\right)\\x=-2-\sqrt{6}\\x=4\left(l\right)\\x=0\end{matrix}\right.\)

2.

- Với \(x\ge-\frac{1}{4}\Leftrightarrow4x+1=x^2+2x-4\)

\(\Leftrightarrow x^2-2x-5=0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{6}\\x=1-\sqrt{6}\left(l\right)\end{matrix}\right.\)

- Với \(x< -\frac{1}{4}\)

\(\Leftrightarrow-4x-1=x^2+2x-4\)

\(\Leftrightarrow x^2+6x-3=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3+2\sqrt{3}\left(l\right)\\x=-3-2\sqrt{3}\end{matrix}\right.\)

Answer 2

3.

- Với \(x\ge\frac{5}{3}\)

\(\Leftrightarrow3x-5=2x^2+x-3\)

\(\Leftrightarrow2x^2-2x+2=0\left(vn\right)\)

- Với \(x< \frac{5}{3}\)

\(\Leftrightarrow5-3x=2x^2+x-3\)

\(\Leftrightarrow2x^2+4x-8=0\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)

4. Do hai vế của pt đều không âm, bình phương 2 vế:

\(\Leftrightarrow\left(x^2-2x+8\right)^2=\left(x^2-1\right)^2\)

\(\Leftrightarrow\left(x^2-2x+8\right)^2-\left(x^2-1\right)^2=0\)

\(\Leftrightarrow\left(2x^2-2x+7\right)\left(-2x+9\right)=0\)

\(\Leftrightarrow-2x+9=0\Rightarrow x=\frac{9}{2}\)

Answer 3

5.

- Với \(x\ge\frac{2}{3}\)

\(\Leftrightarrow x^2+5x-3x+2-5=0\)

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

- Với \(x< \frac{2}{3}\)

\(\Leftrightarrow x^2+5x+3x-2-5=0\)

\(\Leftrightarrow x^2+8x-7=0\Rightarrow\left[{}\begin{matrix}x=-4+\sqrt{23}\left(l\right)\\x=-4-\sqrt{23}\end{matrix}\right.\)

6.

- Với \(x\ge1\)

\(\Leftrightarrow x^2-5x+5-1=0\Leftrightarrow x^2-5x+4=0\Rightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

- Với \(x< 1\)

\(\Leftrightarrow x^2+5x-5-1=0\Leftrightarrow x^2+5x-6=0\Rightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=-6\end{matrix}\right.\)

7.

Hai vế pt ko âm, bình phương:

\(\Leftrightarrow\left(3x^2-2\right)^2=\left(6-x^2\right)^2\)

\(\Leftrightarrow\left(3x^2-2\right)^2-\left(6-x^2\right)^2=0\)

\(\Leftrightarrow\left(2x^2+4\right)\left(4x^2-8\right)=0\)

\(\Leftrightarrow x^2=2\Rightarrow x=\pm\sqrt{2}\)