Question

Bài 14. Giải phương trình:

1. \(\sqrt{4x^2 + 2x + 1} = 2x - 1.\)

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

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

4. \(\sqrt{(x + 1)^2} (2x + 3) = x + 1.\)

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

6. \((x - 4)^2 (x^2 + 1) = x - 4.\)

Answer 1

1: \(\sqrt{4x^2+2x+1}=2x-1\)

=>\(\begin{cases}2x-1\ge0\\ 4x^2+2x+1=\left(2x-1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge\frac12\\ 4x^2-4x+1=4x^2+2x+1\end{cases}\)

=>\(\begin{cases}x\ge\frac12\\ -6x=0\end{cases}\Rightarrow\begin{cases}x\ge\frac12\\ x=0\end{cases}\)

=>x∈∅

2: ĐKXĐ: \(2x^2+3x+1\ge0\)

=>(2x+1)(x+1)>=0

=>\(\left[\begin{array}{l}x\ge-\frac12\\ x\le-1\end{array}\right.\)

Ta có: \(\sqrt{2x^2+3x+1}=1-x\)

=>\(\begin{cases}1-x\ge0\\ 2x^2+3x+1=\left(1-x\right)^2\end{cases}\Rightarrow\begin{cases}x\le1\\ 2x^2+3x+1=x^2-2x+1\end{cases}\)

=>\(\begin{cases}x\le1\\ x^2+5x=0\end{cases}\Rightarrow\begin{cases}x\le1\\ x\left(x+5\right)=0\end{cases}\)

=>x∈{0;-5}

3: Ta có: \(\sqrt{2x^2+x+1}=3x-1\)

=>\(\begin{cases}3x-1\ge0\\ \left(3x-1\right))^2=2x^2+x+1\end{cases}\Rightarrow\begin{cases}x\ge\frac13\\ 9x^2-6x+1=2x^2+x+1\end{cases}\)

=>\(\begin{cases}x\ge\frac13\\ 7x^2-7x=0\end{cases}\Rightarrow\begin{cases}x\ge\frac13\\ 7x\left(x-1\right)=0\end{cases}\Rightarrow\begin{cases}x\ge\frac13\\ x\in\left\lbrace0;1\right\rbrace\end{cases}\)

=>x=1

4:

ĐKXĐ: x>=-3/2

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

=>\(\begin{cases}x+1\ge0\\ \left(x+1\right)^2\cdot\left(2x+3\right)=\left(x+1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-1\\ \left(x+1\right)^2\cdot\left(2x+2\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-1\\ \left(x+1\right)^2\cdot2\left(x+1\right)=0\end{cases}\Rightarrow\begin{cases}x\ge-1\\ 2\left(x+1\right)^3=0\end{cases}\)

=>x+1=0

=>x=-1(nhận)

5: ĐKXĐ: x>=-2/3

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

=>\(\begin{cases}2x-1\ge0\\ \left(2x-1\right)^2\cdot\left(3x+2\right)=\left(2x-1\right)^2\end{cases}\Rightarrow\begin{cases}2x\ge1\\ \left(2x-1\right)^2\cdot\left(3x+1\right)=0\end{cases}\)

=>\(\begin{cases}x\ge\frac12\\ x\in\left\lbrace\frac12;-\frac13\right\rbrace\end{cases}\Rightarrow x=\frac12\)

6: \(\left(x-4\right)^2\cdot\left(x^2+1\right)=x-4\)

=>\(\left(x-4\right)^2\cdot\left(x^2+1\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left\lbrack\left(x-4\right)\left(x^2+1\right)-1\right\rbrack=0\)

=>\(\left(x-4\right)\left(x^3+x-4x^2-5\right)=0\)

=>x=4 hoặc x≃4,06