Giải phương trình:
a, (x2+3x+1)(x2+3x+2)-6=0
b,\(\sqrt{x+3}-\sqrt{x-2}=1\)
c, x2+6x+11=(x+6)\(\sqrt{x^2+11}\)
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Bài 1:
a.
$(4x^2+4x+1)-x^2=0$
$\Leftrightarrow (2x+1)^2-x^2=0$
$\Leftrightarrow (2x+1-x)(2x+1+x)=0$
$\Leftrightarrow (x+1)(3x+1)=0$
$\Rightarrow x+1=0$ hoặc $3x+1=0$
$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$
b.
$x^2-2x+1=4$
$\Leftrightarrow (x-1)^2=2^2$
$\Leftrightarrow (x-1)^2-2^2=0$
$\Leftrightarrow (x-1-2)(x-1+2)=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-1$
c.
$x^2-5x+6=0$
$\Leftrightarrow (x^2-2x)-(3x-6)=0$
$\Leftrightarrow x(x-2)-3(x-2)=0$
$\Leftrightarrow (x-2)(x-3)=0$
$\Leftrightarrow x-2=0$ hoặc $x-3=0$
$\Leftrightarrow x=2$ hoặc $x=3$
2c.
ĐKXĐ: $x\neq 0$
PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$
$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$
$\Leftrightarrow x=-4$ (tm)
2d.
ĐKXĐ: $x\neq 2$
PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$
$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$
$\Rightarrow 3x-5=3-x$
$\Leftrightarrow 4x=8$
$\Leftrightarrow x=2$ (không tm)
Vậy pt vô nghiệm.
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
a. ĐKXĐ: $x\geq 2$ hoặc $x=1$
PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$
$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)
b.
PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$
$\Leftrightarrow |x-2|=|2x-3|$
\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)
c. ĐKXĐ: $x=2$ hoặc $x\geq 3$
PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)
d.
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt vô nghiệm
\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
=>-6x+16=0
=>-6x=-16
hay x=8/3(nhận)
c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)
\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow2x^2+4x-2x^2+2=0\)
=>4x+2=0
hay x=-1/2(nhận)
a) Đặt \(x^2+3x+1=y\)
=> y(y+1) - 6 = 0
=> \(y^2+y-6=0\)
=> \(\left[\begin{array}{nghiempt}y=2\\y=-3\end{array}\right.\)
Với y = 2 ta có:
\(x^2+3x+1=2\)
=> \(\left[\begin{array}{nghiempt}x=\frac{-3+\sqrt{13}}{2}\\x=\frac{-3-\sqrt{13}}{2}\end{array}\right.\)
Với y = -3 ta có:
\(x^2+3x+1=-3\)
=>\(\left[\begin{array}{nghiempt}x=1\\x=-4\end{array}\right.\)
Có j không hiểu có thể hỏi lại mk
Chúc bạn làm bài tốt
b) \(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x-2}\right)^2=1^2\)
\(\Leftrightarrow x+3+x-2-2\sqrt{\left(x+3\right)\cdot\left(x-2\right)}=1\)
\(\Leftrightarrow2x+1-1=2\sqrt{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow2x=2\sqrt{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow x=\sqrt{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow x^2=\left(\sqrt{\left(x+3\right)\left(x-2\right)}\right)^2\)
\(\Leftrightarrow x^2=x^2+x-6\)
\(\Leftrightarrow x-6=0\)
\(\Leftrightarrow x=6\)