bài 1 : giải phương trình:
a. \(\sqrt{x+2\sqrt{ }x-1}=2\)
b. \(\sqrt{x^2-4x+4}=\sqrt{4x^212x+9}\)
c.\(\sqrt{x+4\sqrt{ }x-4}=2\)
d. \(\sqrt{x^2-6x+9}=2\)
e. \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
f. \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
Giải các phương trình:
a) \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
b) \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)
c) \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)
d) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
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: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
\(\Leftrightarrow x^2-3x+2=x-1\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=-x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)
\(\Leftrightarrow x^2-5x+6=x-2\)
\(\Leftrightarrow x^2-6x+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
giải phương trình:
a) \(\sqrt{x-1}=\sqrt{3x-2}+\sqrt{5x-1}\)
b) \(\sqrt{4x+1}-\sqrt{3x+4}=1\)
c) \(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
Giải phương trình:
a)\(\sqrt{\sqrt{5}-\sqrt{3x}}=\sqrt{8+2\sqrt{15}}\)
b)\(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
c) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)
d) \(\sqrt{x^2-6x+9}+x=11\)
e) \(\sqrt{3x^2-4x+3}=1-2x\)
f) \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
g) \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
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
Giải phương trình:
a) \(\sqrt{x^2+4}=\sqrt{2x+3}\)
b) \(\sqrt{x^2-6x+9}=2x-1\)
c) \(\sqrt{4x+12}=\sqrt{9x+17}-5\)
d) \(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
Bài 1 GIẢI PHƯƠNG TRÌNH:
a) \(\sqrt{x-5}=\sqrt{3-x}\)
b) \(\sqrt{4-5x}=\sqrt{2-5x}\)
c) x2+4x+5=2\(\sqrt{2x+3}\)
d) \(\sqrt{x^2-2x+1}=\sqrt{4x^2-4x+1}\)
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
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) \(\sqrt{x-5}=\sqrt{3-x}\)
⇔\(\left(\sqrt{x-5}\right)^2=\left(\sqrt{3-x}\right)^2\)
⇔\(x-5=3-x\)
⇔\(x=4\)
b) \(\sqrt{4-5x}=\sqrt{2-5x}\)
⇔\(\left(\sqrt{4-5x}\right)^2=\left(\sqrt{2-5x}\right)^2\)
⇔\(4-5x=2-5x\)
⇔\(2=0\) (Vô lí)
Bài 1: giải p.trình
a,\(\sqrt{x^2-4x+4}=1\)
b,\(\sqrt{1-4x+4x^2}=5\)
c,\(\sqrt{a\left(1-2x+x^2\right)}-6=0\)
d,\(\sqrt{9x^2}=2x+1\)
e,\(\sqrt{9-6x+x^2}=x\)
a, ĐKXĐ: \(x^2-4x+4\ge0\Rightarrow\left(x-2\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{x^2-4x+4}=1\\ \Rightarrow x-2=1\\ \Rightarrow x=3\)
b,\(ĐKXĐ:1-4x+4x^2\ge0\Rightarrow\left(1-2x\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{1-4x+4x^2}=5\\ \Rightarrow\left|1-2x\right|=5\\ \Rightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
d, ĐKXĐ: \(\left\{{}\begin{matrix}9x^2\ge0\\2x+1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x\ge0\)
\(\sqrt{9x^2}=2x+1\\ \Rightarrow\left|3x\right|=2x+1\\ \Rightarrow\left[{}\begin{matrix}3x=2x+1\\3x=-2x+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
c, ĐKXĐ: \(1-2x+x^2\ge0\Rightarrow\left(1-x\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{1-2x+x^2}-6=0\\ \Rightarrow\left|1-x\right|=6\\ \Rightarrow\left[{}\begin{matrix}1-x=-6\\1-x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
e, \(\left\{{}\begin{matrix}9-6x+x^2\ge0\\x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3-x\right)^2\ge0\left(luônđúng\right)\\x\ge0\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(\sqrt{9-6x+x^2}=x\\ \Rightarrow\left|3-x\right|=x\\ \Rightarrow\left[{}\begin{matrix}3-x=-x\\3-x=x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3=0\left(vôlí\right)\\x=1,5\end{matrix}\right.\)
a) \(\sqrt{x^2-4x+4}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b) \(\sqrt{1-4x+4x^2}=5\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\Leftrightarrow\left|1-2x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
c) \(\sqrt{x\left(1-2x+x^2\right)}-6=0\)
\(\Leftrightarrow\left(\sqrt{x\left(1-x\right)^2}\right)^2=36\Leftrightarrow x\left(1-x\right)^2=36\)
\(\Leftrightarrow x-2x^2+x^3-36=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+2x+9\right)=0\)
\(\Leftrightarrow x=4\)(do \(x^2+2x+9=\left(x+1\right)^2+8>0\))
d) \(\sqrt{9x^2}=2x+1\)
\(\Leftrightarrow3\left|x\right|=2x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\\-3x=2x+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{5}\end{matrix}\right.\)
e) \(\sqrt{9-6x+x^2}=x\left(1\right)\left(đk:x\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(3-x\right)^2}=x\Leftrightarrow\left|3-x\right|=x\)
TH1: \(0\le x\le3\)
\(\left(1\right)\Leftrightarrow3-x=x\Leftrightarrow x=\dfrac{3}{2}\)
TH2: \(x>3\)
\(\left(1\right)\Leftrightarrow x-3=x\Leftrightarrow-3=0\left(vn\right)\)
Giải phương trình:
a. \(3\sqrt{8x}-\sqrt{32x}+\sqrt{50x}=21\)
b. \(\sqrt{25x+50}+3\sqrt{4x+8}-2\sqrt{16x+32}=15\)
c. \(\sqrt{\left(x-2\right)^2}=12\)
d. \(\sqrt{x^2-6x+9}-3=5\)
e.\(\sqrt{\left(2x-1\right)^2}-x=3\)
f. \(\sqrt{3x-6}-x=-2\)
h. \(\sqrt{3-2x}-2=x\)
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)
f.
ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{3(x-2)}-(x-2)=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{3}-\sqrt{x-2})=0$
$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{3}-\sqrt{x-2}=0$
$\Leftrightarrow x=2$ hoặc $x=5$ (tm)
h. ĐKXĐ: $x\leq \frac{3}{2}$
PT $\Leftrightarrow \sqrt{3-2x}=x+2$
\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ 3-2x=(x+2)^2=x^2+4x+4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2+6x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-3+2\sqrt{2}\) (tm)
Vậy.......
* Giải phương trình:
a. \(x+\sqrt{x^2-4x+4}=\dfrac{1}{2}\)
b. \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)
b: Ta có: \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)
\(\Leftrightarrow\sqrt{x^2-1}=2\)
\(\Leftrightarrow x^2-1=4\)
hay \(x\in\left\{\sqrt{5};-\sqrt{5}\right\}\)
a. \(x+\sqrt{x^2-4x+4}=\dfrac{1}{2}\)
<=> \(x+\sqrt{\left(x-2\right)^2}=\dfrac{1}{2}\)
<=> \(x+\left|x-2\right|=\dfrac{1}{2}\)
<=> \(\left[{}\begin{matrix}x+x-2=\dfrac{1}{2}\\x+\left[-\left(x-2\right)\right]=\dfrac{1}{2}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=\dfrac{5}{2}\\x-x+2=\dfrac{1}{2}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\0=\dfrac{-3}{2}\left(VLí\right)\end{matrix}\right.\)
Vậy nghiệm của PT là \(S=\left\{\dfrac{5}{4}\right\}\)
b. \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)
<=> \(\sqrt{9\left(x^2-1\right)}+\sqrt{4\left(x^2-1\right)}=\sqrt{16\left(x^2-1\right)}+2\)
<=> \(3\sqrt{x^2-1}+2\sqrt{x^2-1}-4\sqrt{x^2-1}=2\)
<=> \(\left(3+2-4\right)\sqrt{x^2-1}=2\)
<=> \(\sqrt{x^2-1}=2\)
<=> x2 - 1 = 4
<=> x2 = 5
<=> x = \(\sqrt{5}\)
Giải phương trình:
a) \(\sqrt{4-3x}=8\)
b) \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=-1\)
c) \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)
Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)
a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)
b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)
\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))
\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)
c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)
a) Ta có: \(\sqrt{4-3x}=8\)
\(\Leftrightarrow4-3x=64\)
\(\Leftrightarrow3x=4-64=-60\)
hay x=-20
b) Ta có: \(\sqrt{4x-8}-12\cdot\sqrt{\dfrac{x-2}{9}}=-1\)
\(\Leftrightarrow2\cdot\sqrt{x-2}-12\cdot\dfrac{\sqrt{x-2}}{3}=-1\)
\(\Leftrightarrow-2\cdot\sqrt{x-2}=-1\)
\(\Leftrightarrow\sqrt{x-2}=\dfrac{1}{2}\)
\(\Leftrightarrow x-2=\dfrac{1}{4}\)
hay \(x=\dfrac{9}{4}\)