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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)
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: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
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, \(\left\{{}\begin{matrix}9-6x+x^2\ge0\\x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3-x\right)^2\ge0\left(luônđúng\right)\\x\ge0\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(\sqrt{9-6x+x^2}=x\\ \Rightarrow\left|3-x\right|=x\\ \Rightarrow\left[{}\begin{matrix}3-x=-x\\3-x=x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3=0\left(vôlí\right)\\x=1,5\end{matrix}\right.\)
Câu b bạn có bị lỗi dấu căn không mà sao nó kéo dài cả 2 vế pt vậy :v
\(a,\sqrt{x^2-6x+9}+x=11\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=11-x\)
\(\Leftrightarrow\left|x-3\right|=11-x\\ TH_1:x\ge3\\ x-3=11-x\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\left(tm\right)\)
\(TH_2:x< 3\\ -x+3=11-x\\ \Leftrightarrow-x+x=11-3\\ \Leftrightarrow0=8\left(VL\right)\)
Vậy \(S=\left\{7\right\}\)
\(c,\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\) \(\left(dk:x\ge-1\right)\)
\(\Leftrightarrow\sqrt{4^2}.\sqrt{\left(x+1\right)}-\sqrt{3^2}.\sqrt{\left(x+1\right)}=4\left(1\right)\)
Đặt \(a=\sqrt{x+1}\left(a\ge0\right)\)
Pt trở thành : \(4a-3a=4\Leftrightarrow a=4\left(tmdk\right)\)
\(\Rightarrow\sqrt{x+1}=4\\ \Rightarrow\left(\sqrt{x+1}\right)^2=16\\ \Rightarrow\left|x+1\right|=16\)
\(TH_1:x\ge-1\\ x+1=16\Leftrightarrow x=15\left(tm\right)\\ TH_2:x< -1\\ -x-1=16\Leftrightarrow x=-17\left(tm\right)\)
Nhưng loại TH2 vì dk ban đầu là \(x\ge-1\)
Vậy \(S=\left\{15\right\}\)
\(d,\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\left(dk:x\ge-1\right)\\ \Leftrightarrow\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}-\sqrt{x+1}=0\)
Đặt \(\sqrt{x+1}=a\left(a\ge0\right)\)
Tới đây bạn làm tương tự câu c nha.
a,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)
TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)
TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)
Vậy x=0,5...
d, đk \(x\ge-1\)
=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)
\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)
a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow\left|3x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b) Ta có: \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)
\(\Leftrightarrow\left|x-3\right|=4-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
Lời giải:
a.
PT $\Leftrightarrow |2x+1|=|x-1|$
$\Leftrightarrow 2x+1=x-1$ hoặc $2x+1=-(x-1)$
$\Leftrightarrow x+2=0$ hoặc $3x=0$
$\Leftrightarrow x=-2$ hoặc $x=0$ (tm)
b.
PT $\Leftrightarrow 9x^2-6x+1=x^2-4x+4$
$\Leftrightarrow 8x^2-2x-3=0$
$\Leftrightarrow (4x-3)(2x+1)=0$
$\Leftrightarrow 4x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=\frac{3}{4}$ hoặc $x=\frac{-1}{2}$ (tm)
a: =>|2x+1|=|x-1|
=>2x+1=x-1 hoặc 2x+1=-x+1
=>x=-2 hoặc x=0
b: =>|3x-1|=|x-2|
=>3x-1=x-2 hoặc 3x-1=-x+2
=>2x=-1 hoặc 4x=3
=>x=-1/2 hoặc x=3/4
a, \(\sqrt{\left(x+2\right)^2}=2x+1\Leftrightarrow x+2=2x+1\Leftrightarrow-x=-1\Leftrightarrow x=1\)
b, \(\sqrt{4x^2-4x+1}=\sqrt{x^2-2x+1}\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-1\right)^2}\Leftrightarrow2x-1=x-1\Leftrightarrow x=2\)
c, \(\sqrt{x^2-6x+9}=5\Leftrightarrow\sqrt{\left(x-3\right)^2}=5\Leftrightarrow x=8\)
d, \(\sqrt{4x^2-12x+9}=\sqrt{9x^2-24x+16}\Leftrightarrow\sqrt{\left(2x-3\right)^2}=\sqrt{\left(3x-4\right)^2}\)
\(\Leftrightarrow2x-3=3x-4\Leftrightarrow-x=-1\Leftrightarrow x=1\)
a) \(\sqrt{\left(x+2\right)^2}=2x+1\)
<=> \(\left|x+2\right|=2x+1\)
<=> \(\orbr{\begin{cases}x+2=2x+1\left(đk:x\ge-2\right)\\-x-2=2x+1\left(Đk:x< -2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}-x=-1\\-3x=3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\left(tm\right)\\x=-1\left(ktm\right)\end{cases}}\)
Vậy S = {1}
b) \(\sqrt{x^2-6x+9}=5\)
<=> \(\sqrt{\left(x-3\right)^2}=5\)
<=> \(\left|x-3\right|=5\)
<=> \(\orbr{\begin{cases}x-3=5\left(đk:x\ge3\right)\\3-x=5\left(đk:x< 3\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=8\left(tm\right)\\x=-2\left(tm\right)\end{cases}}\)
Vậy S = {-2; 8}
c) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-2x+1}\)
<=> \(\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-1\right)^2}\)
<=> \(\left|2x-1\right|=\left|x-1\right|\)
<=> \(\orbr{\begin{cases}2x-1=x-1\\2x-1=1-x\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\3x=2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
Vậy S = {0; 2/3}
d) \(\sqrt{4x^2-12x+9}=\sqrt{9x^2-24x+16}\)
<=> \(\sqrt{\left(2x-3\right)^2}=\sqrt{\left(3x-4\right)^2}\)
<=> \(\left|2x-3\right|=\left|3x-4\right|\)
<=> \(\orbr{\begin{cases}2x-3=3x-4\\2x-3=4-3x\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=\frac{7}{5}\end{cases}}\)
Vậy S = {1; 7/5}