Tìm x:
\(\sqrt{9x^2}=2x+1\)
K
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LJ
2
6 tháng 9 2021
\(a,\sqrt{9x^2}=2x+1\\ \Leftrightarrow\left[{}\begin{matrix}3x=2x+1,\forall x\ge0\\-3x=2x+1,\forall x< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1,\forall x\ge0\left(N\right)\\x=-1,\forall x< 0\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-1,\forall x+3\ge0\\x+3=1-3x,\forall x+3< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2,\forall x\ge-3\left(N\right)\\x=-\dfrac{1}{2},\forall x< -3\left(L\right)\end{matrix}\right.\Leftrightarrow x=2\)
\(c,\sqrt{x^2-2x+4}=2x-3\left(x\in R\right)\\ \Leftrightarrow x^2-2x+4=\left(2x-3\right)^2\\ \Leftrightarrow x^2-2x+4=4x^2-12x+9\\ \Leftrightarrow3x^2-10x+5=0\\ \Delta=100-4\cdot3\cdot5=40\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10-\sqrt{40}}{6}\\x=\dfrac{10+\sqrt{40}}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{10}}{3}\\x=\dfrac{5+\sqrt{10}}{3}\end{matrix}\right.\)
KK
6 tháng 9 2021
\(a.\sqrt{9x^2}=2x+1\)
<=> \(\sqrt{9}x=2x+1\)
<=> 3x = 2x + 1
<=> 3x - 2x = 1
<=> x = 1
HN
30 tháng 6 2021
a) ⇔ |2x+3| = 8
⇒ \(\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}2x=5\\2x=-11\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
Vậy...
LT
30 tháng 6 2021
b) ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow3\sqrt{x}-7\sqrt{x}+6\sqrt{x}=8\)
\(\Leftrightarrow2\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\) (Vì \(x\ge0\) )
Vậy x = 16
c) ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{9\left(x-1\right)}=12\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
\(\Leftrightarrow x=17\)(TM)
Vậy x = 17
26 tháng 6 2021
`a)sqrt{x^2-2x+1}=2`
`<=>sqrt{(x-1)^2}=2`
`<=>|x-1|=2`
`**x-1=2<=>x=3`
`**x-1=-1<=>x=-1`.
Vậy `S={3,-1}`
`b)sqrt{x^2-1}=x`
Điều kiện:\(\begin{cases}x^2-1 \ge 0\\x \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}x^2 \ge 1\\x \ge 0\\\end{cases}\)
`<=>x>=1`
`pt<=>x^2-1=x^2`
`<=>-1=0` vô lý
Vậy pt vô nghiệm
`c)sqrt{4x-20}+3sqrt{(x-5)/9}-1/3sqrt{9x-45}=4(x>=5)`
`pt<=>sqrt{4(x-5)}+sqrt{9*(x-5)/9}-sqrt{(9x-45)*1/9}=4`
`<=>2sqrt{x-5}+sqrt{x-5}-sqrt{x-5}=4`
`<=>2sqrt{x-5}=4`
`<=>sqrt{x-5}=2`
`<=>x-5=4`
`<=>x=9(tmđk)`
Vậy `S={9}.`
`d)x-5sqrt{x-2}=-2(x>=2)`
`<=>x-2-5sqrt{x-2}+4=0`
Đặt `a=sqrt{x-2}`
`pt<=>a^2-5a+4=0`
`<=>a_1=1,a_2=4`
`<=>sqrt{x-2}=1,sqrt{x-2}=4`
`<=>x_1=3,x_2=18`,
`e)2x-3sqrt{2x-1}-5=0`
`<=>2x-1-3sqrt{2x-1}-4=0`
Đặt `a=sqrt{2x-1}(a>=0)`
`pt<=>a^2-3a-4=0`
`a-b+c=0`
`<=>a_1=-1(l),a_2=4(tm)`
`<=>sqrt{2x-1}=4`
`<=>2x-1=16`
`<=>x=17/2(tm)`
Vậy `S={17/2}`
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
d.
ĐKXĐ: $x\geq 2$. Đặt $\sqrt{x-2}=a(a\geq 0)$ thì pt trở thành:
$a^2+2-5a=-2$
$\Leftrightarrow a^2-5a+4=0$
$\Leftrightarrow (a-1)(a-4)=0$
$\Rightarrow a=1$ hoặc $a=4$
$\Leftrightarrow \sqrt{x-2}=1$ hoặc $\sqrt{x-2}=4$
$\Leftrightarrow x=3$ hoặc $x=18$ (đều thỏa mãn)
e. ĐKXĐ: $x\geq \frac{1}{2}$
Đặt $\sqrt{2x-1}=a(a\geq 0)$ thì pt trở thành:
$a^2+1-3a-5=0$
$\Leftrightarrow a^2-3a-4=0$
$\Leftrightarrow (a+1)(a-4)=0$
Vì $a\geq 0$ nên $a=4$
$\Leftrightarrow \sqrt{2x-1}=4$
$\Leftrightarrow x=\frac{17}{2}$
MH
27 tháng 9 2023
a) \(\sqrt{2x}=12\left(đk:x\ge0\right)\)
\(2x=144\)
\(x=72\)
b) \(\sqrt{9x^2-6x}+1=10\)\(\left(Đk:x\le0;x\ge\dfrac{2}{3}\right)\)
\(\sqrt{9x^2-6x}=9\)
\(9x^2-6x=81\)
\(\left(3x-1\right)^2=82\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{82}+1}{3}\\x=\dfrac{1-\sqrt{82}}{3}\end{matrix}\right.\)
c) \(x^2\sqrt{5}-\sqrt{125}=0\)
\(x^2\sqrt{5}=5\sqrt{5}\)
\(x^2=5\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
MH
4 tháng 10 2021
c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
4 tháng 10 2021
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
28 tháng 8 2023
\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;-1\right\}\)
\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)
\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)
Vậy \(S=\left\{28\right\}\)
12 tháng 5 2022
a: ĐKXĐ: (x-1)(x-3)>=0
=>x>=3 hoặc x<=1
b: ĐKXĐ: (x-4)(x-3)>=0
=>x>=4 hoặc x<=3
c: ĐKXĐ: (x-5)(x-4)>=0
=>x>=5 hoặc x<=4
\(\sqrt{9x^2}=2x+1\)
ĐK : \(x\ge0\)
Bình phương 2 vế
=> \(9x^2=\left(2x+1\right)^2\)
\(\Leftrightarrow9x^2=4x^2+4x+1\)
\(\Leftrightarrow9x^2-4x^2-4x-1=0\)
\(\Leftrightarrow5x^2-4x-1=0\)
\(\Leftrightarrow5x^2-5x+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\5x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{5}\end{cases}}\)
Vậy ....