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18 tháng 10 2020

a) ĐK : \(x\ge1\)

pt <=> \(\sqrt{3^2\left(x-1\right)}-\frac{1}{2}\sqrt{2^2\left(x-1\right)}=2\)

<=> \(\left|3\right|\sqrt{x-1}-\frac{1}{2}\cdot\left|2\right|\sqrt{x-1}=2\)

<=> \(3\sqrt{x-1}-1\sqrt{x-1}=2\)

<=> \(2\sqrt{x-1}=2\)

<=> \(\sqrt{x-1}=1\)

<=> \(x-1=1\)=> \(x=2\)( tm )

b) \(3x-\sqrt{49-14x+x^2}=15\)

<=> \(\sqrt{x^2-14x+49}=3x-15\)

<=> \(\sqrt{\left(x-7\right)^2}=3x-15\)

<=> \(\left|x-7\right|=3x-15\)(1)

Với x < 7

(1) <=> 7 - x = 3x - 15

     <=> -x - 3x = -15 - 7

     <=> -4x = -22

     <=> x = 11/2 ( tm )

Với x ≥ 7

(1) <=> x - 7 = 3x - 15

      <=> x - 3x = -15 + 7

      <=> -2x = -8

      <=> x = 4 ( ktm )

Vậy x = 11/2

18 tháng 10 2020

a) \(ĐKXĐ:x\ge1\)

\(\sqrt{9x-9}-\frac{1}{2}\sqrt{4x-4}=2\)

\(\Leftrightarrow\sqrt{9.\left(x-1\right)}-\frac{1}{2}.\sqrt{4\left(x-1\right)}=2\)

\(\Leftrightarrow3\sqrt{x-1}-\frac{1}{2}.2\sqrt{x-1}=2\)

\(\Leftrightarrow3\sqrt{x-1}-\sqrt{x-1}=2\)

\(\Leftrightarrow2\sqrt{x-1}=2\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=2\)( thỏa mãn ĐKXĐ )

Vậy phương trình có nghiệm là \(x=2\)

b) \(3x-\sqrt{49-14x+x^2}=15\)

\(\Leftrightarrow3x-\sqrt{\left(7-x\right)^2}=15\)

\(\Leftrightarrow3x-\left|7-x\right|=15\)

+) TH1: Nếu \(7-x< 0\)\(\Leftrightarrow x>7\)

thì \(3x-\left(x-7\right)=15\)

\(\Leftrightarrow3x-x+7=15\)\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\)( không thỏa mãn )

+) TH2: Nếu \(7-x\ge0\)\(\Leftrightarrow x\le7\)

thì \(3x-\left(7-x\right)=15\)

\(\Leftrightarrow3x-7+x=15\)

\(\Leftrightarrow4x=22\)\(\Leftrightarrow x=\frac{22}{4}\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=\frac{22}{4}\)

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

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 \(x\in\left\{\sqrt{5};-\sqrt{5}\right\}\)

30 tháng 9 2021

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 = \(\sqrt{5}\)

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

=>\(\dfrac{3}{2}\sqrt{x-3}=3\)

=>\(\sqrt{x-3}=2\)

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

=>\(\dfrac{3}{2}\sqrt{x-5}=3\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

29 tháng 6 2021

`a)sqrt{x^2-6x+9}=2`

`<=>sqrt{(x-3)^2}=2`

`<=>|x-3|=2`

`**x-3=2`

`<=>x=5`

`**x-3=-2`

`<=>x=1`

Vậy `S={1,5}`

`b)sqrt{4x-20}+sqrt{x-5}-1/3sqrt{9x-45}=4`

đk:`x>=5`

`pt<=>2sqrt{x-5}+sqrt{x-5}-1/3*3*sqrt{x-5}=4`

`<=>2sqrt{x-5}=4`

`<=>sqrt{x-5}=2`

`<=>x-5=4<=>x=9`

Vậy `S={9}`

AH
Akai Haruma
Giáo viên
29 tháng 6 2021

Lời giải:

a.

PT $\Leftrightarrow \sqrt{(x-3)^2}=2$

$\Leftrightarrow |x-3|=2$

$\Leftrightarrow x-3=\pm 2$

$\Leftrightarrow x=1$ hoặc $x=5$

b. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4(x-5)}+\sqrt{x-5}-\frac{1}{3}\sqrt{9(x-5)}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)

26 tháng 10 2023

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)

AH
Akai Haruma
Giáo viên
24 tháng 8 2021

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 $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$

$\Leftrightarrow |x-2|=|2x-3|$

\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)

AH
Akai Haruma
Giáo viên
24 tháng 8 2021

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 $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$

$\Leftrightarrow |2x-1|=|x-3|$

\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

23 tháng 8 2021

d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)

e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)

c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)

\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)

\(\Leftrightarrow x-4=0\)

hay x=4

27 tháng 7 2021

Sửa lại câu c) đặt \(\sqrt{x}+1=\)\(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)

27 tháng 7 2021

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\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)