K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

18 tháng 9 2021

Mn giúp e với ak

18 tháng 9 2021

a) \(\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(x^2-2.x.3+3^2\right)}\)

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

22 tháng 7 2023

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

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

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

22 tháng 7 2023

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

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

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

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

 

2 tháng 11 2021

a) \(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\Leftrightarrow\left|x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)

b) ĐKXĐ: \(x\ge-5\)

\(pt\Leftrightarrow2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5}=4\)

\(\Leftrightarrow2\sqrt{x+5}=4\Leftrightarrow\sqrt{x+5}=2\)

\(\Leftrightarrow x+5=4\Leftrightarrow x=-1\left(tm\right)\)

17 tháng 10 2023

b: Sửa đề: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)(1)

ĐKXĐ: \(x>=5\)

\(\left(1\right)\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

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

=>x-5=4

=>x=9(nhận)

c: ĐKXĐ: \(\dfrac{3x-2}{x+1}>=0\)

=>\(\left[{}\begin{matrix}x>=\dfrac{2}{3}\\x< -1\end{matrix}\right.\)

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

=>\(\dfrac{3x-2}{x+1}=9\)

=>9(x+1)=3x-2

=>9x+9=3x-2

=>6x=-11

=>\(x=-\dfrac{11}{6}\left(nhận\right)\)

d: ĐKXĐ: \(\left\{{}\begin{matrix}5x-4>=0\\x+2>0\end{matrix}\right.\Leftrightarrow x>=\dfrac{4}{5}\)

\(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\)

=>\(\sqrt{\dfrac{5x-4}{x+2}}=2\)

=>\(\dfrac{5x-4}{x+2}=4\)

=>5x-4=4x+8

=>x=12(nhận)

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

HQ
Hà Quang Minh
Giáo viên
28 tháng 7 2023

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH1\(x\ge3\)

\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)

TH2\(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH3\(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH4\(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)

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

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)

15 tháng 10 2021