K
Khách

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

17 tháng 7 2019

a) \(\sqrt{1-4x+4x^2}=5\)

<=> \(\sqrt{4x^2-4x+1}=5\)

<=> 4x2 - 4x + 1 = 52

<=> 4x2 - 4x + 1 = 25

<=> 4x2 - 4x + 1 - 25 = 0

<=> 4x2 - 4x - 24 = 0

<=> 4(x + 2)(x - 3) = 0

<=> x = -2 hoặc x = 3

 => x = -2 hoặc x = 3

b) \(\sqrt{4-5x}=12\)

<=> \(\sqrt{-5x+4}=12\)

<=> -5x + 4 = 122

<=> -5x + 4 = 144

<=> -5x = 144 - 4

<=> -5x = 140

<=> x = -28

=> x = -28

\(a,\sqrt{1-4x+4x^2}=5\)

\(\Rightarrow4x^2-4x+1=25\)

\(\Rightarrow4x^2-4x-24=0\)

\(\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)

\(b,\sqrt{4-5x}=12\)

\(\Rightarrow4-5x=144\)

\(\Rightarrow5x=-140\)

\(\Rightarrow x=-28\)

a) Ta có: \(\sqrt{4-5x}=12\)

\(\Leftrightarrow4-5x=144\)

\(\Leftrightarrow5x=-140\)

hay x=-28

b) Ta có: \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)

\(\Leftrightarrow\sqrt{3x}+10=10+4\sqrt{6}\)

\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)

\(\Leftrightarrow3x=96\)

hay x=32

c) Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

24 tháng 11 2021

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)

28 tháng 10 2020

a) \(\sqrt{4-5x}=12\)

ĐK : x ≤ 4/5

Bình phương hai vế

⇔ \(4-5x=144\)

⇔ \(-5x=140\)

⇔ \(x=-28\)( tm )

b) \(\sqrt{1-4x+4x^2}=5\)

⇔ \(\sqrt{\left(1-2x\right)^2}=5\)

⇔ \(\left|1-2x\right|=5\)

⇔ \(\orbr{\begin{cases}1-2x=5\\1-2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)

ĐK : x ≥ -5

⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)

⇔ \(\left|2\right|\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot\left|3\right|\sqrt{x+5}=6\)

⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)

⇔ \(\frac{5}{4}\sqrt{x+5}=6\)

⇔ \(\sqrt{x+5}=\frac{24}{5}\)

⇔ \(x+5=\frac{576}{25}\)

⇔ \(x=\frac{451}{25}\)( tm )

d)\(\sqrt{x-2}\le3\)

ĐK : x ≥ 2

⇔ \(x-2\le9\)

⇔ \(x\le11\)

Kết hợp với điều kiện => Nghiệm của bpt là 2 ≤ x ≤ 11

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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))

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

 

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∈\(R\)

15 tháng 10 2023

a: ĐKXĐ: x-5>=0

=>x>=5

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

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

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

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x-1>=0

=>x>=1

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)

=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)

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

=>\(\sqrt{x-1}=-2\)(vô lý)

Vậy: Phương trình vô nghiệm

c: ĐKXĐ: x-2>=0

=>x>=2

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)

=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

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

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

=>x-2=16

=>x=18(nhận)

d: ĐKXĐ: x+3>=0

=>x>=-3

\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)

=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)

=>\(4\sqrt{x+3}=0\)

=>x+3=0

=>x=-3(nhận)

15 tháng 10 2023

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

\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)

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

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

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

\(\left|x-5\right|=4\)

=> \(x-5=\pm4\)

\(x=\pm4+5\)

\(x=9;x=1\)

Vậy x=9; x=1

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: 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)

NV
7 tháng 10 2019

Bạn tự tìm ĐKXĐ.

a/ \(\sqrt{4-5x}=12\Rightarrow4-5x=144\Rightarrow x=-28\)

b/ \(10+\sqrt{3x}=\left(2+\sqrt{6}\right)^2=10+4\sqrt{6}\)

\(\Rightarrow\sqrt{3x}=4\sqrt{6}\Rightarrow\sqrt{x}=4\sqrt{2}\)

\(\Rightarrow x=32\)

c/ \(2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Rightarrow\sqrt{x+5}=2\Rightarrow x+5=4\Rightarrow x=-1\)

d/ \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

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

e/ \(\sqrt{\frac{4x+3}{x+1}}=3\Leftrightarrow\frac{4x+3}{x+1}=9\)

\(\Rightarrow4x+3=9x+9\Rightarrow5x=-6\Rightarrow x=-\frac{6}{5}\)

f/ \(\sqrt{x-2}\le3\Rightarrow x-2\le9\Rightarrow2\le x\le11\)