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8 tháng 4 2023

\(\sqrt{x-2}+1=2x-\dfrac{20}{x+2}\left(1\right)\)

Đk: \(x\ge2\)

\(\left(1\right)\Leftrightarrow\sqrt{x-2}-1=2x-\dfrac{20}{x+2}-2\)

\(\Leftrightarrow\dfrac{\left(x-2\right)-1}{\sqrt{x-2}+1}=\dfrac{2x\left(x+2\right)-2\left(x+2\right)-20}{x+2}\)

\(\Leftrightarrow\dfrac{\left(x-2\right)-1}{\sqrt{x-2}+1}=\dfrac{2x^2+2x-24}{x+2}\)

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

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

\(\left(2\right)\Leftrightarrow2\left(x+4\right)\sqrt{x-2}+2\left(x+4\right)=x+2\)

\(\Leftrightarrow2\left(x+4\right)\sqrt{x-2}+x+6=0\left(3\right)\)

Ta có \(x\ge2>0\Rightarrow2\left(x+4\right)\sqrt{x-2}+x+6>0\)

Vì vậy phương trình (3) vô nghiệm. Khi đó phương trình (2) cũng vô nghiệm.

Vậy phương trình (1) có nghiệm duy nhất là \(x=3\)

 

 

8 tháng 2 2021

ĐKXĐ : \(\left\{{}\begin{matrix}2x-1>0\\y+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\y>-2\end{matrix}\right.\)

PT ( I ) \(\Leftrightarrow\left(\sqrt{\dfrac{2x-1}{y+2}}+\sqrt{\dfrac{y+2}{2x-1}}\right)^2=4\)

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

\(\Leftrightarrow\dfrac{2x-1}{y+2}+\dfrac{y+2}{2x-1}=2\)

Từ PT ( II ) ta được : \(x=12-y\)

- Thế x vào PT trên ta được : \(\dfrac{2\left(12-y\right)}{y+2}+\dfrac{y+2}{2\left(12-y\right)}=2\)

\(\Leftrightarrow4\left(y-12\right)^2+\left(y+2\right)^2=4\left(12-y\right)\left(y+2\right)\)

\(\Leftrightarrow4\left(y^2-24y+144\right)+y^2+4y+4=4\left(12y+24-y^2-2y\right)\)

\(\Leftrightarrow4y^2-96y+576+y^2+4y+4-40y-96+4y^2=0\)

\(\Leftrightarrow9y^2-132y+484=0\)

\(\Leftrightarrow y=\dfrac{22}{3}\left(TM\right)\)

- Thay lại vào PT ta được : \(x=\dfrac{14}{3}\)

Vậy phương trình có nghiệm là \(S=\left\{\left(\dfrac{22}{3};\dfrac{14}{3}\right);\left(\dfrac{14}{3};\dfrac{22}{3}\right)\right\}\)

20 tháng 4 2022

a, \(\dfrac{1}{2}\sqrt{x-5}-\sqrt{4x-20+3}=0\left(dkxd:x\ge5\right)\)

\(< =>\dfrac{\sqrt{x-5}}{2}=\sqrt{4x-17}\)

\(< =>\dfrac{x-5}{4}=4x-17\)

\(< =>x-5=16x-68\)

\(< =>15x=68-5=63\)

\(< =>x=\dfrac{63}{15}=\dfrac{21}{5}\)(ktm)

b, \(\sqrt{2x+1}-2\sqrt{x}+1=0\left(dkxd:x\ge0\right)\)

\(< =>\sqrt{2x+1}+1=2\sqrt{x}\)

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

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

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

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

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}-1=2\\\sqrt{2x+1}-1=-2\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{2x+1}=-1\left(loai\right)\end{matrix}\right.\)

\(< =>2x+1=9< =>2x=8< =>x=4\)(tmdk)

25 tháng 11 2023

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

=>\(-2\sqrt{3x}=-4\)

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

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

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

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)

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

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

14 tháng 12 2021

\(ĐK:-1\le x< 0;x\ge1\\ PT\Leftrightarrow x+2\sqrt{x-\dfrac{1}{x}}=3+\dfrac{1}{x}\\ \Leftrightarrow x-\dfrac{1}{x}+2\sqrt{x-\dfrac{1}{x}}-3=0\)

Đặt \(\sqrt{x-\dfrac{1}{x}}=a\ge0\)

\(PT\Leftrightarrow a^2+2a-3=0\\ \Leftrightarrow\left(a-1\right)\left(a+3\right)=0\\ \Leftrightarrow a=1\left(a\ge0\right)\\ \Leftrightarrow x-\dfrac{1}{x}=1\\ \Leftrightarrow x^2-x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\end{matrix}\right.\)

5 tháng 9 2023

1) \(\sqrt{x^2+1}=\sqrt{5}\)

\(\Leftrightarrow x^2+1=5\)

\(\Leftrightarrow x^2=5-1\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2=2^2\)

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

2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\)

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=3+1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=\dfrac{4}{2}\)

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

3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))

\(\Leftrightarrow43-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1=43-x\)

\(\Leftrightarrow x^2-x-42=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))

\(\Leftrightarrow\sqrt{4x-3}=x-2\)

\(\Leftrightarrow4x-3=\left(x-2\right)^2\)

\(\Leftrightarrow x^2-4x+4=4x-3\)

\(\Leftrightarrow x^2-8x+7=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))

\(\Leftrightarrow\sqrt{x}+3=2\sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)

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

\(\Leftrightarrow x=1^2\)

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

5 tháng 9 2023

1)

\(\sqrt{x^2+1}=\sqrt{5}\\ \Leftrightarrow x^2+1=5\\ \Leftrightarrow x^2=5-1=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy PT có nghiệm `x=2` hoặc `x=-2`

2)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)

Vậy PT có nghiệm `x=2`

3)

\(ĐKXĐ:x\le43\)

PT trở thành:

\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm `x=-6` hoặc `x=7`

4)

ĐKXĐ: \(x\ge\dfrac{3}{4}\)

PT trở thành:

\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)

5) 

ĐKXĐ: \(x\ge0\)

PT trở thành:

\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)

Khi đó:

(1)\(\Leftrightarrow3t^2+8t+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)

Vậy PT vô nghiệm.

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

15 tháng 10 2023

ĐKXĐ: \(\left\{{}\begin{matrix}x+2>=0\\2x+1>=0\\x< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\x< >0\end{matrix}\right.\)

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

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

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

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

=>x-1=0

=>x=1