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28 tháng 3 2021

ĐK: \(-1\le x< 0;x\ge1\)

TH1: \(-1\le x< 0\Rightarrow VP< 0;VT\ge0\Rightarrow\) vô nghiệm

TH2: \(x\ge1\)

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

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

\(\Leftrightarrow x^2-x+1-2\sqrt{x^2-x}=0\)

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

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

\(\Leftrightarrow x^2-x-1=0\)

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

\(\Leftrightarrow x=\dfrac{1+\sqrt{5}}{2}\)

Vậy ...

1 tháng 2 2022

mình nhầm mẫu nhé :v mình làm lại 

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

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

Đề sai rồi bạn

23 tháng 7 2023

\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\) (ĐK: \(x\ge0,x\ne1\))

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

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

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

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

\(\Leftrightarrow0=-2\) (vô lý)

⇒ Phương trình vô nghiệm

23 tháng 7 2023

\(đk:x\ge0;x\ne1\)

\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ \Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\\ \Rightarrow x-2\sqrt{x}+\sqrt{x}-2=x-\sqrt{x}\\ \Rightarrow-\sqrt{x}-2+\sqrt{x}=0\\ \Rightarrow-2=0\left(voli\right)\)

Vậy phương trình vô nghiệm

3 tháng 11 2018

Ta có :

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

Tương tự :

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

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

....

\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)

Từ những ý trên , pt trở thành :

\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)

\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)

\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)

\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)

\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)

\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)

\(\Leftrightarrow121x-900480=0\)

\(\Leftrightarrow x=\dfrac{900480}{121}\)

29 tháng 10 2021

\(ĐK:x\ge1\\ PT\Leftrightarrow x-\sqrt{x-\dfrac{1}{x}}=\sqrt{1-\dfrac{1}{x}}\\ \Leftrightarrow x^2+x-\dfrac{1}{x}-2x\sqrt{x-\dfrac{1}{x}}=1-\dfrac{1}{x}\\ \Leftrightarrow x^2+x-1=2x\sqrt{x-\dfrac{1}{x}}\\ \Leftrightarrow x^4+x^2+1+2x^3-2x-2x^2=4x^3-4x\\ \Leftrightarrow x^4-2x^3-x^2+2x+1=0\\ \Leftrightarrow\left(x^2-x-1\right)^2=0\\ \Leftrightarrow x^2-x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(ktm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(x=\dfrac{1+\sqrt{5}}{2}\)

23 tháng 7 2021

a, \(\sqrt[3]{\dfrac{2x}{x+1}}.\sqrt[3]{\dfrac{x+1}{2x}}=2\)

⇔ \(\left\{{}\begin{matrix}1=2\\x\ne0\&x\ne-1\end{matrix}\right.\)

Phương trình vô nghiệm

b, x = \(\dfrac{8}{125}\)

30 tháng 7 2017

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

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

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{x-1}{x}=0\)

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

Pt \(\dfrac{1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{1}{x}=0\) vô n0

=> x - 1 = 0

<=> x = 1 (nhận)

22 tháng 6 2021

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

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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\4x+3=4x^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\\left(2x-3\right)\left(2x+1\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x=-\dfrac{1}{2}\)

Vậy...

22 tháng 6 2021

Sao không nhân 2 cho đỡ khổ phân số =))?