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 :

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

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

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

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

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

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

\(\Leftrightarrow121x-900480=0\)

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

Bài 1

b) ta có pt \(\sqrt{25-x^2}-\sqrt{9-x^2}=2\)

Đặt \(\sqrt{25-x^2}=a;\sqrt{9-x^2}=b\left(a,b\ge0\right)\Rightarrow a-b=2\)

Mà \(a^2-b^2=25-x^2-9+x^2=16\Leftrightarrow\left(a-b\right)\left(a+b\right)=16\Leftrightarrow a+b=8\)

ta có a-b=2;a+b=8=> a=5;b=3

a) ta có pt \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\Leftrightarrow x-\dfrac{4}{x}+\sqrt{2x-\dfrac{5}{x}}-\sqrt{x-\dfrac{1}{x}}=0\)

đặt \(\sqrt{2x-\dfrac{5}{x}}=a;\sqrt{x-\dfrac{1}{x}}=b\Rightarrow a^2-b^2=2x-\dfrac{5}{x}-x+\dfrac{1}{x}=x-\dfrac{4}{x}\)

nên pt \(\Leftrightarrow a^2-b^2+a-b=0\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

Lời giải:

ĐK: \(x\geq \frac{-1}{4}\)

Biến đổi:

\(x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}=\left(x+\frac{1}{4}\right)+2.\frac{1}{2}\sqrt{x+\frac{1}{4}}+(\frac{1}{2})^2\)

\(=(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2\)

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

Do đó pt ban đầu tương đương với:

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

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

\(\Leftrightarrow (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2=2\)

\(\Rightarrow \sqrt{x+\frac{1}{4}}+\frac{1}{2}=\sqrt{2}\) (TH bằng $-\sqrt{2}$ ta có thể loại luôn vì biểu thức không âm)

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

\(\Rightarrow x=2-\sqrt{2}\) (t/m)

Vậy..............

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

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

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

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

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)

\(\dfrac{3\sqrt{x}}{2}-\dfrac{2\sqrt{x}-7}{3}=\sqrt{x}-1\)

\(\Leftrightarrow9\sqrt{x}-15-4\sqrt{x}+14=6\sqrt{x}-6\left(x\ge0\right)\)

\(\Leftrightarrow5\sqrt{x}-1=6\sqrt{x}-6\)

\(\Leftrightarrow x=25\left(TM\right)\)

KL.....