ĐK \(x\ge-\frac{1}{2}\)

Đặt như trên... (\(a\ge\sqrt{\frac{1}{2}};b\ge0\)) ta có hệ:

\(\hept{\begin{cases}2a^2b=a+b^3\\2a^2-b^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(b^2+1\right)b=a+b^3\\2a^2=b^2+1\end{cases}}\)

Xét pt trình đầu của hệ \(\Leftrightarrow a=b\). Thay b bởi a ở pt dưới ta được:

\(2a^2-a^2-1=0\Leftrightarrow\orbr{\begin{cases}a=1\left(TM\right)\\a=-\frac{1}{2}\left(KTM\right)\end{cases}}\). Với a = 1 thì ta có:

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

Vậy...

\(x^2-2\left(m-1\right)x+m-3=0\)

\(\Delta'=b'^2-ac=\left[-\left(m-1\right)^2\right]-1.\left(m-3\right)=m^2-2m+1-m+3=m^2-3m+4\)

\(=m^2-2.m.\frac{3}{2}+\frac{9}{4}+\frac{7}{4}=\left(m-\frac{3}{2}\right)^2+\frac{7}{4}\)

Vì\(\left(m-\frac{3}{2}\right)^2\ge0\forall m\)

\(\Rightarrow\left(m-\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\forall m\)

\(\Rightarrow\Delta'>0\forall m\)

Vậy...

\(1+\left(\frac{a+2\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\cdot\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)

\(=1+\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}\left(1+\sqrt{a}+a\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

\(=1+\left(\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}}{\left(1-\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

\(=1+\left(\frac{\left(1-\sqrt{a}\right)}{\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}}{\left(1-\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

\(=1+\left(\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

\(=1+\frac{1-2\sqrt{a}+a-\sqrt{a}-a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

\(=1+\frac{1-2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

\(=1+\frac{1-2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-2\sqrt{a}}\)

\(=1+\frac{\sqrt{a}}{\left(1+\sqrt{a}\right)}\)

\(=\frac{1+\sqrt{a}+\sqrt{a}}{1+\sqrt{a}}\)

\(=\frac{1+2\sqrt{a}}{1+\sqrt{a}}\)

Câu hỏi của Hiền Nguyễn Thị

sao ko làm đi :v

ĐKXĐ : \(x\ge1;x\ne2;x\ne3\)

\(P=\left[\frac{\sqrt{x}+\sqrt{x-1}}{1}-\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-3}\right].\frac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)

\(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\left(\sqrt{x}-\sqrt{2}\right)}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)

\(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\)

\(P=\frac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\)