Bài 2. 

\(x^2-2mx+4=0\)(2) 

a) Với \(m=3\)(2) trở thành: 

\(x^2-6x+4=0\)

\(\Leftrightarrow x^2-6x+9=5\)

\(\Leftrightarrow\left(x-3\right)^2=5\)

\(\Leftrightarrow x=\pm\sqrt{5}+3\)

b) Để (1) có hai nghiệm phân biệt \(x_1\), \(x_2\)thì \(\Delta'>0\).

\(\Delta'=m^2-4>0\Leftrightarrow\orbr{\begin{cases}m>2\\m< -2\end{cases}}\)

Theo Viete:

\(\hept{\begin{cases}x_1+x_2=2m\\x_1x_2=4\end{cases}}\)

\(\left(x_1+1\right)^2+\left(x_2+1\right)^2=x_1^2+x_2^2+2\left(x_1+x_2\right)+2\)

\(=\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)+2\)

\(=4m^2-8+4m+2=2\)

\(\Leftrightarrow m^2+m-2=0\)

\(\Leftrightarrow\orbr{\begin{cases}m=1\left(l\right)\\m=-2\left(l\right)\end{cases}}\)

Bài 1. 

a) Với \(m=-3\): (1) trở thành:

\(x^2+6x-7=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-7\\x=1\end{cases}}\)

b) Để (1) có hai nghiệm phân biệt \(x_1\), \(x_2\)thì \(\Delta'>0\).

\(\Delta'=m^2-\left(2m-1\right)=m^2-2m+1=\left(m-1\right)^2\)

\(\left(m-1\right)^2>0\Leftrightarrow m\ne1\).

Theo Viete ta có: 

\(\hept{\begin{cases}x_1+x_2=2m\\x_1x_2=2m-1\end{cases}}\)

\(2\left(x_1^2+x_2^2\right)-5x_1x_2=2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2\)

\(=2\left[\left(2m\right)^2-2\left(2m-1\right)\right]-5\left(2m-1\right)=8m^2-18m+9=27\)

\(\Leftrightarrow8m^2-18m-18=0\)

\(\Leftrightarrow\orbr{\begin{cases}m=3\\m=-\frac{3}{4}\end{cases}}\)(thỏa mãn)

\(x-\sqrt{x}=m\Leftrightarrow x-\sqrt{x}-m=0\)

\(\Delta=1-4\left(-m\right)=1+4m\)

Để phương trình trên có nghiệm khi \(\Delta\ge0\)

\(1+4m\ge0\Leftrightarrow m\ge-\frac{1}{4}\)

sửa đề : 

\(N=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)đk : x >= 0 

\(=\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\right)=\frac{x-1}{\sqrt{x}}:\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

n = (\(\sqrt{yyx}\)-r554-56

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

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

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

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

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

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

Ta có: \(P=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)

sử dụng bđt cô-si có: \(a^2+\frac{1}{16a^2}\ge\frac{1}{2};b^2+\frac{1}{16b^2}\ge\frac{1}{2};\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}=\frac{4}{2ab}\)

Lại có: \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{4}{a^2+b^2}\)

\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge4\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)\ge4\frac{4}{a^2+b^2+2ab}=\frac{16}{\left(a+b\right)^2}=16\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge8\)

\(\Rightarrow P\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{2}=\frac{17}{2}\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)