\(a)\) Để pt có hai nghiệm phân biệt \(x_1,x_2\) thì \(\Delta'=\left(1-m\right)^2-m^2+3m=1-2m+m^2-m^2+3m=m+1>0\)\(\Leftrightarrow\)\(m>-1\)

Vậy để pt có hai nghiệm phân biệt \(x_1,x_2\) thì \(m>-1\)

\(b)\) Ta có : \(T=x_1^2+x_2^2-\left(m-1\right)\left(x_1+x_2\right)+m^2-3m\)

\(T=\left(x_1+x_2\right)^2-2x_1x_2+\left(1-m\right)\left(x_1+x_2\right)+m^2-3m\)

Theo định lý Vi-et ta có : \(\hept{\begin{cases}x_1+x_2=2\left(1-m\right)\\x_1x_2=m^2-3m\end{cases}}\)

\(\Rightarrow\)\(T=4\left(1-m\right)^2-2\left(m^2-3m\right)-2\left(1-m\right)\left(1-m\right)+m^2-3m\)

\(T=4m^2-8m+4-2m^2+6m-2m^2+4m-2+m^2-3m\)

\(T=m^2-m+2=\left(m^2-m+\frac{1}{4}\right)+\frac{7}{4}=\left(m-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(m=\frac{1}{2}\) ( thoả mãn ) 

Vậy GTNN của \(T=\frac{7}{4}\) khi \(m=\frac{1}{2}\)

a)\(\)https://www.cymath.com/answer?q=2sqrt(27)-6sqrt(4%2F3)%2B3%2F5sqrt(75)

\(M=2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}=2\sqrt{3^2.3}-6\sqrt{\frac{2^2.3}{3^2}}+\frac{3}{5}\sqrt{5^2.3}=.\) 

        \(=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)  

\(P=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}.Với...0< x< 1\Leftrightarrow\)  \(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{(x-1)}.\frac{\left(1-x\right)}{2x}=\frac{-1}{x}.\)

nãy giải rồi

\(\hept{\begin{cases}7x-3y=4\\4x+y=5\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}7x-3y=4\\12x+3y=15\end{cases}}\)

Cộng vế ta được :

\(7x-3y+12x+3y=4+15\)

\(\Leftrightarrow19x=19\)

\(\Leftrightarrow x=1\)

Khi đó : \(7-3y=4\Leftrightarrow y=1\)

Vậy \(x=y=1\)

a) \(x^4-9x^2+20=0\)

\(\Leftrightarrow x^4-4x^2-5x^2+20=0\)

\(\Leftrightarrow x^2\left(x^2-4\right)-5\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x^2-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\in\left\{\pm2\right\}\\x\in\left\{\pm\sqrt{5}\right\}\end{cases}}\)

Vậy....

\(x^4-9x^2+20=0\)

\(\Leftrightarrow x^4-4x^2-5x^2+20=0\)

\(\Leftrightarrow x^2\left(x^2-4\right)-5\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-5=0\\x^2-4=0\end{cases}}\Leftrightarrow x\in\left\{\pm2\right\}\)

Ta có:\(7\left(\frac{1}{a^2}+...\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2015\)

Mà \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le2015\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{6045}\)

\(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+...\)

Mà \(\left(2+1\right)\left(2a^2+b^2\right)\ge\left(2a+b\right)^2\)(bất dẳng thức buniacoxki)

=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)

Lại có \(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)

=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\le\frac{\sqrt{6045}}{3}\)

Vậy \(MaxP=\frac{\sqrt{6045}}{3}\)khi \(a=b=c=\frac{\sqrt{6045}}{2015}\)

Ta có : \(x^2+2x-m=0\)

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

\(\Leftrightarrow\left(x+1\right)^2=m+1\)

Phương trình vô nghiệm \(\Leftrightarrow m+1< 0\Leftrightarrow m< -1\)

Nhận xét : Nếu \(m+1\ge-1\), phương trình có nghiệm

\(x^2-5x+7=0\)

\(\Rightarrow x(x-5)+7=0\)

\(\Rightarrow x(x+5)=-7\)

Làm nốt :v

\(x^2-5x+7=0\)

\(\Leftrightarrow x^2-2.\frac{5}{4}x+\frac{25}{4}-\frac{25}{4}+7=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2+\frac{3}{4}=0\)

Phương trình vô nghiệm vì : \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\inℝ\)nên \(\left(x-\frac{5}{2}\right)^2+\frac{3}{4}>0\) với \(x\inℝ\)