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Xét \(\Delta=\left(2m-1\right)^2-8\left(m-1\right)=4m^2-12m+9=\left(2m-3\right)^2\ge0\)
=> PT luôn có 2 nghiệm x1,x2 với mọi m
Theo hệ thức Viet ta có \(\hept{\begin{cases}x_1+x_2=\frac{1-2m}{2}\\x_1x_2=\frac{m-1}{2}\end{cases}}\)
\(\Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2=\left(\frac{1-2m}{2}\right)^2-\frac{3\left(m-1\right)}{2}\)
\(=\frac{1-4m+4m^2-6m+6}{4}=\frac{4m^2-10m+7}{4}\)
\(=\frac{\left(2m-\frac{5}{2}\right)^2+\frac{3}{4}}{4}\ge\frac{3}{16}\)
Dấu "=" xảy ra khi \(2m=\frac{5}{2}\Rightarrow m=\frac{5}{4}\Rightarrow\frac{a}{b}=\frac{5}{4}\)
\(\Rightarrow4a=5b\Rightarrow2a=\frac{5b}{2}\)
lúc đó \(P=\frac{5b}{2}+2b=\frac{9b}{2}\)
Khi \(x=1,44\): \(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8,44}{1,2}=\frac{211}{30}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)(ĐK: \(x\ge0,x\ne9\))
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(S=\frac{1}{B}+A=\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{x+7}{\sqrt{x}}=\frac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+\frac{4}{\sqrt{x}}+1\)
\(\ge2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}+1=5\)
Dấu \(=\)khi \(\sqrt{x}=\frac{4}{\sqrt{x}}\Leftrightarrow x=4\)(thỏa mãn)
a) Áp dụng đl Vi-ét vào pt ta có:
x1+x2=-1.5
x1 . x2= -13
C=x1(x2+1)+x2(x1+1)
= 2x1x2 + x1+x2
= 2.(-13) -1.5
= -26 -1.5
= -27.5
a, Theo Vi et : \(\hept{\begin{cases}x_1+x_2=-\frac{b}{a}=-\frac{3}{2}\\x_1x_2=\frac{c}{a}=-13\end{cases}}\)
Ta có : \(C=x_1\left(x_2+1\right)+x_2\left(x_1+1\right)=x_1x_2+x_1+x_1x_2+x_2\)
\(=-13-\frac{3}{2}-13=-26-\frac{3}{2}=-\frac{55}{2}\)
ta có :
\(\sqrt{2}a^2+a-1=0\Leftrightarrow\sqrt{2}a^2=1-a\) nên ta có \(a\le1\)
\(\Rightarrow2a^4=a^2-2a+1\)Vậy \(C=\frac{2a-3}{\sqrt{2\left(a^2-4a+4\right)}+2a^2}=\frac{2a-3}{2a^2+\sqrt{2}\left(2-a\right)}=\frac{2a-3}{\sqrt{2}\left(\sqrt{2}a^2-a+2\right)}\)
\(=\frac{2a-3}{\sqrt{2}\left(1-a-a+2\right)}=\frac{2a-3}{\sqrt{2}\left(3-2a\right)}=-\frac{1}{\sqrt{2}}\)