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Để phương trình có nghiệm x1;x2 thì :
\(\Delta'=\left(m+4\right)^2-\left(m^2-8\right)\)
\(=\left(m^2+8m+16\right)-m^2+8\)
\(=8m+24\ge0\Leftrightarrow m\ge-3\)
Theo hệ thức Viet,ta có :
\(\left\{{}\begin{matrix}x1+x2=2\left(m+4\right)\\x1.x2=m^2-8\end{matrix}\right.\)
a) \(A=x1^2+x2^2-x1-x2=\left(x1+x2\right)^2-\left(x1+x2\right)-2x1x2=4\left(m+4\right)^2-2\left(m+4\right)-2\left(m^2-8\right)\)
\(A=2m^2+30m+66=0\)
\(A=\left(4m+3\right)^2-\frac{519}{8}\ge-\frac{519}{8}\)
b) \(B=2\left(m+4\right)-3\left(m^2-8\right)\)
\(B=-3m^2+2m+32\)
\(B=\frac{97}{3}-\left(3x-1\right)^2\le\frac{97}{3}\Leftrightarrow x=\frac{1}{3}\)
c) \(C=x1^2+x2^2-x1x2=\left(x1+x2\right)^2-3x1x2\)
\(C=4\left(m+4\right)^2-3\left(m^2-8\right)\)
\(C=-3m^2+4m+28\)
\(C=\frac{88}{3}-\left(3x-2\right)^2\le\frac{88}{3}\Leftrightarrow x=\frac{2}{3}\)
\(\Delta^`\ge0\)
\(\Leftrightarrow m^2-\left(m^2-2\right).2\ge0\)
\(\Leftrightarrow4-m^2\ge0\)
\(\Leftrightarrow4\ge m^2\)
\(\Leftrightarrow4\ge m^2\)
\(\Leftrightarrow-2\le m\le2\)
Theo hệ thức Viet có:
\(\hept{\begin{cases}x_1+x_2=m\\x_1.x_2=\frac{m^2-2}{2}\end{cases}}\)
\(\Rightarrow A=\left|2x_1.x_2-x_1-x_2-4\right|=\left|m^2-m-6\right|=\left|\left(m-\frac{1}{2}\right)^2-6,25\right|\)
Có:
\(\left(m-\frac{1}{2}\right)^2\le\left(-2-\frac{1}{2}\right)^2=6,25\)
\(\Rightarrow A=\left|\left(m-\frac{1}{2}\right)^2-6,25\right|=6,25-\left(m-\frac{1}{2}\right)^2\le6,25\)
\(A=6,25\Leftrightarrow m=\frac{1}{2}\left(tm\right)\)
KL:..............................................
\(M=\left(\frac{x-1+\sqrt{xy}+\sqrt{y}}{\sqrt{x}+1}+1\right)\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\sqrt{y}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+1\right)\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+\sqrt{y}-1\right)}{\sqrt{x}+1}+1\right)\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}-1+1\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)
Câu 2:
a/ Bạn tự giải
b/ \(\Delta'=\left(m-1\right)^2-m+5=m^2-3m+6=\left(m-\frac{3}{2}\right)^2+\frac{15}{4}>0\)
Pt luôn có 2 nghiệm phân biệt với mọi m
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=m-5\end{matrix}\right.\)
\(P=x_1^2+x_2^2+2x_1x_2-2x_1x_2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=4\left(m-1\right)^2-2\left(m-5\right)\)
\(=4m^2-10m+14\)
\(=\left(2m-\frac{5}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\)
\(\Rightarrow P_{min}=\frac{31}{4}\) khi \(2m-\frac{5}{2}=0\Leftrightarrow m=\frac{5}{4}\)
\(\Delta=\left(2m+5\right)^2-4\left(2m+1\right)=4m^2+12m+21=\left(2x+3\right)^2+12>0\)
Phương trình luôn có 2 nghiệm pb
Để biểu thức đề bài có nghĩa \(\Rightarrow\left\{{}\begin{matrix}x_1\ge0\\x_2\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2m+5>0\\x_1x_2=2m+1\ge0\end{matrix}\right.\) \(\Rightarrow m\ge-\frac{1}{2}\)
\(P=\left|\sqrt{x_1}-\sqrt{x_2}\right|\Rightarrow P^2=x_1+x_2-2\sqrt{x_1x_2}\)
\(P^2=2m+5-2\sqrt{2m+1}\)
\(P^2=2m+1-2\sqrt{2m+1}+1+4\)
\(P^2=\left(\sqrt{2m+1}-1\right)^2+4\ge4\)
\(\Rightarrow P\ge2\Rightarrow P_{min}=2\) khi \(\sqrt{2m+1}=1\Rightarrow m=0\)
\(P=a-2\sqrt{a}=\left(a-2\sqrt{a}+1\right)-1=\left(\sqrt{a}-1\right)^2-1\ge-1\)
\(MinP=-1\Leftrightarrow\sqrt{a}=1\Leftrightarrow a=1\left(tmĐKXĐ\right)\)
\(\Rightarrow a_0=1\)
Vậy \(E=1^2+1=2\)