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1)
xét a+b+c = (a+b+c)(\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)) = \(\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=\)
\(\frac{a^2}{b+c}+\frac{a\left(b+c\right)}{b+c}+\frac{b^2}{a+c}+\frac{b\left(a+c\right)}{a+c}+\frac{c^2}{a+b}+\frac{c\left(a+b\right)}{a+b}=Q+a+b+c\)
<=> a+b+c =Q + a+b+c => Q=0
2) = (x+ y)2 + (x+ 1)2 +y(x+ 1) +x + y + 1 =0 <=> (x+ y)(x+ y+ 1) + (x+ 1)(x+ y+ 1) + 1= 0 <=> (x+ y+ 1)(2x+ y+ 1) = -1
=> \(\hept{\begin{cases}x+y+1=1\\2x+y+1=-1\end{cases}}\)hoặc \(\hept{\begin{cases}x+y+1=-1\\2x+y+1=1\end{cases}}\)<=> \(\hept{\begin{cases}x=-2\\y=2\end{cases}}\)hoặc \(\hept{\begin{cases}x=2\\y=-4\end{cases}}\)
c) Có \(P=\frac{ax+b}{x^2+1}=-1+\frac{x^2+ax+b+1}{x^2+1}\);
\(P=\frac{ax+b}{x^2+1}=4-\frac{4x^2-ax-b+4}{x^2+1}\)
Để Min P = 1 và Max P = 4 thì
\(\hept{\begin{cases}x^2+ax+b+1=\left(x+c\right)^2\\4x^2-ax-b+4=\left(2x+d\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\left(a-2c\right)+\left(b+1-c^2\right)=0\left(1\right)\\x\left(-a-4d\right)+\left(-b+4-d^2\right)=0\left(2\right)\end{cases}}\)
(1) = 0 khi \(\hept{\begin{cases}a=2c\\b=c^2-1\end{cases}}\)(3)
(2) = 0 khi \(\hept{\begin{cases}a=-4d\\b=4-d^2\end{cases}}\)(4)
Từ (3) (4) => d = 1 ; c = -2 ; b = 3 ; a = -4
Vậy \(P=\frac{-4x+3}{x^2+1}\)
ĐK \(x\ge y\)
Đặt \(\sqrt{x+y}=a;\sqrt{x-y}=b\left(a;b\ge0\right)\)
HPT <=> \(\hept{\begin{cases}a^4+b^4=82\\a-2b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(2b+1\right)^4+b^4=82\\a=2b+1\end{cases}}\Leftrightarrow\hept{\begin{cases}17b^4+32b^3+24b^2+8b-81=0\\a=2b+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}17b^4-17b^3+49^3-49b^2+73b^2-73b+81b-81=0\\a=2b+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(b-1\right)\left(17b^3+49b^2+73b+81\right)=0\left(1\right)\\a=2b+1\end{cases}}\)
Giải (1) ; kết hợp điều kiện => b = 1
=> Hệ lúc đó trở thành \(\hept{\begin{cases}b=1\\a=2b+1\end{cases}}\Leftrightarrow\hept{\begin{cases}b=1\\a=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+y}=3\\\sqrt{x-y}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=9\\x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=10\\x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=4\end{cases}}\)
Vậy hệ có 1 nghiệm duy nhất (x;y) = (5;4)
\(\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}=\frac{1}{a^2+a^2+b^2}+\frac{1}{b^2+b^2+c^2}+\frac{1}{c^2+c^2+a^2}\)
\(< =\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{1}{9}\left(\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{1}{9}\left(\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)\)(bđt svacxo)
\(=\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)=\frac{1}{9}\cdot3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(=\frac{1}{9}\cdot3\cdot\frac{1}{3}=\frac{1}{9}\cdot1=\frac{1}{9}\)
\(\Rightarrow\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}< =\frac{1}{9}\)(đpcm)
dấu = xảy ra khi \(\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}=\frac{1}{9}\Rightarrow a=b=c=3\)
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 : \(C=x_1\left(x_2+1\right)+x_2\left(x_1+1\right)=x_1x_2+x_1+x_1x_2+x_2\)
\(=-13-\frac{3}{2}-13=-26-\frac{3}{2}=-\frac{55}{2}\)
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 \(P=\frac{5b}{2}+2b=\frac{9b}{2}\)
Theo Viet: \(\left\{{}\begin{matrix}a+b=1\\ab=-\frac{1}{2}\end{matrix}\right.\)
\(M=\frac{\left(a+2\right)\left(a-1\right)+\left(b+2\right)\left(b-1\right)}{\left(a-1\right)\left(b-1\right)}=\frac{a^2+b^2+a+b-4}{ab-\left(a+b\right)+1}=\frac{\left(a+b\right)^2-2ab+a+b-4}{ab-\left(a+b\right)+1}=\frac{1+1+1-4}{-\frac{1}{2}-1+1}=2\)
\(N=\frac{a^2+b^2}{\left(ab\right)^2}=\frac{\left(a+b\right)^2-2ab}{\left(ab\right)^2}=\frac{1+1}{\frac{1}{4}}=8\)
\(P=\frac{a^2\left(2a+1\right)+b^2\left(2b+1\right)}{\left(2a+1\right)\left(2b+1\right)}=\frac{2\left(a^3+b^3\right)+a^2+b^2}{4ab+2\left(a+b\right)+1}=\frac{2\left(a+b\right)^3-6ab\left(a+b\right)+\left(a+b\right)^2-2ab}{4ab+2\left(a+b\right)+1}=...\)
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