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Câu hỏi của Tăng Thiện Đạt - Toán lớp 8 - Học toán với OnlineMath
\(ab+bc+ca=3abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
\(Q=\frac{a^2+c^2-c^2}{a\left(c^2+a^2\right)}+\frac{b^2+a^2-a^2}{a\left(a^2+b^2\right)}+\frac{c^2+b^2-b^2}{b\left(b^2+c^2\right)}\)
\(Q=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\left(\frac{a}{a^2+b^2}+\frac{b}{b^2+c^2}+\frac{c}{c^2+a^2}\right)\)
\(Q\ge3-\left(\frac{a}{2ab}+\frac{b}{2bc}+\frac{c}{2ca}\right)=3-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)
\(Q_{min}=\frac{3}{2}\) khi \(a=b=c=1\)
\(7\left(a+b\right)^2-9\left(a-b\right)^2=7\left(a^2+2ab+b^2\right)-9\left(a^2-2ab+b^2\right)\)
\(=-2a^2-2b^2+32ab\)
Từ bđt \(2ab\le a^2+b^2\Rightarrow\)\(32ab\le16\left(a^2+b^2\right)\Rightarrow-2a^2-2b^2+32ab\le14\left(a^2+b^2\right)\)
\(\Rightarrow A\le\frac{14\left(a^2+b^2\right)}{2014\left(a^2+b^2\right)}=\frac{7}{1007}\)
\("="\Leftrightarrow a=b\)
\(ac=-6< 0\Rightarrow\) pt luôn có 2 nghiệm trái dấu
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=1-m\\x_1x_2=-6\end{matrix}\right.\)
\(B=\left(x_1-3\right)\left(x_1+3\right)\left(x_2-2\right)\left(x_2+2\right)\)
\(=\left(x_1-3\right)\left(x_2-2\right)\left(x_1+3\right)\left(x_2+2\right)\)
\(=\left(x_1x_2-2x_1-3x_2+6\right)\left(x_1x_2+2x_1+3x_2+6\right)\)
\(=-\left(2x_1+3x_2\right)\left(2x_1+3x_2\right)=-\left(2x_1+3x_2\right)^2\le0\)
\(\Rightarrow B_{max}=0\) khi \(2x_1+3x_2=0\)
Kết hợp Viet ta được hệ: \(\left\{{}\begin{matrix}x_1+x_2=1-m\\2x_1+3x_2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=3-3m\\x_2=2m-2\end{matrix}\right.\)
Mà \(x_1x_2=-6\Leftrightarrow\left(3-3m\right)\left(2m-2\right)=-6\)
\(\Leftrightarrow\left(m-1\right)^2=1\Rightarrow\left[{}\begin{matrix}m=0\\m=2\end{matrix}\right.\)
\(\frac{\left(1-2a\right)\left(1-2b\right)}{\left(1-a\right)\left(1-b\right)}-\frac{4\left(1-a-b\right)^2}{\left(2-a-b\right)^2}=\frac{\left(1-2a\right)\left(1-2b\right)\left(2-a-b\right)^2-4\left(1-a\right)\left(1-b\right)\left(1-a-b\right)^2}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(=\frac{2a^3-2a^2b-3a^2-2ab^2+6ab+2b^3-3b^2}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(=\frac{\left(2a^3-4a^2b+2ab^2\right)+\left(2a^2b-4ab^2+2b^3\right)-3\left(a^2-2ab+3b^2\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(=\frac{2a\left(a^2-2ab+b^2\right)+2b\left(a^2-2ab+b^2\right)-3\left(a^2-2ab+b^2\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(=\frac{\left(a-b\right)^2\left(2a+2b-3\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)
\(A=\sqrt{\left(x-2\right)\left(x-1\right)x\left(x+1\right)+5}\)
\(=\sqrt{\left(x^2-x-2\right)\left(x^2-x\right)+5}\)
Đặt \(t=x^2-x\) ta đc:
\(A=\sqrt{\left(t-2\right)t+5}=\sqrt{t^2-2t+5}\)
\(=\sqrt{\left(t-1\right)^2+4}\ge\sqrt{4}=2\)
Dấu = khi \(t=1\Leftrightarrow x^2-x=1\Leftrightarrow x=\pm\frac{1}{2}+\frac{\sqrt{5}}{2}\)
Vậy....
b)\(B=\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}\)
\(=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}\)
\(=\left|x-2\right|+\left|x+3\right|\)
Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-2\right|+\left|x+3\right|=\left|x-2\right|+\left|-x-3\right|\ge\left|x-2+\left(-x\right)-3\right|=5\)
Dấu = khi \(\left(x-2\right)\left(x+3\right)\ge0\)\(\Rightarrow-3\le x\le2\)
\(\Rightarrow\hept{\begin{cases}-3\le x\le2\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy....
B1:
1. \(\sqrt{12.5}\cdot\sqrt{0.2}\cdot\sqrt{0.1}\) \(=\sqrt{12.5\cdot0.2\cdot0.1}\) \(=\sqrt{0.25}=0.5\)
2.\(\sqrt{48.4}\cdot\sqrt{5}\cdot\sqrt{0.5}\) = \(\sqrt{48.4\cdot5\cdot0.5}\) =\(\sqrt{121}=11\)
B2:
a, \(\left(\sqrt{7}+\sqrt{3}\right)^2=7+2\cdot\sqrt{7}\cdot\sqrt{3}+3=7+2\cdot\sqrt{21}+3\)\(=10+2\sqrt{21}\)
b,\(\left(\sqrt{11}-\sqrt{5}\right)^2=11-2\sqrt{55}+5=16-2\sqrt{55}\)
c,\(\left(\sqrt{x}+\sqrt{y}\right) ^2=x+2\sqrt{xy}+y\)
d.\(\left(\sqrt{13}+\sqrt{7}\right)^2=13+2\sqrt{7}+7=20+2\sqrt{7}\)
e,\(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\)
f,\(\left(\sqrt{3}-1\right)^2=3-2\sqrt{3}+1=4-2\sqrt{3}\)