Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Từ giả thiết ta dễ thấy dấu "=" xảy ra khi a=1, b=3, c=5
Áp dụng BĐT Cauchy Schawrz, ta có:
\(a^2+\frac{b^2}{3}+\frac{c^2}{5}\ge\frac{\left(a+b+c\right)^2}{1+3+5}\Rightarrow2\sqrt{a^2+\frac{b^2}{3}+\frac{c^2}{5}}\ge\frac{2\left(a+b+c\right)}{3}\)
\(\frac{1}{a}+\frac{9}{b}+\frac{25}{c}\ge\frac{\left(1+3+5\right)^2}{a+b+c}\Rightarrow3\sqrt{\frac{1}{a}+\frac{9}{b}+\frac{25}{c}}\ge\frac{27}{\sqrt{a+b+c}}\)
Từ đó, suy ra
\(A\ge\frac{2\left(a+b+c\right)}{3}+\frac{27}{\sqrt{a+b+c}}=\frac{a+b+c}{6}+\frac{a+b+c}{2}+\frac{27}{2\sqrt{a+b+c}}+\frac{27}{2\sqrt{a+b+c}}\ge\frac{9}{6}+3\sqrt[3]{\frac{729}{8}}=15\)
Dấu "=" xảy ra khi a=1, b=3, c=5
Mong là không có gì sai sót!
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
Ta có: \(4ab\le2a^2+2b^2\)
=> \(\sqrt{2a^2+7b^2+16ab}\le\sqrt{4a^2+9b^2+12ab}=\sqrt{\left(2a+3b\right)^2}=2a+3b\)
=> \(\frac{a^2}{\sqrt{2a^2+7b^2+16ab}}\ge\frac{a^2}{2a+3b}\)
Chứng minh tương tự
=> \(T\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\)
Áp dụng bđt bunhia dạng phân thức
=> \(T\ge\frac{\left(a+b+c\right)^2}{2a+3b+2b+3c+2c+3a}=\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=1\)
=> \(MinT=1\)xảy ra khi a=b=c=5/3
\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)
\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)
\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)
\(4,A=x+\sqrt{x}+1\)
\(A=\left(\sqrt{x}\right)^2+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
Dấu "=" xảy ra khi :
\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)
Vậy Min A = 3/4 khi căn x = -1/2
ĐKXĐ:\(\hept{\begin{cases}a,b,c\ge0\\a,b,c\ne\frac{25}{4}\end{cases}}\)
\(\frac{a}{2\sqrt{b}-5}+2\sqrt{b}-5\ge2\sqrt{a}\left(cosi\right)\)
Làm tương tự rồi cộng 3 vế lại nha bn