\(\frac{18\sqrt{14}-60}{2\left(3\sqrt{7}-5\sqrt{2}\right)}\) giải giúp nhé
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\(\frac{6\sqrt{6}-27}{2\sqrt{2}-3\sqrt{3}}=\frac{3\sqrt{3}\left(2\sqrt{2}-3\sqrt{3}\right)}{2\sqrt{2}-3\sqrt{3}}=3\sqrt{3}\)
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Đặt :
\(A=\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}.\)
Áp dụng bất đăng thức AM - GM :
\(\frac{a^3}{a^2+ab+b^2}\ge\frac{a^3}{a^2+\frac{a^2+b^2}{2}+b^2}=\frac{a^3}{\frac{3}{2}\left(a^2+b^2\right)}\)
Chứng minh tương tự : \(\hept{\begin{cases}\frac{b^3}{b^2+bc+c^2}\ge\frac{b^3}{\frac{3}{2}\left(b^2+c^2\right)}\\\frac{c^3}{c^2+ac+a^2}\ge\frac{c^3}{\frac{3}{2}\left(c^2+a^2\right)}\end{cases}}\)
Cộng theo vế :
\(A\ge\frac{2}{3}\left(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\right)\)
Tiếp tục áp dụng AM - GM ta có :
\(\frac{a^3}{a^2+b^2}=\frac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\frac{ab^2}{a^2+b^2}\ge a-\frac{ab^2}{2ab}=a-\frac{b}{2}\)
Chứng minh tương tự : \(\hept{\begin{cases}\frac{b^3}{b^2+c^2}\ge b-\frac{c}{2}\\\frac{c^3}{c^2+a^2}\ge c-\frac{a}{2}\end{cases}}\)
Cộng theo vế :
\(A\ge\frac{2}{3}\left(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\right)\ge\frac{2}{3}\left(a+b+c-\frac{a+b+c}{2}\right)=\frac{a+b+c}{3}\)
Nguồn : https://h7.net/hoi-dap/toan-9/chung-minh-rang-a-3-a-2-ab-b-2-b-3-b-2-bc-c-2-c-3-c-2-ac-a-2-a-b-c-3-faq413350.html
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\(x=2\sqrt{2}+1\Leftrightarrow x-1=2\sqrt{2}\Rightarrow x^2-2x+1=8\)
\(\Leftrightarrow x^2=2x+7\Leftrightarrow x^3=2x^2+7x\)
\(A=2x^3-x^2-20x+1992\)
\(=4x^2+14x-x^2-20x+1992\)
\(=3\left(x^2-2x\right)+1992\)
\(=3.7+1992\)
\(=2013\)
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\(2\ge2x+3y\ge2\sqrt{2x.3y}\Rightarrow xy\le\frac{1}{6}.\)
\(A=\frac{4}{4x^2+9y^2}+\frac{9}{xy}=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{26}{3xy}\ge\frac{\left(2+2\right)^2}{4x^2+9y^2+12xy}+\frac{26}{3xy}\)
\(\ge\frac{4^2}{2^2}+\frac{26}{3.\frac{1}{6}}=56\)
Dấu \(=\)khi \(\hept{\begin{cases}\frac{2}{4x^2+9y^2}=\frac{2}{12xy}\\2x=3y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\end{cases}}\)
Áp dụng bất đẳng thức Cô – si cho hai số dương, ta có: 2x + 3y ≥ 2
⇔ 2 ≤ 2x + 3y
Mà 2x + 3y ≤ 2
Do đó ≤ 1 6xy ≤ 1. Kết hợp kết quả ở câu 1 ta có:
A = = 4(
) +
≥ 4
+
= 16
≥ 16.
= 56
Dấu “ = “ xảy ra ⇔ ⇔
Vậy giá trị nhỏ nhất của biểu thức A là 56.
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Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\Leftrightarrow a-2\sqrt{ab}+b\ge0\Leftrightarrow a+b\ge2\sqrt{ab}\)
Dấu \(=\)khi \(a=b\ge0\).
Tương tự ta cũng có: \(b+c\ge2\sqrt{bc},c+a\ge2\sqrt{ca}\).
Cộng lại vế theo vế ta được:
\(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\Leftrightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Dấu \(=\)khi \(a=b=c\).
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sửa đề : \(E=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)ĐK : \(x>0;x\ne1\)
\(=\left(\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}+1}{x-1}\right)\)
\(=\frac{\left(x-1\right)^2}{\sqrt{x}\left(x-1\right)}=\frac{x-1}{\sqrt{x}}\)
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\(f\left(x\right)=x^2-\left(2m-1\right)x+m^2-m-6=0\)
\(\Delta=\left(2m-1\right)^2-4\left(m^2-m-6\right)>0\)
Suy ra phương trình luôn có hai nghiệm phân biệt
Để \(-5< x_1< x_2< 5\) thì :
\(\hept{\begin{cases}f\left(-5\right)>0\\f\left(5\right)>0\\-5< \frac{2m-1}{2}< 5\end{cases}}\Leftrightarrow\hept{\begin{cases}m^2+9m+14>0\\m^2-11m+24>0\\-\frac{9}{2}< m< \frac{11}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x< -7\left(h\right)x>-2\\x< 3\left(h\right)x>8\\-\frac{9}{2}< m< \frac{11}{2}\end{cases}}\Leftrightarrow-2< x< 3\)
Chữa:
\(\hept{\begin{cases}m< -7\left(h\right)m>-2\\m< 3\left(h\right)m>8\\-\frac{9}{2}< m< \frac{11}{2}\end{cases}}\Leftrightarrow-2< m< 3\)
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a) \(\sqrt{15}< \sqrt{16}=4\)
\(\sqrt{2}< \sqrt{4}=2\)
\(\Rightarrow\sqrt{15}+\sqrt{2}< \sqrt{16}+\sqrt{4}=4+2=6\)
\(\sqrt{14}< \sqrt{16}=4\)
\(\sqrt{3}< \sqrt{9}=3\)
\(\Rightarrow\sqrt{14}+\sqrt{3}< \sqrt{16}+\sqrt{9}=4+3=7\)
\(\Leftrightarrow\sqrt{15}+\sqrt{2}< \sqrt{14}+\sqrt{3}\)( nhận thấy 6 < 7)
b) \(\sqrt{29}-\sqrt{28}=\frac{29-28}{\sqrt{29}+\sqrt{28}}=\frac{1}{\sqrt{29}+\sqrt{28}}\)
\(\sqrt{28}-\sqrt{27}=\frac{28-27}{\sqrt{28}+\sqrt{27}}=\frac{1}{\sqrt{28}+\sqrt{27}}\)
Mà \(\sqrt{29}+\sqrt{28}>\sqrt{28}+\sqrt{27}\)
\(\Rightarrow\frac{1}{\sqrt{29}+\sqrt{28}}< \frac{1}{\sqrt{28}+\sqrt{27}}\)
\(\Rightarrow\sqrt{29}-\sqrt{28}< \sqrt{28}-\sqrt{27}\)
\(\frac{2\left(9\sqrt{14}-30\right)}{2\left(3\sqrt{7}-5\sqrt{2}\right)}\)
\(\frac{3\sqrt{2}\left(3\sqrt{7}-5\sqrt{2}\right)}{\left(3\sqrt{7}-5\sqrt{2}\right)}\)
\(=3\sqrt{2}\)