\(\dfrac{1}{a-b}.\sqrt{3^2\left(a-b\right)^2}\) vs a>b
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7 tháng 7 2021
Giải:
a) \(A=\dfrac{10^{1990}+1}{10^{1991}+1}\) và \(B=\dfrac{10^{1991}+1}{10^{1992}+1}\)
Ta có:
\(A=\dfrac{10^{1990}+1}{10^{1991}+1}\)
\(10A=\dfrac{10^{1991}+10}{10^{1991}+1}\)
\(10A=\dfrac{10^{1991}+1+9}{10^{1991}+1}\)
\(10A=1+\dfrac{9}{10^{1991}+1}\)
Tương tự :
\(B=\dfrac{10^{1991}+1}{10^{1992}+1}\)
\(10B=\dfrac{10^{1992}+10}{10^{1992}+1}\)
\(10B=\dfrac{10^{1992}+1+9}{10^{1992}+1}\)
\(10B=1+\dfrac{9}{10^{1992}+1}\)
Vì \(\dfrac{9}{10^{1991}+1}>\dfrac{9}{10^{1992}+1}\) nên \(10A>10B\)
\(\Rightarrow A>B\left(đpcm\right)\)
Chúc bạn học tốt!
8 tháng 7 2021
a) Ta có: \(B=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
=1-a
8 tháng 7 2021
b) Ta có: \(2x^2-5x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};2\right\}\)
N
22 tháng 6 2017
ta có :\(a^2-ab+b^2=\left(a+b\right)^2-3ab\ge\left(a+b\right)^2-\dfrac{3}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)(theo BĐT AM-GM)
\(\Rightarrow P\ge\sum\dfrac{a+b}{2\sqrt{ab+1}}\)
ÁP dụng BĐT AM-GM:
\(\dfrac{a+b}{2\sqrt{ab+1}}+\dfrac{b+c}{2\sqrt{bc+1}}+\dfrac{c+a}{2\sqrt{ca+1}}\ge3\sqrt[3]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}=\dfrac{3}{2}.\dfrac{1}{\sqrt[3]{\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}\)
Mà \(\sqrt[3]{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\le\dfrac{1}{3}\left(ab+bc+ca+3\right)\)
\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\left(ab+bc+ca+3\right)}}\)(*)
ta liên tưởng đến BĐT phụ:\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
Cm: phân tích :\(VT=xy\left(x+y\right)+yz\left(y+z\right)+zx\left(x+z\right)+2xyz\)
\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(z+x\right)+3xyz-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)
mà \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz\)
nên \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
Áp dụng:
\(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
mặt khác,theo AM-GM,dễ dàng chứng minh được \(a+b+c\ge\dfrac{3}{2}\)
nên \(1\ge\dfrac{8}{9}.\dfrac{3}{2}\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le\dfrac{3}{4}\)
từ (*)\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\dfrac{3}{4}+3}}=\dfrac{3}{\sqrt{5}}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{2}\)
25 tháng 2 2022
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(\Rightarrow A=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}}\)
13 tháng 7 2018
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 2 2019
Biến đổi tương đương:
\(\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{2}\right)^3\Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\)
\(\Leftrightarrow4a^3+4b^3\ge a^3+3a^2b+3ab^2+b^3\)
\(\Leftrightarrow a^3+b^3-a^2b-ab^2\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng do \(\left\{{}\begin{matrix}a\ge0;b\ge0\Rightarrow a+b\ge0\\\left(a-b\right)^2\ge0\forall a;b\end{matrix}\right.\) )
Vậy BĐT được chứng minh, dấu "=" xảy ra khi \(a=b\)
28 tháng 4 2021
a) Ta có: \(A=\sqrt{23+6\sqrt{10}}-\sqrt{23-6\sqrt{10}}\)
\(=\sqrt{18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5}-\sqrt{18-2\cdot3\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(3\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{5}\right)^2}\)
\(=3\sqrt{2}+\sqrt{5}-3\sqrt{2}+\sqrt{5}\)
\(=2\sqrt{5}\)
28 tháng 4 2021
b) Ta có: \(B=\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+1\right)\left(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-1\right)\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+1\right)\left(\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-1\right)\)
\(=\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)\)
=2-1=2
\(\dfrac{1}{a-b}\cdot\sqrt{3^2\left(a-b\right)^2}\left(a>b\right)\\ =\dfrac{1}{a-b}\cdot3\left|a-b\right|=\dfrac{3\left(a-b\right)}{a-b}=3\)