cho a,b,c\(\in N\)sao cho \(\sqrt{a-6\sqrt{2}}\)=\(\sqrt{b}+3\sqrt{c}\)
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8 tháng 2 2019
Từ giả thiết: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}\)
Xét hạng tử: \(\frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}\)
Từ đó: \(N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}\)
\(=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)
Mặt khác: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13\)
Suy ra: \(N=\frac{4}{9-13}=-1\). Kết luận: N = -1.
25 tháng 9 2019
Từ giả thiết: \sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}a+b+c=7⇔c=7−a−b
Xét hạng tử: \frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}ab+c−61=ab+7−a−b−61=(a−1)(b−1)1
Từ đó: N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}N=(a−1)(b−1)1+(b−1)(c−1)1+(c−1)(a−1)1
=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}=(a−1)(b−1)(c−1)a+b+c−3=abc−(ab+bc+ca)+(a+b+c)−1a+b+c−3
=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=3−(ab+bc+ca)+7−17−3=9−(ab+bc+ca)4
Mặt khác: \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13ab+bc+ca=2(a+b+c)2−(a+b+c)=13
Suy ra: N=\frac{4}{9-13}=-1N=9−134=−1. Kết luận: N = -1.
10 tháng 8 2017
ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
a. Ta có \(P=\frac{3a+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-2}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+2}-1\)
\(=\frac{3a+3\sqrt{a}-3-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{3a+3\sqrt{a}-3-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\)
b. Để \(\left|P\right|=2\Rightarrow\orbr{\begin{cases}P=2\\P=-2\end{cases}}\)
Với \(P=2\Rightarrow\sqrt{a}+1=2\sqrt{a}-2\Rightarrow\sqrt{a}=3\Rightarrow a=9\)
Với \(P=-2\Rightarrow\sqrt{a}+1=2-2\sqrt{a}\Rightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)
c. Ta có \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Để \(P\in N\Rightarrow P\in Z\Rightarrow\sqrt{a}-1\in\left\{-2;-1;1;2\right\}\)
| \(\sqrt{a}-1\) | \(-2\) | \(-1\) | \(1\) | \(2\) |
| \(\sqrt{a}\) | \(-1\) | \(0\) | \(2\) | \(3\) |
| \(a\) | \(0\) | \(4\) | \(9\) | |
| \(\left(l\right)\) | \(\left(tm\right)\) | \(\left(tm\right)\) | \(\left(tm\right)\) |
Vậy \(x\in\left\{0;4;9\right\}\)thì \(P\in N\)
S
0
HV
20 tháng 7 2019
Với \(a,b,c\ge0\). Khi đó ta có
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{ab+bc+ca}\)
Chứng minh: \(\left(ab+bc+ca\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a^2+b^2+c^2+abc\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge a^2+b^2+c^2\)\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{ab+bc+ac}\)
Với \(a,b,c\ge0\) ta có
\(\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(b+a\right)\left(c+a\right)}}+\sqrt{\frac{ca}{\left(c+b\right)\left(c+a\right)}}\ge1\)
Áp dụng bất đẳng thức AM-GM ta có:
\(\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}=\Sigma\sqrt{\frac{ab\left(2ab+2bc+2ac\right)^2}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ca\right)^2}}\)
\(\ge\Sigma\sqrt{\frac{ab\left[a\left(b+c\right)+b\left(a+c\right)\right]^2}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ac\right)^2}}\)
\(\ge\Sigma\sqrt{\frac{ab.4a\left(b+c\right)b\left(a+c\right)}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ca\right)^2}}=\Sigma\frac{ab}{ab+bc+ca}\)
Từ đó ta có \(\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\ge\frac{ab+bc+ca}{ab+bc+ca}=1\)
chứng minh bài toán:
Đặt \(\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ac}}=t\ge1\)
Ta có: \(\left(\Sigma\sqrt{\frac{a}{b+c}}\right)^2=\Sigma\frac{a}{b+c}+2\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\ge\frac{a^2+b^2+c^2}{ab+bc+ac}+2=t^2+2\)
Từ đây ta chứng minh \(\sqrt{t^2+2}+\frac{3\sqrt{3}}{t}\ge\frac{7\sqrt{2}}{2}\)
Áp dụng bất đẳng thức bunhiacopxki ta có:
\(\sqrt{t^2+2}+\frac{3\sqrt{3}}{t}=\frac{\sqrt{\left(t^2+2\right)\left(6+2\right)}}{2\sqrt{2}}+\frac{3\sqrt{3}}{t}\ge\frac{t\sqrt{6}+2}{2\sqrt{2}}+\frac{3\sqrt{3}}{t}=\left(\frac{t\sqrt{3}}{2}+\frac{3\sqrt{3}}{t}\right)+\frac{\sqrt{2}}{2}\)
Áp dụng bất đẳng thức Cauchy ta đc:
\(\left(\frac{t\sqrt{3}}{2}+\frac{3\sqrt{3}}{t}\right)+\frac{\sqrt{2}}{2}\ge3\sqrt{2}+\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2}\)
Vậy ta có đpcm
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
\(A=\frac{x^4+\left(x+\frac{1}{2}\right)^2+\frac{7}{4}}{\left(x^2+1\right)\left(x^2+3x+6\right)}>0\)
\(A-2=\frac{-x^4-6x^3-13x^2-5x-10}{\left(x^2+1\right)\left(x^2+3x+6\right)}=\frac{-\left(x^2+3x\right)^2-4\left(x+\frac{5}{8}\right)^2-\frac{135}{16}}{\left(x^2+1\right)\left(x^2+3x+6\right)}< 0\)
\(\Rightarrow A< 2\Rightarrow0< A< 2\Rightarrow A=1\)
\(\Rightarrow x^4+x^2+x+2=x^4+3x^3+7x^2+3x+6\)
\(\Leftrightarrow3x^3+6x^2+2x+4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x^2+2\right)=0\Rightarrow x=-2\)
2.
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\)
\(P=\frac{x^2}{x^2+3xy}+\frac{y^2}{y^2+3yz}+\frac{z^2}{z^2+3zx}\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+xy+yz+zx}\)
\(P\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{1}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=\frac{4}{3}\)