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Day la bdt Svacso dau bang xay ra <=> \(\frac{a}{x}=\frac{b}{y}\)
Quy đồng full
\(\frac{a^2y+b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
\(\Leftrightarrow a^2xy+a^2y^2+b^2x^2+b^2xy\ge\left(a^2+2ab+b^2\right)xy\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2\ge0\)
\(\Leftrightarrow\left(ay-bx\right)^2\ge0\)
lun đúng
Đặt A=\(\dfrac{b+c+5}{1+a}+\dfrac{c+a+4}{2+b}+\dfrac{a+b+3}{3+c}\)
Ta có :A+3=\(\left(\dfrac{b+c+5}{1+a}+1\right)+\left(\dfrac{c+a+4}{2+b}+1\right)+\left(\dfrac{a+b+3}{3+a}+1\right)\)
=\(\dfrac{a+b+c+6}{1+a}+\dfrac{a+b+c+6}{2+b}+\dfrac{a+b+c+6}{3+c}\)
=\(\left(a+b+c+6\right)\left(\dfrac{1}{1+a}+\dfrac{1}{2+b}+\dfrac{1}{3+c}\right)\)
=\([\left(a+1\right)+\left(b+2\right)+\left(c+3\right)|\left(\dfrac{1}{a+1}+\dfrac{1}{b+2}+\dfrac{1}{c+3}\right)\)
Áp dụng bất đẳng thức AM-GM dạng \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)( với x,y,z>0)
Ta có :A+3\(\ge9\)\(\Rightarrow A\ge6\)
Dấu "=" xảy ra khi a=3,b=2,c=1
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge 3\left(\frac{x}{y}+\frac{y}{x}\right)\) <=>\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4 - 3\left(\frac{x}{y}+\frac{y}{x}\right)\ge0\)
Vì \(\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge 2\)
và \(\left(\frac{x}{y}+\frac{y}{x}\right)\ge 2\)
nên BĐT tương đương 2+ 4- 3x2 \(\ge 0\)
<=> 0\(\ge 0\)
Dấu = xảy ra khi x=y
Đặt \(\frac{x}{y}+\frac{y}{x}=a\) ta có \(lal=l\frac{x}{y}+\frac{y}{x}l=l\frac{x}{y}l+l\frac{y}{x}l\ge2\) ( cô - si )
=> \(a\ge2ora\le-2\)
BĐT <=> \(a^2-2+4\ge3a\Leftrightarrow a^2-3a+2\ge0\Leftrightarrow\left(a-1\right)\left(a-2\right)\ge0\)
(+) với \(a\ge2\) => \(a-1>a-2\ge0\Leftrightarrow\left(a-1\right)\left(a-2\right)\ge0\)
(+) với \(a\le-2\Rightarrow a-2\le0;a-1\le0\Rightarrow\left(a-2\right)\left(a-1\right)\ge0\)
Vậy BĐT trên luôn đúng
a) Ta có: \(P=\left(\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{\sqrt{a}}{a-1}\right):\left(\frac{2}{a}-\frac{2-a}{a\sqrt{a}+a}\right)\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right):\left(\frac{2\left(\sqrt{a}+1\right)}{a\left(\sqrt{a}+1\right)}-\frac{2-a}{a\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{a+\sqrt{a}+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\frac{2\sqrt{a}+2-2+a}{a\left(\sqrt{a}+1\right)}\)
\(=\frac{a+2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a\left(\sqrt{a}+1\right)}{a+2\sqrt{a}}\)
\(=\frac{a}{\sqrt{a}-1}\)
b)
ĐKXĐ: \(a\notin\left\{1;0\right\}\)
Để P-2 là số dương thì P-2>0
⇔\(\frac{a}{\sqrt{a}-1}-2>0\)
\(\Leftrightarrow\frac{a}{\sqrt{a}-1}-\frac{2\left(\sqrt{a}-1\right)}{\sqrt{a}-1}>0\)
\(\Leftrightarrow\frac{a-2\sqrt{a}+2}{\sqrt{a}-1}>0\)
mà \(a-2\sqrt{a}+2=\left(\sqrt{a}-1\right)^2+1>0\forall a\)
nên \(\sqrt{a}-1>0\)
\(\Leftrightarrow\sqrt{a}>1\)
\(\Leftrightarrow a>1\)(tm)
Vậy: Khi a>1 thì P-2 là số dương
A=\((\frac{\sqrt{a}\left(\sqrt{a}+1\right)+\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}):\left(\frac{2\left(\sqrt{a}+1\right)-\left(2-a\right)}{a\left(\sqrt{a}+1\right)}\right)\)
\(A=\left(\frac{a+\sqrt{a}+\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\left(\frac{2\sqrt{a}+2-2+a}{a\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{a+2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{a\left(\sqrt{a}+1\right)}{2\sqrt{a}-a}\)
\(A=\frac{a}{\sqrt{a}-1}\)
Bài 1 :
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b, ĐKXĐ : \(-x^2+10x-25\ge0\)
=> \(x^2-10x+25\le0\)
=> \(\left(x-5\right)^2\le0\)
=> \(x-5\le0\)
=> \(x\le5\)
Bài 2 :
a, Ta có : \(A=\sqrt{\left(2\sqrt{2}-5\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
=> \(A=5-2\sqrt{2}+\sqrt{5}-2=3-2\sqrt{2}+\sqrt{5}\)
b, Ta có : \(B=\sqrt{9+4\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
=> \(B=\sqrt{4+2.2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\)
=> \(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
=> \(B=2+\sqrt{5}-\sqrt{5}+1=3\)
c, Ta có : \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
=> \(C=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
=> \(C=\frac{\sqrt{1+2\sqrt{3}+3}}{\sqrt{2}}+\frac{\sqrt{1-2\sqrt{3}+3}}{\sqrt{2}}\)
=> \(C=\frac{\sqrt{\left(1+\sqrt{3}\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(1-\sqrt{3}\right)^2}}{\sqrt{2}}\)
=> \(C=\frac{1+\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)
<=> \(x^3=\frac{1}{4-\sqrt{15}}+3\left(\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\right)\left(\frac{1}{\sqrt[3]{4-\sqrt{15}}}.\sqrt[3]{4-\sqrt{15}}\right)\)
\(+4-\sqrt{15}\)
<=> \(x^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+3x\)
<=> \(x^3-3x+2006=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+2006\)
<=> \(x^3-3x+2006=\frac{4+\sqrt{15}}{16-15}+4-\sqrt{15}+2006\)
<=> \(x^3-3x+2006=2014\)
Ta có: \(\text{Σ}_{cyc}\left(a-b\right)^2\ge0\forall a,b,c\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^2}{3}\ge\left(ab+bc+ca\right)\)
Dấu "=" khi a = b = c
Đây là bất đằng thức gì vậy bạn ?