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a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
1)
a. \(P=\left(\dfrac{1}{\sqrt{a}-3}+\dfrac{1}{\sqrt{a}+3}\right)\left(1-\dfrac{3}{\sqrt{a}}\right)\)
\(\Leftrightarrow\left(\dfrac{\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\dfrac{\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{3}{\sqrt{a}}\right)\)\(\Leftrightarrow\dfrac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}.\dfrac{\sqrt{a}-3}{\sqrt{a}}\)
\(\Leftrightarrow\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a-3}\right)\left(\sqrt{a}+3\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}+3}\)
b.
Câu 2:
\(x^2-2\left(m-3\right)x-1=0\)
a=1; b=-2m+6; c=-1
Vì ac<0 nên phương trình luôn có hai nghiệm phân biệt
Ta có: \(A=x_1^2+x_2^2-x_1x_2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2-x_1x_2\)
\(=\left(x_1+x_2\right)^2-3x_1x_2\)
\(=\left(2m-6\right)^2-3\cdot\left(-1\right)\)
\(=4m^2-24m+36+3\)
\(=\left(2m-6\right)^2+3\ge3\)
Dấu '=' xảy ra khi m=3
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
Giải:
Ta có:
\(P=\dfrac{a^3}{\sqrt{1+b^2}}+\dfrac{b^3}{\sqrt{1+c^2}}+\dfrac{c^3}{\sqrt{1+a^2}}\)
\(\Leftrightarrow P+3=\dfrac{a^3}{\sqrt{1+b^2}}+b^2+\dfrac{b^3}{\sqrt{1+c^2}}+c^2\dfrac{c^3}{\sqrt{1+a^2}}+a^2\)
\(\Leftrightarrow P+\dfrac{6}{4\sqrt{2}}=\dfrac{a^3}{2\sqrt{1+b^2}}+\dfrac{a^2}{2\sqrt{1+b^2}}+\dfrac{1+b^2}{4\sqrt{2}}+\dfrac{b^3}{2\sqrt{1+c^2}}+\dfrac{b^2}{2\sqrt{1+c^2}}+\dfrac{1+c^2}{4\sqrt{2}}+\dfrac{c^3}{2\sqrt{1+a^2}}+\dfrac{c^2}{2\sqrt{1+a^2}}+\dfrac{1+a^2}{4\sqrt{2}}\)
\(\ge3\sqrt[3]{\dfrac{a^6}{16\sqrt{2}}}+3\sqrt[3]{\dfrac{b^6}{16\sqrt{2}}}+3\sqrt[3]{\dfrac{c^6}{16\sqrt{2}}}\)
\(\Rightarrow P+\dfrac{3}{2\sqrt{2}}\ge\dfrac{3}{2\sqrt[3]{2\sqrt{2}}}\left(a^2+b^2+c^2\right)=\dfrac{9}{2\sqrt[6]{8}}\)
\(\Rightarrow P\ge\dfrac{9}{2\sqrt[6]{2^3}}-\dfrac{3}{2\sqrt{2}}=\dfrac{9}{2\sqrt{2}}-\dfrac{3}{2\sqrt{2}}=\dfrac{3}{\sqrt{2}}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng BĐt cauchy-schwarz:(dạng phân thức + đa thức )
\(P=\dfrac{a^3}{\sqrt{1+b^2}}+\dfrac{b^3}{\sqrt{1+c^2}}+\dfrac{c^3}{\sqrt{1+a^2}}=\dfrac{a^4}{a\sqrt{1+b^2}}+\dfrac{b^4}{b\sqrt{1+c^2}}+\dfrac{c^4}{c\sqrt{1+a^2}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\sqrt{1+b^2}+b\sqrt{1+c^2}+c\sqrt{1+a^2}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{\left(a^2+b^2+c^2\right)\left(3+a^2+b^2+c^2\right)}}=\dfrac{9}{\sqrt{18}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}\)
dấu = xảy ra khi a=b=c=1