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a: \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}=\sqrt{a}-\sqrt{b}\)
b: \(VT=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{2-\left(\sqrt{3}-1\right)}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{3-\sqrt{3}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)\left(\sqrt{3}-1\right)+2\left(\sqrt{2}-1\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\left(\sqrt{6}-\sqrt{2}+\sqrt{3}-1+\sqrt{6}+\sqrt{2}-\sqrt{3}-1\right)}{\sqrt{3}\cdot2}\)
\(=\dfrac{2\left(2\sqrt{6}-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}-2}{\sqrt{3}}\)
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
Lời giải:
Đặt \(\left ( \frac{\sqrt{a^2+b^2}}{c},\frac{\sqrt{b^2+c^2}}{a}, \frac{\sqrt{c^2+a^2}}{b} \right )=(x,y,z)\)
BĐT cần chứng minh tương đương với:
\(x+y+z\geq 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)\((*)\)
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Từ cách đặt $x,y,z$ ta có:
\(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}=1\)
Áp dụng BĐT Bunhiacopxky:
\(\frac{x^2+1}{x^2}+\frac{y^2+1}{y^2}+\frac{z^2+1}{z^2}=\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\left(\frac{x^2+1}{x^2}+\frac{y^2+1}{y^2}+\frac{z^2+1}{z^2}\right)\)
\(\geq \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
\(\Leftrightarrow 3\geq 2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
\(\Leftrightarrow xyz\geq \frac{2}{3}(x+y+z)\)
\(\Rightarrow xyz(x+y+z)\geq \frac{2}{3}(x+y+z)^2\)
Áp dụng BĐT AM_GM ta lại có:
\((x+y+z)^2\geq 3(xy+yz+xz)\). Do đó:
\(xyz(x+y+z)\geq 2(xy+yz+xz)\)
\(\Leftrightarrow x+y+z\geq 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Đúng theo \((*)\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c\)
áp dụng bat dang thuc bunhiacóki
ta có \(\dfrac{\sqrt{a^2+b^2}}{c}\ge\dfrac{a+b}{\sqrt{2}c}\)
ttu vt \(\ge\dfrac{1}{\sqrt{2}}\left(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\right)\)
=\(\dfrac{a}{\sqrt{2}}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{b}{\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+\dfrac{c}{\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) (1)
áp dung bdt \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
ta có (1) \(\ge\dfrac{a}{\sqrt{2}}.\dfrac{4}{b+c}\)
tiếp tục áp dụng bunhia ta có \(\dfrac{a}{\sqrt{2}}.\dfrac{4}{b+c}\ge\dfrac{a}{\sqrt{2}}.\dfrac{4}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{2a}{\sqrt{b^2+c^2}}\)
ttuong tu ta có \(vt\ge2\left(\dfrac{a}{\sqrt{b^2+c2}}+\dfrac{b}{\sqrt{a^2+c^2}}+\dfrac{c}{\sqrt{a^2+b^2}}\right)\left(dpcm\right)\)
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:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
Bài 1:
a)Với x > 0;x ≠ 4 ta có:
\(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}+2\right)-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)\)
\(=\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4}{x-4}\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=b-a\)
Bài 2:
a)Với a > 0;a ≠ 1;a ≠ 2 ta có
\(P=\left(\dfrac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}\)
b)Ta có:
\(P=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2a-4}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
P nguyên khi \(2-\dfrac{8}{a+2}\) nguyên⇒\(\dfrac{8}{a+2}\) nguyên⇒\(a+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(TH1:a+2=1\Rightarrow a=-1\left(loai\right)\)
\(TH2:a+2=-1\Rightarrow a=-3\left(loai\right)\)
\(TH3:a+2=2\Rightarrow a=0\left(loai\right)\)
\(TH4:a+2=-2\Rightarrow a=-4\left(loai\right)\)
\(TH5:a+2=4\Rightarrow a=2\left(loai\right)\)
\(TH6:a+2=-4\Rightarrow a=-6\left(loai\right)\)
\(TH7:a+2=8\Rightarrow a=6\left(tm\right)\)
\(TH8:a+2=-8\Rightarrow a=-10\left(loai\right)\)
Vậy a = 6
b)Áp dụng BĐT AM-GM ta có:
\(\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}\ge2\sqrt{\dfrac{\sqrt{a}}{\sqrt{b}}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}}=2\)
Xảy ra khi \(a=b\)
c)Áp dụng BĐT \(x^2+y^2\ge2xy\) có:
\(VT=\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\)
\(\ge2\sqrt{\left(a+b\right)\cdot2\sqrt{ab}}=2\sqrt{2\left(a+b\right)\cdot\sqrt{ab}}=VP\)
Xảy ra khi \(a=b\)
a)\(\dfrac{a^2+3}{\sqrt{a^2+3}}=\sqrt{a^2+3}\ge\sqrt{3}< 2\)\
sai đề