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bài 3:
a, đặt x12=y9=z5=kx12=y9=z5=k
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: x5=y7=z3=x225=y249=z29x5=y7=z3=x225=y249=z29
A/D tính chất dãy tỉ số bằng nhau ta có:
x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
Câu 1:
\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\\ \Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}=0\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}+3-a-b-c=0\\ \Leftrightarrow\left[\left(x-a\right)-2\sqrt{x-a}+1\right]+\left[\left(y-b\right)-2\sqrt{y-b}+1\right]+\left[\left(z-c\right)-2\sqrt{z-c}+1\right]=0\\ \Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}-1=0\\\sqrt{y-b}-1=0\\\sqrt{z-c}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}=1\\\sqrt{y-b}=1\\\sqrt{z-c}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-a=1\\y-b=1\\z-c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=a+1\\y=b+1\\z=c+1\end{matrix}\right.\)Vậy \(\left\{x;y;z\right\}=\left\{a+1;b+1;c+1\right\}\)
Câu 2:
\(\text{ a) Ta có }:\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\\ =\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\left(1\right)\)
\(\text{Lại có: }\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}-\sqrt{n}\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow2\left(\sqrt{n+1}-n\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b) Áp dụng bất đảng thức ở câu a:
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\\ >2\left(\sqrt{101}-\sqrt{100}\right)+...+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{100}+...+\sqrt{4}-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-1\right)=2\left(10-1\right)=18\left(3\right)\)
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 2\left(\sqrt{100}-\sqrt{99}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-\sqrt{0}\right)\\ =2\left(\sqrt{100}-\sqrt{99}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\sqrt{1}\right)\\ =2\cdot\sqrt{100}=2\cdot10=20\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\Rightarrow18< S< 20\)
Bài 3:
Áp dụng bất đẳng thức AM - GM có:
\(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\sqrt{x.\dfrac{1}{x}}+2\sqrt{y.\dfrac{1}{y}}+2\sqrt{z.\dfrac{1}{z}}\)
\(=2+2+2=6\)
Dấu " = " khi x = y = z = 1
Vậy...
3. Với x,y,z>0 áp dụng BĐT Cauchy ta có
\(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(=\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)+\left(z+\dfrac{1}{z}\right)\)
\(\ge2\sqrt{x.\dfrac{1}{x}}+2\sqrt{y.\dfrac{1}{y}}+2\sqrt{z.\dfrac{1}{z}}=2+2+2=6\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\\z=\dfrac{1}{z}\end{matrix}\right.\Leftrightarrow x=y=z=1\)
1. Với a=b=c=0, ta thấy BĐT trên đúng
Với a,b,c>0 áp dụng BĐT Cauchy cho 3 số dương
\(a^3+a^3+b^3\ge3\sqrt[3]{a^3.a^3.b^3}=3\sqrt[3]{a^6b^3}=3a^2b\) (1)
\(b^3+b^3+c^3\ge3\sqrt[3]{b^3.b^3.c^3}=3\sqrt[3]{b^6c^3}=3b^2c\) (2)
\(c^3+c^3+a^3\ge3\sqrt[3]{c^3.c^3.a^3}=3\sqrt[3]{c^6a^3}=3c^2a\) (3)
Cộng (1), (2), (3) vế theo vế:
\(a^3+b^3+c^3\ge a^2b+b^2c+c^2a>\dfrac{a^2b+b^2c+c^2a}{3}\) (vì a,b,c>0)
Do đó BĐT trên đúng \(\forall a,b,c\ge0\)
a.
\(\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
b.
\(VP=\dfrac{4\left(a+b+c\right)}{2\sqrt{4a\left(a+3b\right)}+2\sqrt{4b\left(b+3c\right)}+2\sqrt{4c\left(c+3a\right)}}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{4a+a+3b+4b+b+3c+4c+c+3a}\)
\(VP\ge\dfrac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Lời giải:
Ta có \(x=\frac{1}{\sqrt{b}+\sqrt{c}}; y=\frac{1}{\sqrt{a}+\sqrt{c}}; z=\frac{1}{\sqrt{b}+\sqrt{a}}\)
\(\Rightarrow \left\{\begin{matrix} \sqrt{b}+\sqrt{c}=\frac{1}{x}\\ \sqrt{c}+\sqrt{a}=\frac{1}{y}\\ \sqrt{b}+\sqrt{a}=\frac{1}{z}\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} \sqrt{a}=\frac{1}{2}(\frac{1}{y}+\frac{1}{z}-\frac{1}{x})\\ \sqrt{b}=\frac{1}{2}(\frac{1}{x}+\frac{1}{z}-\frac{1}{y})\\ \sqrt{c}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})\end{matrix}\right.\)
Khi đó: \(2b=a+c\)
\(\Leftrightarrow \frac{1}{2}(\frac{1}{x}+\frac{1}{z}-\frac{1}{y})^2=\frac{1}{4}(\frac{1}{y}+\frac{1}{z}-\frac{1}{x})^2+\frac{1}{4}(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})^2\)
\(\Leftrightarrow \frac{1}{xz}-\frac{1}{xy}-\frac{1}{yz}=\frac{1}{2yz}-\frac{1}{2xz}-\frac{1}{2xy}+\frac{1}{2xy}-\frac{1}{2yz}-\frac{1}{2xz}\)
\(\Leftrightarrow \frac{1}{xz}-\frac{1}{xy}-\frac{1}{yz}=\frac{-1}{xz}\)
\(\Leftrightarrow \frac{2}{xz}=\frac{1}{xy}+\frac{1}{yz}\)
\(\Leftrightarrow 2y=z+x\)
Ta có đpcm.
Câu 1:
\(A=21\left(a+\frac{1}{b}\right)+3\left(b+\frac{1}{a}\right)=21a+\frac{21}{b}+3b+\frac{3}{a}\)
\(=(\frac{a}{3}+\frac{3}{a})+(\frac{7b}{3}+\frac{21}{b})+\frac{62}{3}a+\frac{2b}{3}\)
Áp dụng BĐT Cô-si:
\(\frac{a}{3}+\frac{3}{a}\geq 2\sqrt{\frac{a}{3}.\frac{3}{a}}=2\)
\(\frac{7b}{3}+\frac{21}{b}\geq 2\sqrt{\frac{7b}{3}.\frac{21}{b}}=14\)
Và do $a,b\geq 3$ nên:
\(\frac{62}{3}a\geq \frac{62}{3}.3=62\)
\(\frac{2b}{3}\geq \frac{2.3}{3}=2\)
Cộng tất cả những BĐT trên ta có:
\(A\geq 2+14+62+2=80\) (đpcm)
Dấu "=" xảy ra khi $a=b=3$
Câu 2:
Bình phương 2 vế ta thu được:
\((x^2+6x-1)^2=4(5x^3-3x^2+3x-2)\)
\(\Leftrightarrow x^4+12x^3+34x^2-12x+1=20x^3-12x^2+12x-8\)
\(\Leftrightarrow x^4-8x^3+46x^2-24x+9=0\)
\(\Leftrightarrow (x^2-4x)^2+6x^2+24(x-\frac{1}{2})^2+3=0\) (vô lý)
Do đó pt đã cho vô nghiệm.
Đặt \(\dfrac{a}{x^3}=\dfrac{b}{y^3}=\dfrac{c}{z^3}=m\)
Ta có:
\(\dfrac{a}{x^2}+\dfrac{b}{y^2}+\dfrac{c}{z^2}=\dfrac{a}{x^3}.x+\dfrac{b}{y^3}.y+\dfrac{c}{z^3}.z=m.x+m.y+m.z=m\left(x+y+z\right)=m\)
\(\Rightarrow\sqrt[3]{\dfrac{a}{x^2}+\dfrac{b}{y^2}+\dfrac{c}{z^2}}=\sqrt[3]{m}\) (1)
Lại có:
\(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{\dfrac{a}{x^3}.x^3}+\sqrt[3]{\dfrac{b}{y^3}.y^3}+\sqrt[3]{\dfrac{c}{z^3}.z^3}=\sqrt[3]{\dfrac{a}{x^3}}.x+\sqrt[3]{\dfrac{b}{y^3}}.y+\sqrt[3]{\dfrac{c}{z^3}}.z=\sqrt[3]{m}.x+\sqrt[3]{m}.y+\sqrt[3]{m}.z=\sqrt[3]{m}\left(x+y+z\right)=\sqrt[3]{m}\left(2\right)\)
Từ (1), (2)
=> \(\Rightarrow\sqrt[3]{\dfrac{a}{x^2}+\dfrac{b}{y^2}+\dfrac{c}{z^2}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\) (đpcm)
Cảm ơn nha :D