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a) \(A=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2=2a+2b\le2\)
Vậy GTLN của A là 2 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)
b) Ta có : \(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2\left(a^2+b^2+6ab\right)\)
Tương tự : \(\left(\sqrt{a}+\sqrt{c}\right)^4\le2\left(a^2+c^2+6ac\right)\)
\(\left(\sqrt{a}+\sqrt{d}\right)^4\le2\left(a^2+d^2+6ad\right)\)
\(\left(\sqrt{b}+\sqrt{c}\right)^4\le2\left(b^2+c^2+6bc\right)\)
\(\left(\sqrt{b}+\sqrt{d}\right)^4\le2\left(b^2+d^2+6bd\right)\)
\(\left(\sqrt{c}+\sqrt{d}\right)^4\le2\left(c^2+d^2+6cd\right)\)
Cộng các vế lại, ta được :
\(B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bd+2cd+2bc\right)=6\left(a+b+c+d\right)^2\)
\(\Rightarrow B\le6\)
Vậy GTLN của B là 6 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}=\sqrt{c}=\sqrt{d}\\a+b+c+d=1\end{cases}}\Leftrightarrow a=b=c=d=\frac{1}{4}\)
\(P=\sqrt{a\left(b+1\right)}+\sqrt{b\left(a+1\right)}\)
\(\Rightarrow P\sqrt{2}=\sqrt{2a\left(b+1\right)}+\sqrt{2b\left(a+1\right)}\)
\(\le\frac{1}{2}\left(2a+b+1\right)+\frac{1}{2}\left(2b+a+1\right)\)
\(\le\frac{1}{2}\left(3a+3b+2\right)\le\frac{1}{2}.\left(3.2+2\right)=4\)
\(\Rightarrow p\le2\sqrt{2}\)
Dấu"=" xảy ra \(\Leftrightarrow a=b=1\)
Vậy Max P \(=2\sqrt{2}\)\(\Leftrightarrow a=b=1\)
Ta có
\(a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow2\left(a+b\right)\ge\left(\sqrt{a}+\sqrt{b}\right)^2\)
\(\Leftrightarrow2\ge\left(\sqrt{a}+\sqrt{b}\right)^2\)
Vậy GTLN là 2 đạt được khi \(a=b=\frac{1}{2}\)
1. Áp dụng BĐT Bunhiakovski
a) \(\sqrt{x-2}+\sqrt{4-x}=\sqrt{\left(\sqrt{x-2}.1+\sqrt{4-x}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(x-2+4-x\right)}=2\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{x-2}=\sqrt{4-x}\) \(\Leftrightarrow\) \(x=3\)
b) \(\sqrt{6-x}+\sqrt{x+2}=\sqrt{\left(\sqrt{6-x}.1+\sqrt{x+2}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(6-x+x+2\right)}=4\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{6-x}=\sqrt{x+2}\) \(\Leftrightarrow\) \(x=2\)
c) \(\sqrt{x}+\sqrt{2-x}=\sqrt{\left(\sqrt{x}.1+\sqrt{2-x}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(x+2-x\right)}=2\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{x}=\sqrt{2-x}\) \(\Leftrightarrow\) \(x=1\)
1.Điều kiện xđ \(x\ge2,x\le4\)
Từ ĐKXĐ ta có
\(x\ge2\Leftrightarrow x-2\ge0\Leftrightarrow\sqrt{x-2}\ge0\left(1\right)\)
\(x\le4\Leftrightarrow4-x\ge0\Leftrightarrow\sqrt{4-x}\ge0\left(2\right)\)
Từ (1),(2) cộng vế theo vế ta có:
\(\sqrt{x-2}+\sqrt{4-x}\ge0+0=0\)
Ta có \(2=a^2+b^2\ge2ab\)
\(\Leftrightarrow ab\le1\)
\(M\le\sqrt{\left(a^2+b^2\right)\left(36ab+45b^2+36ab+45a^2\right)}\)
\(=\sqrt{2\left(72ab+90\right)}\)\(\le\sqrt{2\left(72+90\right)}=\sqrt{324}=18\)
GTLN là 18 đạt được khi a = b = 1
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
Ta có: \(M=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2\)=\(2a+2b\le2\)
\(Max\)\(M=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}+\sqrt{b}\\a+b=1\end{matrix}\right.\)\(\Leftrightarrow a=b=\dfrac{1}{2}\)
\(M=\left(\sqrt[]{a}+\sqrt[]{b}\right)^2;a+b\le1\left(a;b>0\right)\)
Áp dụng Bất đẳng thức Bunhiacopxki cho 2 cặp số \(\left(1;\sqrt[]{a}\right);\left(1;\sqrt[]{b}\right)\)
\(M=\left(1.\sqrt[]{a}+1.\sqrt[]{b}\right)^2\le\left(1^2+1^2\right)\left(a+b\right)\le2\) \(\left(a+b\le1\right)\)
\(\Rightarrow M=\left(\sqrt[]{a}+\sqrt[]{b}\right)^2\le2\)
Dấu "=" xảy ra khi và chỉ khi
\(\dfrac{1}{\sqrt[]{a}}=\dfrac{1}{\sqrt[]{b}}\Leftrightarrow a=b=1\)
\(\Rightarrow GTLN\left(M\right)=2\left(khi.a=b=1\right)\)