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\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
Ta có: \(P=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)
sử dụng bđt cô-si có: \(a^2+\frac{1}{16a^2}\ge\frac{1}{2};b^2+\frac{1}{16b^2}\ge\frac{1}{2};\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}=\frac{4}{2ab}\)
Lại có: \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{4}{a^2+b^2}\)
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge4\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)\ge4\frac{4}{a^2+b^2+2ab}=\frac{16}{\left(a+b\right)^2}=16\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge8\)
\(\Rightarrow P\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{2}=\frac{17}{2}\)
Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)
\(x=\dfrac{1}{2}\cdot\left(\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}\right)=\dfrac{a+b}{2\sqrt{ab}}\)
\(2\sqrt{x^2}-1=2\cdot\dfrac{a+b}{2\sqrt{ab}}-1=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\)
\(x-\sqrt{x^2-1}=\dfrac{a+b}{2\sqrt{ab}}-\sqrt{\dfrac{a^2+2ab+b^2}{4ab}-1}\)
\(=\dfrac{a+b}{2\sqrt{ab}}-\dfrac{a-b}{2\sqrt{ab}}=\dfrac{2b}{2\sqrt{ab}}=\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(G=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}:\dfrac{\sqrt{b}}{\sqrt{a}}=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\cdot\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\dfrac{a+b-\sqrt{ab}}{b}\)
Bài 2:
a: \(\sqrt{4-x^2}>=0\)
Dấu '=' xảy ra khi x=2 hoặc x=-2
b: \(\sqrt{x^2-x+3}=\sqrt{x^2-x+\dfrac{1}{4}+\dfrac{11}{4}}\)
\(=\sqrt{\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}>=\dfrac{\sqrt{11}}{2}\)
Dấu '=' xảy ra khi x=1/2
c: \(x+\sqrt{x}+1>=1\)
=>1/(x+căn x+1)<=1
Dấu '=' xảy ra khi x=0
\(M=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\div\dfrac{\sqrt{x}-1}{2}\)
(ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\))
\(=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right]\times\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(x+2\right)+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\times\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\)
\(M=\dfrac{1}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le1\)
Dấu "=" xảy ra khi x = 0
Cảm ơn nhé! Nhưng tớ làm ra câu a,b rồi :( cậu biết làm c,d không?
2.
a/ Áp dụgn hệ quả bđt cô si,ta có :
\(A=xy+yz+zx\le\dfrac{\left(x+y+z\right)}{3}=\dfrac{a^2}{3}\)
Vậy GTLN A =a^2/3 khi x= y =z =a/3
b/Áp dụng BĐT Cô-Si dạng Engel,ta có :
\(B=\dfrac{x^2}{1}+\dfrac{y^2}{1}+\dfrac{z^2}{z}\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{a^2}{3}\)
Vậy GTNN của B = a^2/2 khi x=y=z =a/3
\(B=\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}+7\ge2\sqrt{\dfrac{3x}{1-x}.\dfrac{4\left(1-x\right)}{x}}+7=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
Vậy min B = \(\left(2+\sqrt{3}\right)^2\) khi \(\dfrac{3x}{1-x}=\dfrac{4\left(1-x\right)}{x}\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)
\(A=\dfrac{b^2}{b-1}=\dfrac{b^2-1+1}{b-1}=b+1+\dfrac{1}{b-1}=b-1+\dfrac{1}{b-1}+2\)
Áp dụng BĐT cosi cho \(b>0\left(b>1\right)\)
\(A=b-1+\dfrac{1}{b-1}+2\ge2\sqrt{\left(b-1\right)\cdot\dfrac{1}{b-1}}+2=2+2=4\)
Dấu \("="\Leftrightarrow\left(b-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}b-1=1\\b-1=-1\left(ktm\right)\end{matrix}\right.\Leftrightarrow b=2\left(tm\right)\)