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A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
Tacó \(\Delta\)=(-7)2-4x1x2=41>0 =>\(\sqrt{_{ }x1}\)=\(\dfrac{7+\sqrt{41}}{2}\)=>\(_{x1}\)=\(\dfrac{\left(7+\sqrt{41}\right)^2}{4}\)=\(\dfrac{45+7\sqrt{41}}{2}\) =>\(\sqrt{_{ }x2}\)=\(\dfrac{7-\sqrt{41}}{2}\)=>\(_{x_2}\)=\(\dfrac{\left(7-\sqrt{41^{ }}\right)^2}{4}\)=\(\dfrac{45-7\sqrt{41}}{2}\) so sánh với điều kiện X>_0
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: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
\(a.\sqrt{2a}.\sqrt{18a}=\sqrt{2a}.3\sqrt{2a}=3.2a=6a\)
\(b.\sqrt{3a.27ab^2}=\sqrt{9a^2b^2.9}=9\text{ |}ab\text{ |}\)
\(c.2y^2.\sqrt{\dfrac{x^4}{4y^2}}=2y^2.\dfrac{x^2}{-2y}=-x^2y\)
\(d.\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}\)
\(e.\sqrt{\dfrac{9a^2}{16}}=\dfrac{3\text{ |}a\text{ |}}{4}\)
\(f.\sqrt{10.16a^2}=-4a\sqrt{10}\)
\(g.\sqrt{a^4\left(3-a\right)^2}=a^2\left(a-3\right)\)
\(h.\sqrt{\dfrac{2a^2b^4}{98}}\sqrt{\dfrac{a^2b^4}{49}}=\dfrac{b^2\text{ |}a\text{ |}}{7}\)
Bài này đưa về giải hệ phương trình
\(\left\{{}\begin{matrix}a-b+4ab=1\\a^2+b^2=2\end{matrix}\right.\) với \(a,b\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b+4ab=1\left(1\right)\\\left(a-b\right)^2+2ab=2\left(2\right)\end{matrix}\right.\)
Từ pt (1) suy ra \(a-b=1-4ab\Rightarrow\left(a-b\right)^2=1+16a^2b^2-8ab\)
Do đó
\(\left(2\right)\Rightarrow1+16a^2b^2-8ab+2ab=2\)
\(\Leftrightarrow16a^2b^2-6ab-1=0\)
Xem đây là pt bậc 2 với ab tìm được \(\left[{}\begin{matrix}ab=\dfrac{1}{2}\\ab=-\dfrac{1}{8}\end{matrix}\right.\)
- TH1: \(ab=\dfrac{1}{2}\Rightarrow a-b=-1\)
Có \(\left\{{}\begin{matrix}a-b=-1\\ab=\dfrac{1}{2}\end{matrix}\right.\) tìm được \(\left\{{}\begin{matrix}a=\dfrac{-1+\sqrt{3}}{2}\\b=\dfrac{1+\sqrt{3}}{2}\end{matrix}\right.\) (thỏa mãn a,b>0)
Từ đó tìm x
Tương tự cho TH còn lại
sao lại đặt bằng x,y mà lại suy ra a,b nhỉ =))