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c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
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a, A\(=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x-1}{\sqrt{x}}\) ĐK x>0 ;\(x\ne1;x\ne-1\)
\(A=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}}{x-1}\)
\(A=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}\)=\(\frac{4x^2}{\left(x-1\right)^2}\)
b, Để A =2 \(\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Rightarrow4x^2=2\left(x-1\right)^2\)
<=> \(4x^2=2x^2-4x+2\)
<=> \(2x^2+4x-2=0\)
<=> \(x^2+2x-1=0\)
\(\Delta=1^2-1.\left(-1\right)\) = 2
=> \(\orbr{\begin{cases}x_1=-1-\sqrt{2}\left(loại\right)\\x_2=-1+\sqrt{2}\left(nhận\right)\end{cases}}\)
Vậy x=\(-1+\sqrt{2}\)thì A =2
c, Thay x =\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)=2
=>A = \(\frac{4.2^2}{\left(2-1\right)^2}=16\)
Vậy A=16 thì x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
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Em không chắc lắm đâu nhé!
Biến đổi \(A=\frac{\left(\frac{a^4}{b^2}\right)}{b\left(c+2a\right)}+\frac{\left(\frac{b^4}{c^2}\right)}{c\left(a+2b\right)}+\frac{\left(\frac{c^4}{a^2}\right)}{a\left(b+2c\right)}\)
\(=\frac{\left(\frac{a^2}{b}\right)^2}{b\left(c+2a\right)}+\frac{\left(\frac{b^2}{c}\right)^2}{c\left(a+2b\right)}+\frac{\left(\frac{c^2}{a}\right)^2}{a\left(b+2c\right)}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel:\(A\ge\frac{\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2}{3\left(ab+bc+ca\right)}\)
Áp dụng BĐT Cauchy-Schwarz cho cái biểu thức trong ngoặc ở trên tử,ta lại được:
\(A\ge\frac{\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2}{3\left(ab+bc+ca\right)}\ge\frac{\left(\frac{\left(a+b+c\right)^2}{a+b+c}\right)^2}{3\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\) (áp dụng BĐT quen thuộc \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) cho cái biểu thức dưới mẫu)
Dấu "=" xảy ra khi a = b =c
Vậy \(A_{min}=1\Leftrightarrow a=b=c\)