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b, Ta có : \(0\le x\le1\)
\(\Rightarrow-2\le x-2\le-1< 0\)
Ta có : \(y=f\left(x\right)=2\left(m-1\right)x+\dfrac{m\left(x-2\right)}{\left(2-x\right)}\)
\(=2\left(m-1\right)x-m< 0\)
TH1 : \(m=1\) \(\Leftrightarrow m>0\)
TH2 : \(m\ne1\) \(\Leftrightarrow x< \dfrac{m}{2\left(m-1\right)}\)
Mà \(0\le x\le1\)
\(\Rightarrow\dfrac{m}{2\left(m-1\right)}>1\)
\(\Leftrightarrow\dfrac{m-2\left(m-1\right)}{2\left(m-1\right)}>0\)
\(\Leftrightarrow\dfrac{2-m}{m-1}>0\)
\(\Leftrightarrow1< m< 2\)
Kết hợp TH1 => m > 0
Vậy ...
\(x^2-2\left(m-1\right)x-m^3+\left(m+1\right)^2=0\)
Để pt có hai nghiệm thỏa mãn
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\x_1+x_2=2\left(m-1\right)\le4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\left(m-2\right)\left(m+2\right)\ge0\\m\le3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m\in\left[-2;0\right]\cup\left(2;+\infty\right)\cup\left\{2\right\}\\m\le3\end{matrix}\right.\)\(\Rightarrow m\in\left[-2;0\right]\cup\left[2;3\right]\)
\(P=x^3_1+x_2^3+x_1x_2\left(3x_1+3x_2+8\right)\)
\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+3x_1x_1\left(x_1+x_2\right)+8x_1x_2\)
\(=8\left(m-1\right)^3+8\left(-m^3+m^2+2m+1\right)\)
\(=-16m^2+40m\)
Vẽ BBT với \(f\left(m\right)=-16m^2+40m\) ;\(m\in\left[-2;0\right]\cup\left[2;3\right]\)
Tìm được \(f\left(m\right)_{min}=-144\Leftrightarrow m=-2\)
\(f\left(m\right)_{max}=16\Leftrightarrow m=2\)
\(\Rightarrow P_{max}=16;P_{min}=-144\)
Vậy....
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a/ ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)
\(\Leftrightarrow x^2+2x-15=x^2-9\)
\(\Leftrightarrow2x=6\Rightarrow x=3\) (ktm)
Vậy pt vô nghiệm
b/ ĐKXĐ: \(x\ne1\)
\(\Leftrightarrow\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2}{x^2+x+1}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow x^2+x+1+2\left(x-1\right)=3x^2\)
\(\Leftrightarrow2x^2-3x+1=0\Rightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=\frac{1}{2}\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne\pm4\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{\left(x-4\right)\left(x+4\right)}+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow5x^2-80+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2+16=5x^2+2x\)
\(\Rightarrow x=8\)
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\(P\ge\frac{\left(x+y\right)^2}{2\left(2x^2+1\right)\left(2y^2+1\right)}+\frac{1}{xy}=\frac{2}{\left(2x^2+1\right)\left(2y^2+1\right)}+\frac{2}{9xy}+\frac{7}{9xy}\)
\(P\ge\frac{8}{4x^2y^2+2x^2+2y^2+4xy+5xy+1}+\frac{7}{9xy}\)
\(P\ge\frac{8}{4\left(\frac{x+y}{2}\right)^4+2\left(x+y\right)^2+\frac{5}{4}\left(x+y\right)^2+1}+\frac{28}{9\left(x+y\right)^2}=\frac{11}{9}\)
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Từ bđt Cauchy : \(a+b\ge2\sqrt{ab}\) ta suy ra được \(ab\le\frac{\left(a+b\right)^2}{4}\)
Áp dụng vào bài toán của bạn :
a/ \(y=\left(x+3\right)\left(5-x\right)\le\frac{\left(x+3+5-x\right)^2}{4}=...............\)
b/ Tương tự
c/ \(y=\left(x+3\right)\left(5-2x\right)=\frac{1}{2}.\left(2x+6\right)\left(5-2x\right)\le\frac{1}{2}.\frac{\left(2x+6+5-2x\right)^2}{4}=.............\)
d/ Tương tự
e/ \(y=\left(6x+3\right)\left(5-2x\right)=3\left(2x+1\right)\left(5-2x\right)\le3.\frac{\left(2x+1+5-2x\right)^2}{4}=.......\)
f/ Xét \(\frac{1}{y}=\frac{x^2+2}{x}=x+\frac{2}{x}\ge2\sqrt{x.\frac{2}{x}}=2\sqrt{2}\)
Suy ra \(y\le\frac{1}{2\sqrt{2}}\)
..........................
g/ Đặt \(t=x^2\) , \(t>0\) (Vì nếu t = 0 thì y = 0)
\(\frac{1}{y}=\frac{t^3+6t^2+12t+8}{t}=t^2+6t+\frac{8}{t}+12\)
\(=t^2+6t+\frac{8}{3t}+\frac{8}{3t}+\frac{8}{3t}+12\)
\(\ge5.\sqrt[5]{t^2.6t.\left(\frac{8}{3t}\right)^3}+12=.................\)
Từ đó đảo ngược y lại rồi đổi dấu \(\ge\) thành \(\le\)
x=yx44444444444444444444444444444
Đặt \(x+\frac{1}{x}=t\)thì \(x^2+\frac{1}{x^2}=t^2-2\)
Lúc đó: \(y=f\left(x\right)=t^2-2+2t+8=\left(t^2+2t+1\right)+5=\left(t+1\right)^2+5\ge5\)
Đẳng thức xảy ra khi \(t=x+\frac{1}{x}=-1\Leftrightarrow x^2+x+1=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}\)\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{\sqrt{3}}{2}i\\x+\frac{1}{2}=-\frac{\sqrt{3}}{2}i\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}i-1}{2}\\x=\frac{-\sqrt{3}i-1}{2}\end{cases}}\)