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x + 2 P x - 2 = x - 1 Q x 2 - 4
⇒ x + 2 . P . x 2 - 4 = x - 2 x - 1 . Q
Hay (x + 2)(x – 2)(x + 2).P = (x – 2)(x – 1).Q
Chọn P = (x – 1) thì Q = x + 2 2
điều kiện \(x\ne2;x\ne-2\)
\(\frac{\left(x+2\right)^2.P}{x^2-4}=\frac{\left(x-1\right).Q}{x^2-4}\)
\(\left(x+2\right)^2.P=\left(x-1\right)Q\)
\(P=x-1\)
\(Q=\left(x+2\right)^2\)
\(\left(x+\frac{3}{5}\right).\left(\frac{-4}{3}-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{5}=0\\\frac{-4}{3}-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{5}\\x=\frac{-4}{3}\end{cases}}}\)
\(\left(x+\frac{3}{5}\right)\left(-\frac{4}{3}-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{5}=0\\-\frac{4}{3}-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{5}\\x=-\frac{4}{3}\end{cases}}}\)
vậy \(x=-\frac{3}{5}\)hoặc \(x=-\frac{4}{3}\).
a) \(\dfrac{\left(x+2\right)P}{x-2}=\dfrac{\left(x-1\right)Q}{x^2-4}\)
\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2\left(x-2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2P=\left(x-1\right)Q\)
\(\Leftrightarrow P=x-1\)
\(Q=\left(x+2\right)^2=x^2+4x+4\)
b)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)
\(\Leftrightarrow P=\left(x+1\right)\left(x-2\right)=x^2-x-2\)
\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
Hì , giải đc rùi nha.
Vì \(x,y\in R\)
\(\Rightarrow\left(x+2\right).\left(y+2\right)=\frac{25}{4}\)
Min \(P=\sqrt{1+x^4}+\sqrt{1+y^4}\)
- Dự đoán \(x=y=\frac{1}{2}\)
- Sử dụng BĐT : \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\) ( Với a,b > 0 )
=> \(1+x^4=16.\frac{1}{16}+a^4=16.\left(\frac{1}{4}\right)^2+a^2\ge\frac{[16.\frac{1}{4}+a^2]^2}{17}\)
\(=\frac{(a^2+4)^2}{17}\)
=> \(1+y^4\ge\frac{\left(y^2+4\right)^2}{17}\)
=> \(P\ge\frac{x^2+y^2+8}{\sqrt{17}}\)
\(\Leftrightarrow P\sqrt{17}=\frac{1}{5}\left(x^2+y^2\right)+\frac{4}{5}\left(x^2+\frac{1}{4}+y^2+\frac{1}{4}\right)+8-\frac{2}{5}\)
\(\ge\frac{2xy}{5}+\frac{4}{5}\left(x+y\right)+8-\frac{2}{5}=\frac{2}{5}[xy+2\left(x+y\right)]+8-\frac{2}{5}\)
Theo giả thiết \(\left(x+2\right)\left(y+2\right)=\frac{25}{4}\)
\(\Leftrightarrow xy+2\left(x+y\right)=\frac{9}{4}\)
\(\Rightarrow P\sqrt{17}\ge\frac{2}{5}.\frac{9}{4}+8-\frac{2}{5}=\frac{17}{2}\)
\(\Leftrightarrow P\ge\frac{\sqrt{17}}{2}\)
Điểm rơi \(x=y=\frac{1}{2}\)
\(6a+3b+2c=abc\Leftrightarrow\dfrac{2}{ab}+\dfrac{3}{ac}+\dfrac{6}{bc}=1\)
Đặt \(\left(\dfrac{1}{a};\dfrac{2}{b};\dfrac{3}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(Q=\dfrac{1}{\sqrt{\dfrac{1}{x^2}+1}}+\dfrac{2}{\sqrt{\dfrac{4}{y^2}+4}}+\dfrac{3}{\sqrt{\dfrac{9}{z^2}+9}}=\dfrac{x}{\sqrt{x^2+1}}+\dfrac{y}{\sqrt{y^2+1}}+\dfrac{z}{\sqrt{z^2+1}}\)
\(Q=\dfrac{x}{\sqrt{x^2+xy+yz+zx}}+\dfrac{y}{\sqrt{y^2+xy+yz+zx}}+\dfrac{z}{\sqrt{z^2+xy+yz+zx}}\)
\(Q=\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{y}{\sqrt{\left(x+y\right)\left(y+z\right)}}+\dfrac{z}{\sqrt{\left(x+z\right)\left(y+z\right)}}\)
\(Q\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)=\dfrac{3}{2}\)
\(Q_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(\left(a;b;c\right)=\left(\sqrt{3};2\sqrt{3};3\sqrt{3}\right)\)
2/ \(\frac{1}{2}x2y5z3=\left(\frac{1}{2}.2.5.3\right)xyz\)\(=15xyz\)
\(\Rightarrow\frac{1}{2}x2y5z3\)có bậc là 3
3/ \(\frac{x}{4}=\frac{9}{x}\Leftrightarrow x^2=9.4\Rightarrow x^2=36\) mà \(x>0\Rightarrow x=6\)
4/ \(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\Rightarrow\left|2x+\frac{1}{2}\right|=\frac{35}{7}=5\Rightarrow\hept{\begin{cases}2x+\frac{1}{2}=5\Rightarrow2x=\frac{9}{2}\Rightarrow x=\frac{9}{4}\\2x+\frac{1}{2}=-5\Rightarrow2x=\frac{-11}{2}\Rightarrow x=\frac{-11}{4}\end{cases}}\)
Coi Q là đa thức bị chia, muốn tìm đa thức bị chia ta lấy đa thức thương nhân với đa thức chia nên Q = 4(2q + 3)(q + 1).