Ta có: \(P\left(x\right)⋮Q\left(x\right)\)

\(\Leftrightarrow2x^3-7x^2+5x+1⋮2x-1\)

\(\Leftrightarrow2x^3-x^2-6x^2+3x+2x-1+2⋮2x-1\)

\(\Leftrightarrow2⋮2x-1\)

\(\Leftrightarrow2x-1\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow2x\in\left\{2;0;3;-1\right\}\)

hay \(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2}\right\}\)

\(P=\dfrac{1}{2xy}+\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\ge\dfrac{1}{\dfrac{2.\left(x+y\right)^2}{4}}+\dfrac{4}{2xy+x^2+y^2}=\dfrac{6}{\left(x+y\right)^2}=6\)

\(P_{min}=6\) khi \(a=b=\dfrac{1}{2}\)

Cách khác:

Đặt $xy=t$. Bằng $AM-GM$ dễ thấy $t\leq \frac{1}{4}$

\(P=\frac{1}{xy}+\frac{1}{(x+y)^2-2xy}=\frac{1}{xy}+\frac{1}{1-2xy}=\frac{1}{t}+\frac{1}{1-2t}\)

\(=\frac{1}{t}-4+\frac{1}{1-2t}-2+6=\frac{(1-4t)(1-3t)}{t(1-2t)}+6\geq 6\) với mọi $t\leq \frac{1}{4}$

Vậy $P_{\min}=6$ khi $x=y=\frac{1}{2}$

a, \(\left(2x-1\right)^2:9=49\)

\(\left(2x-1\right)^2=441\)

\(\Rightarrow\orbr{\begin{cases}2x-1=441\\2x-1=-441\end{cases}\Rightarrow\orbr{\begin{cases}x=221\\x=-220\end{cases}}}\)

b, \(3^x+3^{x+2}=810\)

\(3^x+3^x.3^2=810\)

\(3^x\left(1+3^2\right)=810\)

\(3^x.10=810\)

\(3^x=81=3^4\)

\(\Rightarrow x=4\)

\(\hept{\begin{cases}80⋮x\\56⋮x\end{cases}}\Rightarrow x\inƯC\left(80;56\right)\)

\(80=2^4.5\)

\(56=2^3.7\)

\(ƯCLN\left(80;56\right)=2^3=8\)

\(\RightarrowƯC\left(80;56\right)=Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Mà \(x\ge3\)

\(\Rightarrow x\in\left\{4;8\right\}\)

a) \(\left(2x-1\right)^2\div9=49\)

\(\Leftrightarrow\left(2x-1\right)^2=49\times9\)

\(\Leftrightarrow\left(2x-1\right)^2=441\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(\pm21\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=21\\2x-1=-21\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-10\end{cases}}\)

Vậy \(x\in\left\{-10;11\right\}\)

b) \(3^x+3^{x+2}=810\)

\(\Leftrightarrow3^x+3^x.3^2=810\)

\(\Leftrightarrow3^x\left(1+3^2\right)=810\)

\(\Leftrightarrow3^x.10=810\)

\(\Leftrightarrow3^x=810\div10\)

\(\Leftrightarrow3^x=81\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

Bạn học 7 hằng đẳng thức chưa?

1/x+1/y=1/2 <=> (x+y)/xy=1/2 <=>[(\(\sqrt{x}+\sqrt{y}\))²-2\(\sqrt{xy}\)]/xy=1/2 <=>(\(\sqrt{x}+\sqrt{y}\))²=xy/2+2\(\sqrt{xy}\)=A²

1/2=1/x+1/y\(\ge\)2/\(\sqrt{xy}\)(bdt cosi cho 1/x và 1/y) <=>1/2 \(\ge\frac{2}{\sqrt{xy}}\)<=> \(\sqrt{xy}\ge\)4

Vậy A²\(\ge\)4²/2+2.4=16 <=> A\(\ge\)4( vì A >0)

Dấu = xảy ra khi 1/x=1/y và \(\sqrt{xy}=4\)=> x=y=4

\(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}=\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2\ge\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2\)

=> \(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2\le1\)

=> \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\le1\)

=> \(1\ge\frac{1^2}{\sqrt{x}}+\frac{1^2}{\sqrt{y}}\ge\frac{\left(1+1\right)^2}{\sqrt{x}+\sqrt{y}}=\frac{4}{\sqrt{x}+\sqrt{y}}\)

=> \(\sqrt{x}+\sqrt{y}\ge4\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{y}}\\\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\end{cases}}\Leftrightarrow x=y=4\)

Vậy min A = 4 đạt tại x = y= 4.

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