Đặt giá tiền một quyển vở và một cái bút lần lượt là \(x,y\)(đồng), \(x,y>0\).

Theo bài ra, ta có hệ phương trình: 

\(\hept{\begin{cases}8x+15y=88500\\13x+9y=90000\end{cases}}\Leftrightarrow\hept{\begin{cases}24x+45y=265500\\65x+45y=450000\end{cases}\Leftrightarrow\hept{\begin{cases}41x=184500\\y=\frac{90000-13x}{9}\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4500\\y=3500\end{cases}}\left(tm\right)\)

em chịu thôi

\(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+2a^2+2b^2-2ab+a^2+b^2+2ab+2.2a+2.2b+2^2\)

\(=\left(a+b+2\right)\left(a^2-ab+b^2\right)+\left(a+b+2\right)^2\)

\(=\left(a+b+2\right)\left(a^2-ab+b^2+a+b+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+2=0\\a^2-ab+b^2+a+b+2=0\end{cases}}\)

Ta có: \(a^2-ab+b^2+a+b+2=\frac{1}{2}a^2-ab+\frac{1}{2}b^2+\frac{1}{4}a^2+a+1+\frac{1}{4}b^2+b+1+\frac{1}{4}\left(a^2+b^2\right)\)

\(=\frac{1}{2}\left(a-b\right)^2+\frac{1}{4}\left(a+2\right)^2+\frac{1}{4}\left(b+2\right)^2+\frac{1}{4}\left(a^2+b^2\right)>0,\forall a,b\inℝ\).

Suy ra \(a+b+2=0\Leftrightarrow a+b=-2\)

mà \(ab>0\Rightarrow\hept{\begin{cases}a< 0\\b< 0\end{cases}}\). 

\(Q=\frac{1}{a}+\frac{1}{b}=-\left(\frac{1}{-a}+\frac{1}{-b}\right)\le-\frac{\left(1+1\right)^2}{-a-b}=-2\)

Dấu \(=\)xảy ra khi \(a=b=-1\).

Áp dụng BĐT Cauchy - Schwarz và Cauchy ta có:

\(P=\frac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\)

\(\ge\frac{b^2+c^2}{a^2}+a^2\cdot\frac{9}{b^2+c^2}\) (Cauchy - Schwarz)

\(=\left(\frac{b^2+c^2}{a^2}+\frac{a^2}{b^2+c^2}\right)+8\cdot\frac{a^2}{b^2+c^2}\)

\(\ge2\sqrt{\frac{b^2+c^2}{a^2}\cdot\frac{a^2}{b^2+c^2}}+8\cdot\frac{b^2+c^2}{b^2+c^2}\) (BĐT Cauchy)

\(=2+8=10\)

Dấu "=" xảy ra khi: \(a=b\sqrt{2}=c\sqrt{2}\)

Vậy Min(P) = 10 khi \(a=b\sqrt{2}=c\sqrt{2}\)

Ta có: \(x^2-6x+11=\left(x-3\right)^2+2\ge2;\forall x\)

Lại có: \(\left(\sqrt{x+2}+\sqrt{4-x}\right)^2\le\left[\left(\sqrt{x+2}\right)^2+\left(\sqrt{4-x}\right)^2\right]\left(1+1\right)=2\)( bunhiacopxki )

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-3\right)^2=0\\\sqrt{x-2}=\sqrt{4-x}\end{cases}\Leftrightarrow}x=3\)

Vậy pt có no x=3