\(\text{Ta có:}\)

\(a^{12}+b^{12}-\left(a^{11}+b^{11}\right)\left(a+b\right)+\left(a^{10}+b^{10}\right)ab=0\)

\(\Rightarrow\left(a^{12}+b^{12}\right)\left(ab-a-b+1\right)=0\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}\)

\(+,a=1\Rightarrow b^{10}=b^{11}=b^{12}\Rightarrow b=1\left(\text{vì b dương}\right)\)

\(+,b=1\Rightarrow a^{10}=a^{11}=a^{12}\Rightarrow a=1\left(\text{vì a dương}\right)\)

\(\text{nên: a=b=1}\)

\(\text{Vậy: P=2011a-2012b=2011-2012=-1}\)

Chữ nhỏ quá mik đọc ko ra

công nhận chữ nhỏ thật

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\(a\left(b-c\right)\left(b+c-a\right)^2+c\left(a-b\right)\left(a+b-c\right)^2\)

\(=\left(ab-ac\right)\left(a^2+b^2+c^2+2bc-2ab-2ca\right)\)

\(+\left(ac-bc\right)\left(a^2+b^2+c^2+2ab-2bc-2ca\right)\)

\(=-ac^3+\left(2a^2-ab\right)c^2+\left(ab^2-a^3\right)c+ab^3-2a^2b^2+a^3b\)

\(+\left(a-b\right)c^3+\left(2b^2-2a^2\right)c^2+\left(-b^3-ab^2+a^2b+a^3\right)c\)

\(=-bc^3+\left(2b^2-ab\right)c^2+\left(a^2b-b^3\right)c+ab^3-2a^2b^2+a^3b\)

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

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

Gọi O là giao điểm AC, BD=> O là trung điểm BC

=> Q là trọng tâm tam giác ABC \(\Rightarrow BQ=\frac{2}{3}BO=\frac{1}{3}BD\)

Lần lượt kẻ QK và OH vuông góc BC \(\Rightarrow\frac{QK}{OH}=\frac{BQ}{BO}=\frac{2}{3}\)(định lí Ta-lét)

Ta có: \(S_{BQM}=\frac{1}{2}.QK.BM\)

\(S_{OBC}=\frac{1}{2}.OH.BC=\frac{1}{2}.\left(\frac{3}{2}QK\right).2BM=3\left(\frac{1}{2}QK.BM\right)=3S_{BQM}\)

Lại có:\(S_{OBC}=\frac{1}{2}S_{BCD}=\frac{1}{4}S_{ABCD}=\frac{1}{4}\)

\(\Rightarrow S_{BQM}=\frac{1}{3}S_{OBC}=\frac{1}{12}\)

\(\Leftrightarrow S_{MQDC}=S_{BCD}-S_{BQM}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)

\(\frac{7-x}{2}+\frac{2}{3}.\left(x-7\right)\left(x-3\right)=0\)

\(\Leftrightarrow\frac{x-7}{-2}+\frac{2\left(x-7\right)}{3}\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left[\frac{-1}{2}+\frac{2}{3}\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-7\right)\left(\frac{-1}{2}+\frac{2x}{3}-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-7=0\\\frac{-1}{2}+\frac{2x}{3}-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\-\frac{5}{2}+\frac{2x}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\\frac{2x}{3}=\frac{5}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{15}{4}\end{cases}}}\)