\(Q=\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)\left(x-2\right)+\left(x-1\right)\)

Thay vào ta được:

\(Q=\left(5-1\right)\left(5-2\right)\left(5-3\right)+\left(5-1\right)\left(5-2\right)+\left(5-1\right)\)

\(=4.3.2+4.3+4\)

\(=24+12+4\)

\(=40\)

\(2012^{100}+11\)

\(=\left(2012^4\right)^{25}+11=\left(...6\right)^{25}+\left(...1\right)=\left(...6\right)+\left(...1\right)=\left(...7\right)\)

Mà không có số chính phương nào có tận cùng là 7 \(\Rightarrow2012^{100}+11\)không phải là số chính phương (đpcm)

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

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

b) \(\left(x-y\right)^2+2xy\)

\(=x^2-2xy+y^2+2xy\)\(=x^2+y^2\) => đpcm

c) \(\left(x+y\right)^2-2xy\)

\(=x^2+2xy+y^2-2xy\)\(=x^2+y^2\) => đpcm

\(A=x^2+y^2-6x+4y+20\)

\(=x^2+2x.3+9+y^2+2y.2+4+7\)

\(=\left(x-3\right)^2+\left(y+2\right)^2+7\)

Ta có: \(\hept{\begin{cases}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}}\)\(\Rightarrow A=7\)

Hình tự vẽ nhé

Theo đề ra, ta có: \(P_{AEMF}=2a\Rightarrow2\left(AE+EM\right)=2a=2AB\)

\(\Rightarrow AE+EM=AB=AE+EB\)

\(\Rightarrow EM=EB\)

=> Tam giác EBM vuông cân tại E

\(\Rightarrow\widehat{EBM}=\widehat{ABC}=45^o\)

=> B, M, C thẳng hàng

=> M di động trên BC

b nữa

\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Leftrightarrow10x^2+9x-\left(10x^2+13x-3\right)=8\)

\(\Leftrightarrow-4x+3=8\Leftrightarrow-4x=5\Leftrightarrow x=-\frac{5}{4}\)

(10x+9)x-(5x-1)(2x+3)=8

<=>10x²+9x-(10x²+15x-2x-3)=8

<=>10x²+9x-10x²-15x+2x+3=8

<=>-4x+3=8

<=>-4x=5

<=>x=-5/4

$(10x+9).x-(5x-1).(2x+3)=8<=>10x^2+9x-10x^2-15x+2x+3=8<=>-4x+3=8<=>-4x=5<=>x=\frac{-5}{4}Vậyx=\frac{-5}{4}$

\(5x\left(x-4y\right)-4y\)

Thay vào ta được:

\(5\left(\frac{-1}{5}\right)[\left(\frac{-1}{5}\right)-4\left(\frac{-1}{2}\right)]-4\left(\frac{-1}{2}\right)\)

\(=-[\left(\frac{-1}{5}\right)-2]-2\)

\(=\left(\frac{1}{5}-2\right)-2\)

\(=\frac{11}{5}-2\)

\(=\frac{1}{5}\)

\(5x\left(x-4y\right)-4y=5x^2-20xy-4y\)

thay   x= -1/5; y= -1/2 vào ta có:

\(5\left(-\frac{1}{5}\right)^2-20\left(-\frac{1}{5}\right)\left(-\frac{1}{2}\right)-4\left(-\frac{1}{2}\right)^2=\frac{5}{25}-\frac{20}{10}-\frac{4}{4}=\frac{1}{5}-2-1=\frac{1}{5}-\frac{15}{5}=-\frac{14}{5}\)

\(\left(x-2\right)^2-\left(2x-1\right)^2+\left(3x-1\right)\left(x-5\right)\)

\(=x^2-4x+4-\left(4x^2-4x+1\right)+3x^2-16x+5\)

\(=\left(x^2-4x^2+3x^2\right)-\left(4x-4x+16x\right)+4-1+5\)

\(=-16x+8\)