Ta thấy : \(a_1+a_2+a_3+.....+a_{2015}+a_1=1008.1=1008\)

Mà \(a_1+a_2+a_3+......+a_{2015}=0\)

\(\Rightarrow a_1+\left(a_1+a_2+a_3+....+a_{2015}\right)=1008\Leftrightarrow a_1+0=1008\)                                                                                                                                                                                                           \(\Rightarrow a_1=1008\) 

\(a.A=2x^2+6x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+20\)

  \(A=\left(2x+\frac{3}{2}\right)^2+\frac{71}{4}\ge\frac{71}{4}\)

Vậy MinA = \(\frac{71}{4}\Leftrightarrow\left(2x+\frac{3}{2}\right)^2=0\)

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

\(\Delta ABC\)cân tại A nên\(\widehat{ABC}=\widehat{ACB}=\frac{180^0-\widehat{BAC}}{2}=75^0\)

Trên nửa mặt phẳng bờ BC chứa A lấy E sao cho\(\widehat{B_1}=\widehat{C_1}=45^0\)

=>\(\widehat{ABE}=75^0-45^0=30^0;\Delta EBC\)vuông cân tại E =>\(BE=EC=\frac{BC}{\sqrt{2}}=\sqrt{2}\left(cm\right)\)(định lí Pitago)

\(\Delta ABE,\Delta BAD\)có AB chung ; BE = AD\(\left(=\sqrt{2}cm\right)\);\(\widehat{ABE}=\widehat{BAD}\left(=30^0\right)\)

\(\Rightarrow\Delta ABE=\Delta BAD\left(c.g.c\right)\Rightarrow\widehat{A_1}=\widehat{B_2}\)

Lại có\(\Delta AEB=\Delta AEC\left(c.c.c\right)\)nên\(\widehat{A_1}=\widehat{A_2}=\frac{\widehat{BAC}}{2}=15^0\Rightarrow\widehat{B_2}=15^0\)

\(\Rightarrow\widehat{D_1}=\widehat{BAD}+\widehat{B_2}=45^0\)(\(\widehat{D_1}\)là góc ngoài\(\Delta ABD\)) ;\(\widehat{DBC}=75^0-15^0=60^0\)

\(\Delta BDC\)có\(\widehat{D_1}< \widehat{DBC}< \widehat{DCB}\left(45^0< 60^0< 75^0\right)\)nên BC < DC < BD

bai nay trong sach nang cao toan 7 trang 141

Thay x=1 vào bt

A=a-3-2a+1=4

-a-2=4

-a=6

a=6

Ta có A(1) = 4

<=> a - 2 - 2a +2 = 4

<=> a = -4

;a52 nhe

x³+y³=(x+y)(x²-xy+y²)=(x+y(x²+2xy+y²-3xy)=(x+y)[(x+y)²-3xy]=4[16-6]=40

Đáp số: 40

\(x+y=4\Rightarrow\left(x+y\right)^2=4^2\Leftrightarrow x^2+2xy+y^2=16\Leftrightarrow x^2+2.2+y^2=16\)

\(\Leftrightarrow x^2+4+y^2=16\Leftrightarrow x^2+y^2=12\)

=>\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=4\left(12-2\right)=4.10=40\)