a) \(2a^3+7a^2b+7ab^2+2b^3\)

\(=2\left(a^3+b^3\right)+7ab\left(a+b\right)\)

\(=2\left(a+b\right)\left(a^2-ab+b^2\right)+7ab\left(a+b\right)\)

\(=\left(a+b\right)\left(2a^2-2ab+b^2+7ab\right)\)

\(=\left(a+b\right)\left(2a^2+5ab+2b^2\right)\)

b) \(x^4+2x^2y+y^2-9\)

\(=\left(x^4+2x^2y+y^2\right)-9\)

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

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

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-24\)

\(=\left(x^2+7x+11\right)^2-1-24\)

\(=\left(x^2+7x+11\right)^2-5^2\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) \(x^4-6x^2-7x-6\) 

\(=x^4-3x^3+3x^3-9x^2+3x^2-9x+2x-6\)

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

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

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

\(=\left[x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\right]\left(x-3\right)\)

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

Câu 1

a) \(P=2a^3+7a^2b+7ab^2+2b^3\)

\(=\left(2a^3+2b^3\right)+\left(7a^2b+7ab^2\right)\)

\(=2\left(a^3+b^3\right)+7ab\left(a+b\right)\)

\(=2\left(a+b\right)\left(a^2-ab+b^2\right)+7ab\left(a+b\right)\)

\(=\left(a+b\right)\left[2\left(a^2-ab+b^2\right)+7ab\right]\)

\(=\left(a+b\right)\left(2a^2-2ab+2b^2+7ab\right)\)

\(=\left(a+b\right)\left(2a^2+5ab+2b^2\right)\)

\(=\left(a+b\right)\left(2a^2+4ab+ab+2b^2\right)\)

\(=\left(a+b\right)\left[\left(2a^2+4ab\right)+\left(ab+2b^2\right)\right]\)

\(=\left(a+b\right)\left[2a\left(a+2b\right)+b\left(a+2b\right)\right]\)

\(=\left(a+b\right)\left(a+2b\right)\left(2a+b\right)\)

b) \(x^4+2x^2y+y^2-9\)

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

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

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

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)  (1)

Đặt \(t=x^2+7x+10\)

\(\Rightarrow\left(1\right)=t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=t^2+2t+1-25\)

\(=\left(t+1\right)^2-5^2\)

\(=\left(t+1-5\right)\left(t+1+5\right)\)

\(=\left(t-4\right)\left(t+6\right)\)   (2)

Thế \(t=x^2+7x+10\) vào (2), ta có:

\(\left(2\right)=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)

\(=\left[\left(x^2+x\right)+\left(6x+6\right)\right]\left(x^2+7x+16\right)\)

\(=\left[x\left(x+1\right)+6\left(x+1\right)\right]\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

d) \(x^4-6x^2-7x-6\)

\(=x^4-3x^3+3x^3-9x^2+3x^2-9x+2x-6\)

\(=\left(x^4-3x^3\right)+\left(3x^3-9x^2\right)+\left(3x^2-9x\right)+\left(2x-6\right)\)

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

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

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

\(=\left(x-3\right)\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(x+2\right)\right]\)

\(=\left(x-3\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\right]\)

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

e) \(x^4+2008x^2+2007x+2008\)

Xem lại đề nhé em!

2x² - 18x + 6x -6 = 16 + 25

2x² - 12x -47 =0

\(x=\pm\dfrac{\sqrt{130}+6}{2}\)

Hằng đẳng thức: \(a^2+2ab+b^2=\left(a+b\right)^2\)

Cách chứng minh: \(VT=\left(a^2+ab\right)+\left(ab+b^2\right)=a\left(a+b\right)+b\left(a+b\right)\\ =\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2=VP\)

Áp dụng:

Kiểu đề 1: \(2x\left(x-9\right)+3\left(2x\right)-6=4^2+5^2\\ \Rightarrow2x^2-18x+6x-6=16+25\\ \Rightarrow2x^2-12x-47=0\\ \Rightarrow x^2-6x-\dfrac{47}{2}=0\\ \Rightarrow\left(x^2-2.x.3+3^2\right)-9-\dfrac{47}{2}=0\\ \Rightarrow\left(x-3\right)^2=\dfrac{65}{2}=\left(\dfrac{\pm\sqrt{130}}{2}\right)^2\\\)

\(\Rightarrow\left[{}\begin{matrix}x-3=\dfrac{\sqrt{130}}{2}\\x-3=\dfrac{-\sqrt{130}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{130}}{2}\\x=\dfrac{6-\sqrt{130}}{2}\end{matrix}\right.\)

Kiểu đề 2: \(2x\left(x-9\right)+3\left(2x-6\right)=4^2+5^2\\ \Rightarrow2x^2-18x+6x-18=16+25\\ \Rightarrow2x^2-12x-59=0\\ \Rightarrow x^2-6x-\dfrac{59}{2}=0\\ \Rightarrow\left(x^2-2.x.3+3^2\right)-9-\dfrac{59}{2}=0\\ \Rightarrow\left(x-3\right)^2=\dfrac{77}{2}=\left(\dfrac{\pm\sqrt{154}}{2}\right)^2\\ \)

\(\Rightarrow\left[{}\begin{matrix}x-3=\dfrac{\sqrt{154}}{2}\\x-3=\dfrac{-\sqrt{154}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{154}}{2}\\x=\dfrac{6-\sqrt{154}}{2}\end{matrix}\right.\)

làm phần c thôi nhé

\(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)

Nhận xét:

\(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\\\left(y+\dfrac{1}{2}\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2\ge0,\forall x,y\)

Dấu \("="\) xảy ra khi:

\(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{2};y=-\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2};y=-\dfrac{1}{2}\)

\(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

`(x^2+1)(x^2-1)=0`

`\Leftrightarrow x^4-1=0`

`\Leftrightarrow x^4=1`

`\Leftrightarrow x=\pm1`

`10)`

`(x^2 + 1)(x^2 -1)=0`

`<=>(x^2 +1)(x-1)(x+1)=0`

`<=>` \(\left[{}\begin{matrix}x^2+1=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=1\\x=-1\end{matrix}\right.\)

Vậy `x=-1` hoặc `x=1`

`18)`

`x^2 - 6x+8=-1`

`<=>x^2 - 6x + 9=0`

`<=>x^2 - 2.3x+3^2=0`

`<=>(x-3)^2=0`

`<=>x-3=0`

`<=>x=3`

Vậy `x=3`

Xét tam giác ABC vuông tại A. Đặt \(\widehat{B}=a\left(0^o< a< 90^o\right)\) 

Khi đó ta có \(\tan a=\dfrac{\sin a}{\cos a}=\dfrac{AC}{AB}< 1\) (vì \(\cos a>\sin a\))

\(\Rightarrow AC< AB\)

\(\Rightarrow\widehat{B}< \widehat{C}\) (quan hệ giữa góc và cạnh đối diện trong tam giác)

Lại có \(\widehat{B}+\widehat{C}=90^o>\widehat{B}+\widehat{B}=2\widehat{B}\)  nên \(\widehat{B}=a< 45^o\).

Ta có đpcm.

Cửa hàng đã bán tất cả số chai dầu là:

10 x 6 = 60 (chai)

ĐS: 60 chai 

                                             Bài giải

                      Cửa hàng đó bán được số chai dầu là :

                                        10 x 6 = 60 (chai)

                                                   Đáp số : 60 chai dầu .

 Chọn hệ trục tọa độ Mxyz (M là gốc tọa độ) sao cho Mx trùng với tia MB, My trùng với tia MA và Mz cùng phương với BB' sao cho \(\overrightarrow{BB'}\) hướng theo chiều dương của Mz. 

 Gọi chiều cao lăng trụ là \(h>0\)

 Khi đó \(B\left(a;0;0\right)\), \(C'\left(-a;0;h\right)\), \(A'\left(0;a\sqrt{3};h\right)\)

 Ta có \(\overrightarrow{MC'}=\left(-a;0;h\right),\overrightarrow{BA'}=\left(-a;a\sqrt{3};h\right)\)

\(\Rightarrow\left[\overrightarrow{MC'},\overrightarrow{BA'}\right]=\left(-ah\sqrt{3};0;a^2\sqrt{3}\right)\)

\(\Rightarrow\left|\left[\overrightarrow{MC'},\overrightarrow{BA'}\right]\right|=\sqrt{\left(-ah\sqrt{3}\right)^2+\left(a^2\sqrt{3}\right)^2}=a\sqrt{3h^2+3a^2}\)

Lại có \(\overrightarrow{MB}=\left(a;0;0\right)\)

\(\Rightarrow\left[\overrightarrow{MC'},\overrightarrow{BA'}\right].\overrightarrow{MB}=-a^2h\sqrt{3}\)

\(\Rightarrow d\left(MC',BA'\right)=\dfrac{\left|\left[\overrightarrow{MC'},\overrightarrow{BA'}\right].\overrightarrow{MB}\right|}{\left|\left[\overrightarrow{MC'},\overrightarrow{BA'}\right]\right|}\) \(=\dfrac{a^2h\sqrt{3}}{a\sqrt{3a^2+3h^2}}=\dfrac{ah}{\sqrt{a^2+h^2}}\)

Theo đề bài, ta có: \(\dfrac{ah}{\sqrt{a^2+h^2}}=\dfrac{a}{2}\) 

\(\Leftrightarrow\dfrac{h}{\sqrt{a^2+h^2}}=\dfrac{1}{2}\)

\(\Leftrightarrow2h=\sqrt{a^2+h^2}\) 

\(\Leftrightarrow4h^2=a^2+h^2\)

\(\Leftrightarrow3h^2=a^2\)

\(\Leftrightarrow h=\dfrac{a}{\sqrt{3}}\)

\(\Rightarrow V=S_đ.h=\dfrac{\left(2a\right)^2\sqrt{3}}{4}.\dfrac{a}{\sqrt{3}}=a^3\)

Vậy thể tích lăng trụ bằng \(a^3\)

a)

\(\dfrac{x^4+12x^2-5x}{-x}=-\dfrac{x^4}{x}-\dfrac{12x^2}{x}+\dfrac{-5x}{-x}=-x^3-12x+5\)

b)

\(\dfrac{15x^5y^9-10x^3y^5+25x^4y^4}{5x^2y^2}=\dfrac{15x^5y^9}{5x^2y^2}-\dfrac{10x^3y^5}{5x^2y^2}+\dfrac{25x^4y^4}{5x^2y^2}=3x^3y^7-2xy^3+5x^2y^2\)

`a)`

`(x^4 + 12x^2 -5x):(-x)`

`=[x^4 : (-x)] + [12x^2 : (-x)] - [5x:(-x)]`

`=-x^3 - 12x + 5`

`b)`

`(15 x^5 y^9 - 10 x^3 y^5 + 25 x^4 y^4) : 5x^2 y^2`

`=(15 x^5 y^9 : 5 x^2 y^2) - (10 x^3 y^5 : 5x^2 y^2) + (25 x^4 y^4 : 5 x^2 y^2)`

`=3 x^3 y^7 - 2 x y^3 + 5 x^2 y^2`