Bài 3  Tính nhanh

A, 892^2+892.216+108^2                 B, 36^2+26^2-52.36

  =892^2+2.892.108+108^2                 =36^2-52.62+26^2

  =(892+108)^2           

  =1000^2

  =1000000

Bài 4 Phân tích đa thức sau thành nhân tử   

X^3-2x^2+x

5(x-y)-y(x-y)

36-12x+x^2

4x^2+12x-9

Bài 3:

\(892^2+892.216+108^2=892^2+2.892.108+108^2=\left(892+108\right)^2=1000000\)

\(36^2+26^2-52.36=36^2-2.26.36+26^2=\left(36-26\right)^2=100\)

Bài 4: 

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

\(5.\left(x-y\right)-y.\left(x-y\right)=\left(5-y\right)\left(x-y\right)\)

\(36-12x+x^2=x^2-12x+36=x^2-2x.6+6^2=\left(x-6\right)^2\)

\(4x^2+12x-9=\left(2x\right)^2+2.2x.3+3^2=\left(2x+3\right)^2\)

\(\left(5x-y\right)^2\)

\(=25x^2-10xy+y^2\)

\(\left(5x-y\right)^2\)

\(=25x^2-10xy+y^2\)

My favorite hobby is playing football in spare time. After completing my home work at home, I generally spend my lot of free time in playing football. I was so interested to play football from my childhood however started learning to play well when I was 5 years old. I was in one class when I was 5 years old. My father asked to my class teacher in the PTM about my hobby of football. And my teacher told him that there is a facility of playing sports daily in the school from class 1 so you can admit your child. Now, I really enjoy playing football and paripate in the inter-school competitions.

Tham khảo!!!

You know, I go to school during the weekday, so I have little leisure time.Therefore, I usually watch films on the weekend. I really enjoy watching many types of films, but my favourite is dramas. Twice a week, I go to the cinema to watch there. I sometimes watch films online if I cannot find the films I want in the cinema.I enjoy doing it because I had more experience in life when I watching films which help me less stressful and more self-confident after days of hard study.Thanks to the film, I can broaden my knowledge and I learn new things.

hahaahahhahaha

9 10 J Q K

Trả lời:

Ta có: \(B=3x^2+y^2-12x+2xy+15\)

\(=x^2+2x^2+y^2-12x+2xy+18-3\)

\(=\left(x^2+2xy+y^2\right)+\left(2x^2-12x+18\right)-3\)

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

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

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

Vậy GTNN của B = - 3 khi x = 3; y = - 3.

Bạn tham khảo nhé:

Trên tia đối của KG lấy điểm F sao cho KG=KF.

Ta có: ΔABC đều => ^A=60⁰. Xét ΔADE có: ^A=60⁰, AD=AE

=> ΔADE đều. Mà G là trọng tâm của ΔADE

=> G cũng là giao của 3 đường trung trực trong ΔABC 

=> DG=AG (T/c đường trung trực) (1)

Xét ΔGDK và ΔFCK:

KD=KC

^DKG=^CKF              => ΔGDK=ΔFCK (c.g.c)

KG=KF

=> DG=CF (2 cạnh tương ứng). (2)

Từ (1) và (2) => AG=CF.

Cũng suy ra đc: ^GDK=^FCK (2 góc tương ứng) => ^GDE+^EDK=^FCB+^BCK

Lại có: ED//BC (Vì ΔADE đều) => ^EDK=^BCK (So le trong)

=> ^GDE=^FCB (Bớt 2 vế cho ^EDK, ^BCK) (3)

Xét ΔΔADE: Đều, G trọng tâm => DG cũng là phân giác ^ADE

=> ^GDE=^ADE/2=30⁰. 

Tương tự tính được: ^GAD=30⁰ => ^GDE=^GAD hay ^GDE=^GAB (4)

Từ (3) và (4) => ^GAB=^FCB

Xét ΔAGB và ΔCFB có:

AB=CB

^GAB=^CFB           => ΔAGB=ΔCFB (c.g.c)

AG=CF

=> GB=FB (2 cạnh tương ứng) (5).

=> ^ABG=^CBF (2 góc tương ứng). Lại có:

^ABG+^GBC=^ABC=60⁰. Thay ^ABG=^CBF ta thu được:

^CBF+^GBC=60⁰ => ^GBF=60⁰ (6)

Từ (5) và (6) => ΔGBF là tam giác đều. => ^BGF=60⁰ hay ^BGK=60⁰

K là trung điểm của GF => BK là phân giác ^GBF => ^GBK= ^GBF/2=30⁰

Xét ΔBGK: ^BGK=60⁰, ^GBK=30⁰ => ^BKG=90⁰.

ĐS: ^GBK=30⁰, ^BGK=60⁰, ^BKG=90⁰.