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17 tháng 8 2020

2,3 ko chia đc 785,457

17 tháng 8 2020

12,66 - 2,3 = 10.36

17 tháng 8 2020

                                         Bài làm :

  •  Cách 1:  x2- 6x + 8 

                          = x2 - 2x - 4x + 8

                          = x (x - 2) - 4(x -2)

                          = (x - 4)(x -2)

  • Cách 2: x2 - 6x + 8  

                     = x2 - 6x + 9 - 1

                     = ( x - 3)2 - 1

                     =( x -3 - 1)( x- 3 + 1)

                     = (x - 4)(x -2)

  •  Cách 3: x2 - 6x + 8  

                       = x2 - 16 - 6x + 24

                       =( x - 4)(x + 4 ) - 6 (x - 4)

                       =(x - 4)(x + 4 - 6)

                       = (x - 4)(x -2) 

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


 

17 tháng 8 2020

mình cũng được tròn 3 cách 

c1 \(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

c2 \(x^2-6x+8=\left(x^2-6x+9\right)-1=\left(x-3\right)^2-1=\left(x-4\right)\left(x-2\right)\)

c3 Gỉa sử \(x^2-6x+8=\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)

Cân bằng hệ số ta được \(\hept{\begin{cases}a+b=-6\\ab=8\end{cases}< =>\orbr{\begin{cases}a=-4\\b=-2\end{cases}or\orbr{\begin{cases}a=-2\\b=-4\end{cases}}}}\)

Vậy ta có : \(\left(x+a\right)\left(x+b\right)=\left(x-2\right)\left(x-4\right)\)

17 tháng 8 2020

Đề bài là gì thế bạn nhờ?

17 tháng 8 2020

lm trên symbolab.com

17 tháng 8 2020

\(\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\cos^2x\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\left(2-\sin^2x\right)\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=4\sin^2x-1\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=\left(2\sin x-1\right)\left(2\sin x+1\right)\)

\(\Leftrightarrow2\sin2x+1=2\sin x+1\)

\(\Leftrightarrow\sin2x=\sin x\)

\(\Leftrightarrow\sin2x-\sin x=0\)

\(\Leftrightarrow2\cos\frac{3}{2}-\cos\frac{x}{2}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\cos\frac{3}{2}=0\\\cos\frac{x}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{cases}\left(k\inℤ\right)}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{3}+\frac{2\pi}{3}k\\x=\pi+4k\pi\end{cases}\left(k\inℤ\right)}\)

17 tháng 8 2020

a)

\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(A=100+99+98+97+...+2+1\)

\(A=\frac{100.101}{2}=5050\)

b)

\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(B=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(B=\left(2^8-1\right)...\left(2^{64}+1\right)+1\)

\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)

\(B=2^{128}-1+1=2^{128}\)

c)

\(C=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\)

\(C=2c^2\)

17 tháng 8 2020

thanks bạn nhaaa :3

17 tháng 8 2020

chú ý phải làm dài

17 tháng 8 2020

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17 tháng 8 2020

2 / 3 x 3 / 4 = 2x3 / 3x4 =  6/12 = 1/2 

Học tốt ^_^

17 tháng 8 2020

ah nha

\(\frac{2}{3}\).\(\frac{3}{4}\)=\(\frac{6}{12}\)=\(\frac{1}{2}\)

17 tháng 8 2020

9 GIỜ LÀ 2 KIM ĐỒNG HỒ VUÔNG GÓC VỚI NHAU RỒI

17 tháng 8 2020

30 phút

18 tháng 8 2020

A.will enjoy

A. will enjoy

20 tháng 8 2020

bvzrglvobhROCV^%