MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

Ta có:

N x3 – 3x2 + 3x – 1 = x3 – 3.x2.1 + 3.x.12 – 13 = (x – 1)3

U 16 + 8x + x2 = 42 + 2.4.x + x2 = (4 + x)2 = (x + 4)2

H 3x2 + 3x + 1 + x3 = x3 + 3x2 + 3x + 1 = (x + 1)3 = (1 + x)3

 1 – 2y + y2 = 12 – 2.1.y + y2 = (1 – y)2 = (y – 1)2

Điền vào bảng như sau:

Vậy: Đức tính đáng quý là "NHÂN HẬU"

(Chú ý: Bạn có thể làm theo cách ngược lại, tức là khai triển các biểu thức (x – 1)3, (x + 1)3, (y – 1)2, (x + 4)2 … để tìm xem kết quả ứng với chữ nào và điền vào bảng.)

a, \(25x^2+5xy+\frac{1}{4}y^2=\left(5x\right)^2+2.5x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)

\(=\left(5x+\frac{1}{2}y\right)^2\)

b, \(9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2=\left(3x+2\right)^2\)

c, \(x^2-6x+5-y^2-4y=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)\)

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

d, \(\left(2x-y\right)^2+4\left(x+y\right)^2-4\left(2x-y\right)\left(x+y\right)\)

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

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

Có cái cc

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+y+1\right)^2\)

này mình có vài câu không làm được, xin lỗi bạn nha

\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)

a) ( x   +   1 ) 2 .                     b) ( x   –   4 ) 2 .

c) x 2 4 + x + 1 ;                   d) ( 2 x   –   2 y ) 2 .

A)\(1-2x+x^2\)

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

B)\(4y+4+y^2\)

\(=2^2+4y+y^2\)

\(=\left(2+y\right)^2\)

C)\(\frac{1}{16}+\frac{1}{2}x+x^2\)

\(=\left(\frac{1}{4}\right)^2+\frac{1}{2}x+x^2\)

\(=\left(\frac{1}{4}+x\right)\)

D)\(36x^2+12xy+y^2\)

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

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

dài thế ai trả lời đc hả ?

tu lam di luoi vua thoi

a) \(x^2+4x+4\)

\(=x^2+2\cdot2\cdot x+2^2\)

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

b) \(4x^2-4x+1\)

\(=\left(2x\right)^2-2\cdot2x\cdot1+1^2\)

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

c) \(x^2-x+\dfrac{1}{4}\)

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

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

d) \(4\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)

\(=\left[2\left(x+y\right)-1\right]^2\)

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