\(x^2-6x+9=x^2-2.3x+3^2=\left(x-3\right)^2\)

\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b^2+\left(2b\right)^2=\left(\frac{1}{2}a+2b\right)^2\)

\(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)

\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}\right)^2-2.\frac{1}{3}y^4+\left(y^4\right)^2=\left(\frac{1}{3}-y^4\right)^2\)

a,(x-3)^2

b,(1/4x+2b^2)^2

c,(5+x)^2

d,(1/3-y^4)^2

sorry ,phần b fải là (1/2a+2b^2)^2

Khó nhỉ

a. Đề đúng phải là \(\frac{1}{4}a^2+2ab^2+4b^4\)hoặc \(\frac{1}{4}a^2+2ab+4b^2\)

Ở đây mình giải trường hợp 2, bạn dựa theo để giải trường hợp 1 nhé :))

\(\frac{1}{4}a^2+2ab+4b^2\)

\(=\left(\frac{1}{2}a\right)^2+2ab+\left(2b\right)^2\)

\(=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b+\left(2b\right)^2\)

\(=\left(\frac{1}{2}a+2b\right)^2\)

b. \(25+10x+x^2\)

\(=x^2+2.x.5+5^2\)

\(=\left(x+5\right)^2\)

c. \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)

\(=\left(y^4\right)^2-2.y^4.\frac{1}{3}+\left(\frac{1}{3}\right)^2\)

\(=\left(y^4-\frac{1}{3}\right)^2\)

a) \(9x^2+6x+1=\left(3x+1\right)^2\)

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

c)\(x^2y^4-2xy^2+1=\left(xy^2-1\right)^2\)

d) \(x^2+\frac{2}{3}x+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2\)

a) 9x² + 6x + 1 = ( 3x + 1 )²

b) x² - x + 1/4 = ( x - 1/2)²

c) x² . y⁴ - 2xy² + 1 = ( xy² - 1 ) ²

d) x² + 2/3x + 1/9 = (x+1/3)²

a) \(9x^2+6x+1\)

\(=\left(3x\right)^2+6x+1\)

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

b) \(x^2-x+\frac{1}{4}\)

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

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

c)  \(x^2y^4-2xy^2+1\)

.... Rút gọn lũy thừa cao

d ) Tương tự câu b . Làm theo cách thêm bớt

a:Sửa đề: \(\dfrac{1}{4}a^2+2ab+4b^2\)

\(=\left(\dfrac{1}{2}a\right)^2+2\cdot\dfrac{1}{2}a\cdot2b+\left(2b\right)^2\)

\(=\left(\dfrac{1}{2}a+2b\right)^2\)

b: Sửa đề:\(y^4-\dfrac{1}{3}y^4+\dfrac{1}{36}\)

\(=y^8-2\cdot y^4\cdot\dfrac{1}{6}+\dfrac{1}{36}\)

\(=\left(y^4-\dfrac{1}{6}\right)^2\)

bạn vào loigiaihay rồi chọn toán lớp 8 rồi chọn đẳng thức đáng nhớ

dễ mà áp dụng hết hằng đẳng thức nếu bạn thuộc hằng đẳng thức mik chỉ làm mỗi bài 1 ý nha xong dựa vô mà làm

\(1a.\left(2x+3y\right)^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2\)

                                   \(=4y^2+12xy+9y^2\)

\(2a.x^2-6x+9\)

\(=x^2-2.x.3+3^2\)

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

CẢM ƠN BẠN MINH ÁNH NHÉ

a) \(9x^2+6x+1=\left(3x\right)^2+2.3x.1+1^2=\left(3x+1\right)^2\)

b) \(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2=\left(x-0,5\right)^2\)

c) \(x^2y^4-2xy^2+1=\left(xy^2\right)^2-2.xy^2.1+1^2=\left(xy^2-1\right)^2\)

d) \(x^2+\dfrac{2}{3}x+\dfrac{1}{9}=x^2+2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2=\left(x+\dfrac{1}{3}\right)^2\)

a) \(9x^2+6x+1\)

\(=\left(3x\right)^2+2.3x.1+1^2\)

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

b) \(x^2-x+\dfrac{1}{4}\)

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

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

\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)

Bài 1 em dùng HĐT nha

Bài 2:

a. x² + 4x + 4

= x² + 2.2.x + 2²

= (x + 2)²

b. 9x² - 12x + 4

= (3x)² - 3x.2.2 + 2²

= (3x - 2)²

c. \(\dfrac{x^2}{4}+x+1\)

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

= \(\left(\dfrac{x}{2}+1\right)^2\)