\(a,8-2\sqrt{7}=\sqrt{7}^2-2\sqrt{7}+1^2=\left(\sqrt{7}-1\right)^2\)

\(b,8-2\sqrt{15}=\sqrt{5}^2-2.\sqrt{3}.\sqrt{5}+\sqrt{3}^2=\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(c,8+4\sqrt{3}=2^2+2.2.\sqrt{3}+\sqrt{3}^2=\left(2+\sqrt{3}\right)^2\)

xl chuyển hộ mk "bình phương" thành "lập phương" nha

\(=\left(\dfrac{2}{3}a\right)^3-3.\left(\dfrac{2}{3}\right)^2a^2.2b+3.\dfrac{2}{3}a.4b^2-\left(2b\right)^3=\left(\dfrac{2}{3}a-2b\right)^3\)

b)\(27-10\sqrt{2}=5^2-2.5\sqrt{2}+2=\left(5-\sqrt{2}\right)^2\)

c)\(18-8\sqrt{2}=4^2-2.4\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)

d)\(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

e)\(6\sqrt{5}+14=9+2.3\sqrt{5}+5=\left(3+\sqrt{5}\right)^2\)

f)\(20\sqrt{5}+45=5^2+2.5.2\sqrt{5}+20=\left(5+2\sqrt{5}\right)^2\)

g)\(7-2\sqrt{6}=6-2\sqrt{6}+1=\left(\sqrt{6}-1\right)^2\)

a,\(5+\sqrt{24}=5+\sqrt{6.4}=5+2\sqrt{6}=\left(\sqrt{2}\right)^2+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{2}+\sqrt{3}\right)^2\)

b,\(14+6\sqrt{5}=14+2.3.\sqrt{5}=3^2+2.3\sqrt{5}+\left(\sqrt{5}\right)^2=\left(3+\sqrt{5}\right)^2\)

Ấn đúng cho mình nha ( hãy kết bạn với tui)

(y^4 - 1/3 )^2

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

C₁: 9x² - 6x + 1 

= (3x)² - 3x.2.1 + 1²

= (3x - 1)²

C₂: 9x² - 6x + 1

= 9x² - 3x - 3x + 1 

= 3x(3x - 1) - (3x - 1) 

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

= (3x - 1)²

Bình phương tổng có CT: (a + b)² đấy

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. \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)

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

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