a) \(\left(2+\sqrt{3}\right)^2=4+4\sqrt{3}+3=7+4\sqrt{3}\)

b) \(\left(\sqrt{7}-3\right)^2=7-6\sqrt{7}+9=16-6\sqrt{7}\)

c) \(\left(5+\sqrt{2}\right)^2=25+10\sqrt{2}+2=27+10\sqrt{2}\)

d) \(\left(\sqrt{11}-5\right)^2=11-10\sqrt{11}+25=36-10\sqrt{11}\)

e) \(\left(1+\sqrt{5}\right)^2+\left(3-\sqrt{5}\right)^2=1+2\sqrt{5}+5+9-6\sqrt{5}+5=20-4\sqrt{5}\)

f) \(\left(2+\sqrt{7}\right)^2+\left(\sqrt{7}-3\right)^2=4+4\sqrt{7}+7+7-6\sqrt{7}+9=27-2\sqrt{7}\)

Đặt A=(2+x)⁵(3x-1)⁷

^{khai triển ta có:A=(\(_{k=0}^5\Sigma C_5^k2^{5-k}x^k\)}).(\(^7_{i=0}\Sigma C_7^i\left(3x\right)^i\))

=\(\left(_{k=0}^5\Sigma\right)\left(_{i=0}^7\Sigma\right)\left(C_5^kC^i_7\right)\left(x^k.\left(3x\right)^i\right)\)

=số hạng\(\left(C_5^kC^i_7\right)\left(x^k.\left(3x\right)^i\right)\)chứa x⁵ tại k+i=5

có k\(\in\){0,1,2,...5},i\(\in\){0,1,2,...7}

=>(k,i)={(0,5);(1,4);(2,3);(3,2);(4,1);(5,0)}

=>Hệ số của x⁵ là:\(\left(C_5^0C^5_7\right)3^5\)+\(\left(C_5^1C^4_7\right)\left(3^4\right)\)+\(\left(C_5^2C^3_7\right)\left(3^3\right)\)+\(\left(C_5^3C^2_7\right)\left(3^2\right)\)+

\(\left(C_5^4C^1_7\right)\left(3^1\right)\)+\(\left(C_5^5C^0_7\right)3^0\)=30724

Hok tốt!!!

b) ta có (1+x-x²)⁸=(1+(x-x²))⁸

=\(^8_{k=0}\Sigma.C_8^k\left(x-x^2\right)^k\)=\(^8_{k=0}\Sigma.C_8^k\left(x-1\right)^kx^k\)=\(^8_{k=0}\Sigma.C_8^k\left(x-1\right)^kx^k\)

\(a,=x^2+4x+4\\ b,=x^3+3x^2+3x+1\\ c,=\left(x-3\right)\left(x+3\right)\)

a,\(\left(x+2\right)^2=x^2+2.x.2+2^2=x^2+4x+4\)

b, \(\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\)

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

a,

b, 

c,

Bài 3:

a) \(4x^2+4x+1=\left(2x+1\right)^2\)

b) \(9x^2-12x+4=\left(3x-2\right)^2\)

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

Phần d bài 3 mk chưa lm dc

Bài 3: 

a: \(4x^2+4x+1=\left(2x+1\right)^2\)

b: \(9x^2-12x+4=\left(3x-2\right)^2\)

c: \(\dfrac{1}{4}a^2b^4+ab^2+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)

d: 

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

\(\left(x^3-1\right)\left(x^3+1\right)\)

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

\(=x^3-1\)

Z thôi T nha

a: \(\left(x^2+2y\right)^3\)

\(=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot2y+3\cdot x^2\cdot\left(2y\right)^2+\left(2y\right)^3\)

\(=x^6+6x^4y+12x^2y^2+8y^3\)

b: \(\left(\dfrac{1}{2}x-1\right)^3\)

\(=\left(\dfrac{1}{2}x\right)^3-3\cdot\left(\dfrac{1}{2}x\right)^2\cdot1+3\cdot\dfrac{1}{2}x\cdot1^2-1^3\)

\(=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2+\dfrac{3}{2}x-1\)

\(a,=x^3+3x^2+3x+1\\ b,=8x^3+36x^2+54x+27\\ c,=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\\ d,=x^6-6x^4+12x^2-8\\ e,=8x^3-36x^2y+54xy^2-27y^3\)

a) (x - 1/2x²y)²

= x² - 2x . 1/2 x²y + (1/2x²y)²

= x² - x³y + 1/4 x⁴y²

b) (2xy² - 1)(1 + 2xy²)

= (2xy²)² - 1²

= 4x²y⁴ - 1

c) (x - y + 2)²

= (x - y)² + 2(x - y).2 + 2²

= x² - 2xy + y² + 4x - 4y + 4

= x² + y² - 2xy + 4x - 4y + 4

d) (x + 1/2)(1/2 - x)

= (1/2)² - x²

= 1/4 - x²

e) (x² - 1/3)²

= (x²)² - 2x².1/3 + (1/3)²

= x⁴ - 2/3 x² + 1/9

a)(x + 1)² – (x – 2)²

= (x+1-x+2)(x+1+x-2)

= 3(2x-1)

b)(x – 3)(x – 1) – (2x – 1)²

= x²-4x+3-4x²+4x-1

= -(3x²-2)

c)(x + 3)² - 2(x + 3)(1 – x) + (1 - x)²

= [(x+3)-(1-x)]²

=(2x-2)²=4(x-1)²