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

\(A=\left[\left(3x+1\right)-\left(5x+5\right)\right]^2\)

\(A=\left(-2x-4\right)^2\)

A = (3x + 1)² - 2(3x + 1)(5x + 5) + (5x + 5)²

= [(3x + 1)-(5x + 5)]²

= (3x + 1 - 5x - 5)²

= [(-2x) - 4]²

B = (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

=> (3 - 1)B = (3 - 1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

=>2B = (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

= (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

= (3⁸ - 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

= (3¹⁶ - 1)3¹⁶ +1)(3³² + 1)

= (3³² - 1)(3³² + 1)

= 3⁶⁴ - 1

vì 2B = 3⁶⁴ - 1

=> B = \(\dfrac{3^{64}-1}{2}\)

C = a² + b² + c² + 2ab - 2ac - 2bc + a² + b² + c² - 2ab + 2ac - 2bc - 2( b² - 2bc + c²⁾

= 2a² + 2b² + 2c² - 4bc - 2b² + 4bc - 2c²

= 2a²

nhiều thế, đăng ít một thôi bạn

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)

e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)

\(\Rightarrow E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=\left(2a+2b\right)^2-2\left(a+b+c\right)\left(a+b-c\right)-2\left(a+b\right)^2\)

\(=4\left(a+b\right)^2-2\left(\left(a+b\right)^2-c^2\right)-2\left(a+b\right)^2\)

\(=4\left(a+b\right)^2-2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2=2c^2\)

g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)

\(=\left(2a+2b\right)^2-2\left(a+b+c+d\right)\left(a+b-c-d\right)+\left(2a-2b\right)^2-2\left(a+c-b-d\right)\left(a+d-b-c\right)\)

\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)

\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)

\(=2\left(\left(2a\right)^2-2\left(a+b\right)\left(a-b\right)\right)+2\left(\left(2c\right)^2-2\left(c+d\right)\left(c-d\right)\right)\)

\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)

\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)

bn đăng nhiều quá nên mk làm câu nào hay câu đó nha

mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .

Khó vl , dẹp mẹ điiii

a)     \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)

\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=4\)

b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)

\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)

\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)

\(\Leftrightarrow B=x^3-20x^2+18x+69\)

c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)

\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)

\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)

\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)

\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)

d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)

\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)

\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)

\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)

Chúc bạn học tốt !

Bài 1:

a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)

b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(18^8-\left(18^8-1\right)=1\)

\(c,100^2-99^2+98^2-97^2+...+2^2-1\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)

áp dụng công thức Gauss ta đc đáp án là:10100

d, mk khỏi ghi đề dài dòng:

\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:

\(A=\left(2-1\right)\left(2+1\right)\)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)

\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)

\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)

1c,

\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)

rút gọn biểu thức

2x

=2x(4x²-4x+1)-3x.(x²-9)-4x(x²+2x+1)

=8x³-8x²+2x-3x³-27x-4x³-8x²-4x

=8x³-16x²-7x³-29x

a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)

\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)

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

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

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

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

\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)

\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)

\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)

\(=\frac{a-b}{b+c}\)

Sửa lại: \(\frac{a-c}{b+c}\)

a) \(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)

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

\(=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac\)

\(=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc\)

\(=a^2\)