Bài 2:

Ta có: \(a+b+c=2\)

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

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=4\)

\(\Leftrightarrow2\left(ab+bc+ca\right)=2\)

\(\Rightarrow ab+bc+ca=1\)

Thay vào ta được: \(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\)

Tương tự CM được: \(b^2+1=\left(b+a\right)\left(b+c\right)\) và \(c^2+1=\left(c+a\right)\left(c+b\right)\)

=> \(M=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

=> đpcm

2. Hợp chất : Fe₃O_4  ^; H₂SO₄ ; KHCO₃ ;...

1. Đơn chất : O ; Fe ; H ; C ; Cu 

3. Hỗn hợp : ... Tự tìm :)) 

1, 23.2+12.1+16.3+10(1.2+16)=286 (dvC)

2, 12n+2n+z+x+(12.1+16.1+16.1+1.1)x

=12n+2n+z+x+43x = 14n+z+44x (dvC)

2, phan 3

H2 + O2, H2 & N2, Na + O2, Fe + O, CO2 & O2

Bài 1 :

a) Theo đề bài ta có : p + e + n = 40 ( vì p = e)

                             => 2p + n = 40 (1)

Mặt khác ta có :  p + e - n = 12 

                         => 2p - n = 12 => n = 2p - 12 (2)

Thay (2) vào (1) ta được : 2p + 2p - 12 = 40

=> 4p-  12 = 40

=> 4p = 52

=> p = 13

Thay vào (2) ta lại có :

n = 2.13 - 12 = 14

Vậy p = e = 13 , n = 14

=> X = p + n = 13 + 14 = 27 => X là nguyên tố nhôm ( kí hiệu : Al)

Bài 2 : Nguyên tử khối của O là M_O = 16

Gọi x là nguyên tử khối cần tìm cùa nguyên tử X

Theo đề bài ta có : x = 2.M_O = 2.16 = 32

=> x là lưu huỳnh ( S)

cảm ơn bạn

Bài 1 : 

Ta có : \(VP=\left(a+b\right)^4=\left(a+b\right)\left(a+b\right)^3\)

\(=\left(a+b\right)\left(a^3+3a^2b+3ab^2+b^3\right)=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

=> HĐT ko đc CM 

Bài 2 : 

a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)

\(=x^3+2x^2+4x-2x^2-4x-8-x+1+7=x^3-x=x\left(x^2-1\right)\)

Sửa đề : b, \(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x+1\right)\)

\(=8\left(x^3-1\right)-8x^3+1=8x^3-8-8x^3+1=-7\)

Xin phép chủ nahf cho mjnh sửa đề:D

\(\left(a+b\right)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

a,\(\left(a+b\right)^4\)

\(=\left[\left(a+b\right)^2\right]^2\)

\(=\left(a^2+2ab+b^2\right)^2\)

\(=\left[\left(a^2+2ab\right)+b^2\right]^2\)

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

\(=a^4+4a^3b+4a^2b^2+2a^2b^2+4ab^3+b^4\)

\(=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

Bài 2:

a,\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)

\(=\left(x^3-8\right)-\left(x-1\right)+7\)

b,\(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x-1\right)\)

\(=8\left(x^3-1\right)-\left(8x^3-1\right)\)

\(=8x^3-8-8x^3+1\)

\(=-7\)

(x + 10)(x² - x + 1)  - x(x - 3)(x - 3)(x + 3) = -27

=> x(x² - x + 1) + 10(x² - x + 1) - x(x² - 9)(x - 3) = -27

=> x³ - x² + x + 10x² - 10x + 10 - x[x²(x - 3) - 9(x - 3)] = -27

=> x³ - x² + x + 10x² - 10x + 10 - x(x³ - 3x² - 9x + 27) = -27

=> x³ - x² + x + 10x² - 10x + 10 - x⁴ + 3x³ + 9x² - 27x = -27

=> (x³ + 3x³) + (-x² + 10x² + 9x²) + (x - 10x - 27x) + 10 - x⁴ = -27

=> 4x³ + 18x² - 36x + 10 - x⁴ = -27

=> -(x⁴ - 4x³ - 18x² + 36x - 10) = -27

=> vô nghiệm

\(\left(x+10\right)\left(x^2-x+1\right)-x\left(x-3\right)\left(x-3\right)\left(x+3\right)=-27\)

\(\Leftrightarrow x^3-x^2+x+10x^2-10x+10-x\left(x-3\right)^2\left(x+3\right)=-27\)

\(\Leftrightarrow x^3-11x^2-9x+10-x\left[\left(x^2-6x+9\right)\left(x+3\right)\right]=-27\)

\(\Leftrightarrow x^3-11x^2-9x+10-x\left(x^3+3x^2-6x^2-18x+9x+27\right)=-27\)

\(\Leftrightarrow x^3-11x^2-9x+10-x\left(x^3-3x^2-9x+27\right)=-27\)

\(\Leftrightarrow x^3-11x^2-9x+10-x^4+3x^3+9x^2-27x=-27\)

Vô nghiệm :) chắc rồi 

Ta có: 

\(C=1-x-\frac{x^2}{3}-\frac{x^3}{27}\)

\(C=1-\left(-27\right)-\frac{\left(-27\right)^2}{3}-\frac{\left(-27\right)^3}{27}\) tại x = -27

\(C=1+27-\frac{3^6}{3}+27^2\)

\(C=1+27-243+729\)

\(C=514\)

1) \(x^3+2x-3\)

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

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

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

2) \(x^3-6x+4\)

\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(2x-4\right)\)

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

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

3) \(x^3-2x^2+1\)

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

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

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

4) \(x^3+5x^2-12\)

\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)

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

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

Ta có:

\(-25x^6-y^8+10x^3y^4\)

\(=-\left[\left(5x^3\right)^2-10x^3y^4+\left(y^4\right)^2\right]\)

\(=-\left(5x^3-y^4\right)^2\)