Ta sẽ biến đổi vế phải:

\(a^3+b^3+3ab\left(a+b\right)\)

\(=a^3+3a^2.b+3ab^2+b^3+3a^{2b}+3ab^2\)

\(=a^3+b^3\)

Vậy VT = VP đẳng thức được chứng minh

Minecraft 1.15 

a) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Rightarrow72-20x-36x-84=30x-240-6x+84\)

\(\Rightarrow\left(72-84\right)-\left(20x+36x\right)=\left(30x-6x\right)-240+84\)

\(\Rightarrow-12-56=24x-56x\)

\(\Rightarrow-12+156=24x+56x\)

\(\Rightarrow144=80x\)

\(\Rightarrow x=144:80\)

\(\Rightarrow x=\frac{9}{5}\)

b) \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)

\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow15x+25-8x+12-5x-6x-36-1=0\)

\(\Rightarrow-4x=0\)

\(\Rightarrow-4.0\)

\(\Rightarrow x=0\)

à) bằng 1,8

b) bằng 0

Đặt \(A=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4\)

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

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

Áp dụng hàng đẳng thức \(\left(a^2-b^2+c^2\right)=a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2+2\left(ca\right)^2\):

\(A=-\left[\left(a^2-b^2+c^2\right)^2-4\left(ca\right)^2\right]\)

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

2222222222222a+257222222222222222222222222222222222222222222222222222222222222222222222222222222222222222a=?

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

 mik nghĩ bạn Hồ Khánh Châu giải đúng r

Ta có:\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2\)

\(\Rightarrow\left(a-b\right)^2=\left(a+b\right)^2-4ab=5^2-4.3=25-12=13\)

\(\Rightarrow a-b=\pm\sqrt{13}\)

Thank!!! Mk cũng ra kết quả giống bạn!!

\(\text{5.(3x + 5) - 4.(2x - 3) = 5x + 3.(2x - 12) + 1}\)

<=> 15x + 25 - 8x + 12 = 5x + 6x - 36 + 1

<=> 15x - 8x - 5x - 6x + 25 + 12 + 36 - 1 = 0

<=> - 4x + 72 = 0

<=> 4x = 72

<=> x = 18