2005 + 2003 = 4008

 - Chúc bạn học tốt- 

= 4008

\(x.\left(x+1\right)+3.\left(x+1\right)=0\)

\(\left(x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)

\(3x\left(12x-4\right)-2x\left(18x+3\right)=36\)

\(36x^2-12x-36x^2-6x=36\)

\(-18x=36\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

Tham khảo nhé~

a. 3x²+3xy-x-y

= (3x²+3xy)-(x-y)

= 3x(x+y)-(x+y) (sau dấu trừ đổi dấu)

=(x+y)(3x-1)

(x+1)²+2x(x-2)=3(x+4)(x+1)

<= >(x²+2x+1)+2x²-4x=(3x+12)(x+1)

<=>x²+2x+1+2x²-4x=3x²+3x+12x+12

<=>x²+2x+2x²-4x-3x²-3x-12x=12-1

<=>-17x=11

<=>x=\(\frac{-11}{17}\)

4x²-1+x(2x-1)=0

<=>4x²-1+2x²-x=0

<=> 6x²-x-1=0

<=>(6x²-x)-1=0

<=>x (6x-1) =1

<=>\(6x=2\)

<=>x=\(\frac{2}{6}=\frac{1}{3}\)

^-^

       \(-x^4-x^3-2x^2+x-3\)

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

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

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

2. Đặt c + d = x

Ta có: \(a+b+c+d=0\Rightarrow a+b+x=0\Rightarrow a^3+b^3+c^3+d^3=3abx\)

\(\Rightarrow a^3+b^3+c^3+d^3+3cd\left(c+d\right)=3ab\left(c+d\right)\)

\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(ab-cd\right)\left(c+d\right)\)

Câu 4:

      \(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}+a^{1008}\)

\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}-2a^{1008}b^{1008}-2b^{1008}c^{1008}-2c^{1008}a^{1008}=0\)

\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2=0\)

\(\Rightarrow a^{1008}=b^{1008},b^{1008}=c^{1008},c^{1008}=a^{1008}\)

\(\Rightarrow a=b,b=c,c=a\) (vì a,b,c > 0 nên \(a\ne-b,b\ne-c,c\ne-a\) )

\(\Rightarrow a-b=0,b-c=0,a-c=0\)

Thay vào A ta tính được A = 0

Ta có:

x2+y2+z2=12;x+y+z=6⇒3(x2+y2+z2)−(x+y+z)2=0⇔3(x2+y2+z2)−(x2+y2+z2+2xy+2xz+2yz)=0⇔2x2+2y2+2z2−2xy−2xz−2yz=0⇔(x−y)2+(y−z)2+(z−x)2=0(1)x2+y2+z2=12;x+y+z=6⇒3(x2+y2+z2)−(x+y+z)2=0⇔3(x2+y2+z2)−(x2+y2+z2+2xy+2xz+2yz)=0⇔2x2+2y2+2z2−2xy−2xz−2yz=0⇔(x−y)2+(y−z)2+(z−x)2=0(1)

Mà (x−y)2,(y−z)2,(z−x)2≥0,∀x,y,z(x−y)2,(y−z)2,(z−x)2≥0,∀x,y,z

⇒(x−y)2+(y−z)2+(z−x)2≥0,∀x,y,z⇒(x−y)2+(y−z)2+(z−x)2≥0,∀x,y,z

→(1)→(1) đúng chỉ khi dấu bằng xảy ra

(x−y)2=(y−z)2=(z−x)2=0⇔x−y=y−z=z−x=0⇔x=y=z(x−y)2=(y−z)2=(z−x)2=0⇔x−y=y−z=z−x=0⇔x=y=z

Mà x+y+z=6x+y+z=6⇒x=y=z=2⇒x=y=z=2\, suy ra A=10