\(\left(\dfrac{1}{2}a-\dfrac{2}{3}b\right)^3=\left(\dfrac{1}{2}a\right)^3-3.\left(\dfrac{1}{2}a\right)^2.\dfrac{2}{3}b+3.\dfrac{1}{2}a.\left(\dfrac{2}{3}b\right)^2-\left(\dfrac{2}{3}b\right)^3\)

                        \(=\dfrac{1}{8}a^3-\dfrac{1}{2}a^2b+\dfrac{2}{3}ab^2-\dfrac{8}{27}b^3\)

\(\left(\dfrac{1}{2}a-\dfrac{2}{3}b\right)^2\)

\(=\left(\dfrac{1}{2}a\right)^2-2.\left(\dfrac{1}{2}a\right).\left(\dfrac{2}{3}b\right)+\left(\dfrac{2}{3}b\right)^2\)

\(=\dfrac{1}{4}a^2-\dfrac{2}{3}ab+\dfrac{4}{9}b^2\)

cứu với

Ta có

\(a^{2020}+b^{2020}=a^{2021}+b^{2021}\)

\(\Leftrightarrow a^{2021}-a^{2020}=b^{2020}-b^{2021}\)

\(\Leftrightarrow a^{2020}\left(a-1\right)=b^{2020}\left(1-b\right)\)

\(\Leftrightarrow\dfrac{a-1}{1-b}=\dfrac{b^{2020}}{a^{2020}}=\left(\dfrac{b}{a}\right)^{2020}\) (1) 

Ta có

\(a^{2021}+b^{2021}=a^{2022}+b^{2022}\)

\(\Leftrightarrow\dfrac{a-1}{1-b}=\left(\dfrac{b}{a}\right)^{2021}\) (2)

Từ (1) và (2) \(\Rightarrow\left(\dfrac{b}{a}\right)^{2020}=\left(\dfrac{b}{a}\right)^{2021}\)

\(\Rightarrow\dfrac{b}{a}=1\Rightarrow a=b\)

\(\Rightarrow2.a^{2020}=2.a^{2021}\Leftrightarrow a^{2020}=a^{2021}\Rightarrow a=b=1\)

\(\Rightarrow S=a^{2021}+b^{2021}=1+1=2\)

Are you missed something? \(C\) is a first degree polynomial so we can't find a minimum value of \(C\)

I think you mean \(C=12x^2+6x-10\) 

Am I correct?

Ta có: \(x+y+z+t=0\)

\(\Rightarrow t=-\left(x+y+z\right)\)

\(VT=x^3+y^3+z^3+t^3\)

\(=x^3+y^3+z^3-\left(x+y+z\right)^3\)

\(=x^3+y^3+z^3-\left[x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\right]\)

\(=-3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(VP=3\left[xy+z\left(x+y+z\right)\right]\left(z-x-y-z\right)\)

\(=3\left(xy+yz+zx+z^2\right)\left(-x-y\right)\)

\(=-3\left(y+z\right)\left(x+z\right)\left(x+y\right)\)

\(\Rightarrow VT=VP\)

x+y+z+t=0

<=> t= - (x+y+z)

<=> t³ = - (x+y+z)³

<=> t³ = - x³- y³- z³ - 3(x+y)(y+z)(z+x)

=>  x³+y³+z³+t³ = x³+y³+z³ + (- x³- y³- z³ - 3(x+y)(y+z)(z+x))

=> 3(y+z)(xt-yz) = -3(x+y)(y+z)(z+x)

=>xt-yz= (x+y)(z+x)

=> x²+xy+xz+xt=0

=> x(x+y+z+t)=0 luôn đúng => đpcm 

\(3x\left(2x-4\right)-6x\left(x+5\right)=x-1\)

\(\Leftrightarrow6x^2-12x-6x^2-30x-x+1=0\)

\(\Leftrightarrow-43x+1=0\)

\(\Leftrightarrow x=\dfrac{1}{43}\)

Bài tiếp:

\(a)-10x+5x\left(x^2-7x+2x^2\right)-x^2\left(5x-8\right)+27x^2\)

\(=-10x+5x^3-35x^2+10x^3-5x^3+8x^2+27x^2\)

\(=10x^3-10x\)

Vậy biểu thức phụ thuộc vào biến \(x\).

\(b)4x\left(3x+5\right)-6x\left(2x-3\right)-38x+5\)

\(=12x^2+20x-12x^2+18x-38x+5\)

\(=5\)

Vậy biểu thức không phụ thuộc vào biến \(x\).

<=> 6x² - 12x - 6x² - 30x = x-1

<=> -32x = x-1 

<=> -33x=-1

<=> x=\(\dfrac{1}{33}\)

a, x² + 2x + 1 = (x+1)(x+1)

b,   9x²  + 6xy + y² = (3x + y)(3x+y)

c,     x⁴ -  2x² + 1 =  (x² - 1)(x² -1)

d, 9x² + 16 - 24 x = (3x - 4)(3x -4)

a, x2 + 2x + 1 = (x+1)(x+1)

b,   9x2  + 6xy + y2 = (3x + y)(3x+y)

c,     x4 -  2x2 + 1 =  (x2 - 1)(x2 -1)

d, 9x2 + 16 - 24 x = (3x - 4)(3x -4)