\(G=-3x^2+2x-1=-3\left(x-\frac{1}{3}\right)^2-\frac{2}{3}\le-\frac{2}{3}\)

Vậy Min G = -2/3 <=> x = 1/3

\(E=2x^2+8x+15\)

\(=\left(2x^2+8x+8\right)+7\)

\(=2\left(x^2+2.2.x+2^2\right)+7\)

\(=2\left(x+2\right)^2+7\)

\(2\left(x+2\right)^2\ge0\)

\(\Rightarrow2\left(x+2\right)^2+7\ge7\)

\(\Leftrightarrow Min_E=7\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy ...

\(4x^2-4xy+y-25a^2+10a-136\)

\(\text{Phân tích thành nhân tử}\)

\(-\left(4xy-y-4x^2+25a^2-10a+136\right)\)

k nhé !

Ta có :

\(x^8+x^7+1\)

\(=\left(x^8+x^7+x^6\right)-x^6+1\)

\(=x^6\left(x^2+x+1\right)-\left[\left(x^3\right)^2-1^2\right]\)

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

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

\(=\left(x^2+x+1\right)\left[x^6-\left(x^3+1\right)\left(x-1\right)\right]\)

\(=\left(x^2+x+1\right)\left[x^6-\left(x^4-x^3+x-1\right)\right]\)

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

đề kiểu gì vậy bạn

\(x^2+y^2+z^2+2x-4y+6z=-14\)

\(x^2+y^2+z^2+2x-4y+6z+14=0\)

\(x^2+2x+1+y^2-4y+4+z^2+6z+9=0\)

\(\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)

vậy x + y + z = -1 + 2 + (-3) = -2

mk nghĩ k có n nào thỏa mãn

ồ, hay quá

(1/4)ⁿ = (1/2)²ⁿ = 1/2²ⁿ = n = 1/2

cám ơn bn nhiu

a) x²-2x-y²+2y

=(x²-y²)-(2x-2y)

=(x-y)(x+y)-2(x-y)

=(x-y)(x+y-2)

d) x²-25+y²+2xy

=(x²+y²+2xy)-5²

=(x+y)²-5²

=(x+y+5)(x+y-5)