\(6\left(x^2-5x\right)-9\left(x-5\right)\)

\(=6x\left(x-5\right)-9\left(x-5\right)\)

\(=\left(x-5\right)\left(6x-9\right)\)

p/s : ms học lớp 7 đọc qua đoán thế sai thì thôi =))

6(x^2-5x)-9(x-5)

= 6x(x-5)-9(x-5)

=(x-5)-(6x-9)

=(x-5)-3(2x-3)

\(\left(x+1\right)\left(x+3\right)\left(x+4\right)\left(x+6\right)-7\)

\(=\left(x+1\right)\left(x+6\right)\left(x+3\right)\left(x+4\right)-7\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+12\right)-7\)

Đặt \(x^2+7x+9=t\)

\(=\left(t-3\right)\left(t+3\right)-7\)

\(=t^2-9-7=t^2-16=\left(t-4\right)\left(t+4\right)\)

\(=\left(x^2+7x+9-4\right)\left(x^2+7x+9+4\right)\)

\(=\left(x^2+7x+5\right)\left(x^2+7x+13\right)\)

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

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

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

Đặt \(x^2-3x+1=t\)

\(=\left(t-1\right)\left(t+1\right)-3\)

\(=t^2-1-3=t^2-4\)

\(=\left(t-2\right)\left(t+2\right)\)

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

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

Bạn nào biết chỉ mk với. Mk sẽ đãi hậu hĩnh luôn.

\(a,\frac{3^7.5^4}{25^2}=\frac{3^7.5^4}{\left(5^2\right)^2}=\frac{3^7.5^4}{5^4}=3^7\)

Ta có: \(\left(x-y\right)^2\ge0\left(\forall x;y\right)\)

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

           \(y^2\ge0\left(\forall y\right)\)

\(\Rightarrow A=\left(x-y\right)^2+x^2+y^2-1\ge-1\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=0\)

Vậy A_min = -1 <=> x = y = 0

=a³(b-c)-b³(a-c)+c³(a-c-b+c)

=a³(b-c)-b³(a-c)+c³(a-c)-c³(b-c)

=(a³-c³)(b-c)-(b³-c³)(a-c)

=(a-c)(a²+ac+c²)(b-c)-(b-c)(b²+bc+c²)(a-c)

=(b-c)(a-c)(a²+ac+c²-b²-bc-c²)

=(b-c)(a-c)[(a-b)(a+b)+c(a-b)]

=(a-b)(b-c)(a-c)(a+b+c)

Ta có:\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-c-b+c\right)\)

\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-c\right)-c^3\left(b-c\right)\)

\(=\left(a^3-c^3\right)\left(b-c\right)-\left(b^3-c^3\right)\left(a-c\right)\)

\(=\left(a-c\right)\left(a^2+ac+c^2\right)\left(b-c\right)-\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-c\right)\)

\(=\left(b-c\right)\left(a-c\right)\left(a^2+ac+c^2-b^2-bc-c^2\right)\)

\(=\left(b-c\right)\left(a-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)