\(a.=x\)

\(b.=y^3\)

\(c.=3xy\)

\(d.=-\frac{5}{2}a\)

\(e.=3yz\)

\(f.=-3xy\)

làm cả bước giải ra đc ko ạ?

Bạn viết rõ đề đc ko, nếu x thì viết x , còn nhân thì viết . nha

anh bn à ,chỉ có lm mới có ăn :))))))

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\)\(\left(x^3+3^3\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2-2.x\right)=0\)

\(\Leftrightarrow\)\(x\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=0\\x+3=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=0\\x=-3\\x=2\end{cases}}\)

 x3+27+(x+3)(x−9)=0x3+27+(x+3)(x-9)=0

⇔(x+3)(x2−3x+9)+(x+3)(x−9)=0⇔(x+3)(x2-3x+9)+(x+3)(x-9)=0

⇔(x+3)(x2−3x+9+x−9)=0⇔(x+3)(x2-3x+9+x-9)=0

⇔(x+3)(x2−2x)=0⇔(x+3)(x2-2x)=0

⇔x(x+3)(x−2)=0⇔x(x+3)(x-2)=0

⇒⇒ ⎡⎢⎣x=0x+3=0x−2=0[x=0x+3=0x−2=0

⇒⇒⎡⎢⎣x=0x=−3x=2

a) \(x^2-2x+1=25\)

\(\Rightarrow x^2-2x+1-25=0\)

\(\Rightarrow x^2-2x.1+1^2-5^2=0\)

\(\Rightarrow\left(x-1\right)^2-5^2=0\)

\(\Rightarrow\left(x-1-5\right)\left(x-1+5\right)=0\)

\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=\left(-4\right)\end{cases}}\)

b) \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)30\)

\(\Rightarrow\left(5x\right)^2+2.5x.1+1^2-\left(25x^2-9\right)-30=0\)

\(\Rightarrow25x^2+10x+1-25x^2+9-30=0\)

\(\Rightarrow\left(25x^2-25x^2\right)+10x+\left(1+9-30\right)=0\)

\(\Rightarrow10x-20=0\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)