a) \(x^2-4x=0\)

\(x\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

b) \(4x^2-9=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}\)

c) \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\left(2x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

d) \(x\left(2x+9\right)-4x-18=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)

\(\left(2x-1-x-2\right)\left(2x-1+x+2\right)=0\)

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

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

\(x^2-4x=0\)

\(x.\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\Leftrightarrow x=4\end{cases}}\)

\(4x^2-9=0\)

\(2^2x^2-9=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)

\(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\cdot\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

\(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-\left(4x+18\right)=0\)

\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)

\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x-1=x+2\\2x-1=-x+2\end{cases}\Rightarrow\orbr{\begin{cases}2x=3+x\\2x=-x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-x=3\\2x+x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}}}\)

\(\)

Sửa đề : \(x^2-6x+9-y^2\)

\(=\left(x^2-6x+9\right)-y^2\)

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

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

\(A=x^3+3xy^2-9+y\left(3x^2+y^2\right)\)

\(=x^3+3x^2y+3xy^2+y^3-9\)

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

\(=\left(1,95+0,05\right)^3-9=2^3-9=-1\)

\(B=x^2+\frac{1}{2}x+\frac{1}{16}\)

\(=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)

\(=\left(x+\frac{1}{4}\right)^2\)

\(=\left(9,75+0,25\right)^2=10^2=100\)

  x²+4x+3

=x²+x+3x+3

=x(x+1)+3(x+1)

=(x+1)(x+3)

k cho mk nhé

Đặt \(A=\frac{3}{2}x^2+x+1\)

\(6A=9x^2+6x+6\)

\(6A=\left(9x^2+6x+1\right)+5\)

\(6A=\left(3x+1\right)^2+5\ge5\)

\(A=\frac{\left(3x+1\right)^2+5}{6}\ge\frac{5}{6}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(3x+1\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{-1}{3}\)

Vậy GTNN của \(A\) là \(\frac{5}{6}\) khi \(x=\frac{-1}{3}\)

Chúc bạn học tốt ~ 

2x³-18x=0

->2x(x²-9)=0

->2x(x-3)(x+3)=0

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

->x=0

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

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