24 x ( x + 36 ) = 1080

( x + 36 ) = 1080 : 24

( x + 36 ) = 45

x = 45 - 36 

x = 9

 hok tốt

x + 36 = 1080 : 24

x + 36 = 45

x         = 45 - 36

x         = 9

24×(x+36)=1080

        x+36=1080÷24

         x+36=45

             x=45-36

                x=9

\(\dfrac{25}{24}\times\dfrac{27}{5}\times2\times\dfrac{34}{9}\times\dfrac{36}{17}\)

\(=(\dfrac{25}{24}\times\dfrac{54}{5})\times\left(\dfrac{34}{9}\times\dfrac{36}{17}\right)\)

\(=\dfrac{45}{4}\times2\times4\)

\(=90\)

\(-36+24x-x^2\)

\(=-x^2+24x-36\)

\(=-x^2+24x-144+108\)

\(=-\left(x^2-24x+144\right)+108\)

\(=-\left(x-12\right)^2+108\le108\)

Đẳng thức xảy ra khi \(x=12\)

Đkxđ : \(x\ne3;-3\)

Ta có :

\(\frac{4x^2-24x+36}{x^2-9}\)

\(=\frac{4\left(x^2-6x+9\right)}{x^2-3^2}\)

\(=\frac{4\left(x^2-2.3.x+3^2\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{4\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{4\left(x-3\right)}{x+3}\)

\(4x^2-24x+36=\left(x-3\right)^3\)

\(\Leftrightarrow4x^2-24x+36=x^3-9x^2+27x-27\)

\(\Leftrightarrow-x^3+13x^2-51x+63=0\)

\(\Leftrightarrow\left(-x^3+10x^2-21x\right)+\left(3x^2-30x+63\right)=0\)

\(\Leftrightarrow-x\left(x^2-10x+21\right)+3\left(x^2-10x+21\right)=0\)

\(\Leftrightarrow\left(x^2-10x+21\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left(x^2-3x-7x+21\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left[x\left(x-3\right)-7\left(x-3\right)\right]\left(3-x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-7\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left(3-x\right)^2\left(7-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(3-x\right)^2=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)

Vậy...

\(4x^2-24x+36=\left(x-3\right)^3\)\(\Leftrightarrow4\left(x^2-6x+9\right)=\left(x-3\right)^3\)

\(\Leftrightarrow4\left(x-3\right)^2=\left(x-3\right)^3\)\(\Leftrightarrow4\left(x-3\right)^2-\left(x-3\right)^3=0\)

\(\Leftrightarrow\left(x-3\right)^2\left[4-\left(x-3\right)\right]=0\)\(\Leftrightarrow\left(x-3\right)^2\left(4-x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2\left(7-x\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)^2=0\\7-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-3=0\\x=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}\)

Vậy \(x=3\)hoặc \(x=7\)

Bài 4:

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

\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)

\(x^2-12x+36=\left(x-6\right)^2\)

 \(24x-4\left(2x-\frac{3}{4}\right)-4\left(3+\frac{2x}{2}\right)=36-3\left(x-\frac{3}{2}\right)-3\left(3-\frac{2x}{3}\right)\)

Đề như này đúng không bạn