24x(x+36=1080
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24 x (x + 36) = 1080
x + 36 = 1080 : 24
x + 36 = 45
x = 45 - 36
x = 9
24 x (x + 36) = 1080
x + 36 = 1080 : 24
x + 36 = 45
x = 45 - 36
x = 9
24 x ( x + 36 ) = 1080
( x + 36 ) = 1080 : 24
( x + 36 ) = 45
x = 45 - 36
x = 9
hok tốt
\(\dfrac{25}{24}\times\dfrac{27}{5}\times2\times\dfrac{34}{9}\times\dfrac{36}{17}\)
\(=(\dfrac{25}{24}\times\dfrac{54}{5})\times\left(\dfrac{34}{9}\times\dfrac{36}{17}\right)\)
\(=\dfrac{45}{4}\times2\times4\)
\(=90\)
\(-36+24x-x^2\)
\(=-x^2+24x-36\)
\(=-x^2+24x-144+108\)
\(=-\left(x^2-24x+144\right)+108\)
\(=-\left(x-12\right)^2+108\le108\)
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