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a ) 2x ( x - 5 ) - x ( 3 + 2x ) = 26
2x2 - 10x - 3x - 2x2 = 26
- 13x = 26
x = 26 : ( -13 )
x = -2
b) 49x2 - 81 = 0
( 7x - 9 )( 7x + 9 ) = 0
Th1 :
7x - 9 = 0
7x = 9
x = \(\frac{9}{7}\)
Th2
7x + 9 = 0
7x = -9
x = \(-\frac{9}{7}\)
Vay x = \(\frac{9}{7}\) hoac x = \(-\frac{9}{7}\)
\(x\left(x^2-49\right)=0\\ x\left(x-7\right)\left(x+7\right)=0\\ \left\{{}\begin{matrix}x=0\\x-7=0\\x+7=0\end{matrix}\right.\\ \left\{{}\begin{matrix}x=0\\x=7\\x=\left(-7\right)\end{matrix}\right.\)
A = ( x - 5 ) ( x2 + 26 + 5x - 1 )
A = ( x - 5 ) ( x2 + 5x + 25 )
A = ( x - 5 )3
\(\text{a. A = (x - 5)(x^2 + 26) + (x - 5)(5x - 1)}\\ A=\left(x-5\right)\left(x^2+5x+25\right)\\ A=x^3-5^3\\ A=x^3-125\)
3x2-49x+198=0
=>3x2-27x-22x+198=0
=>3x(x-9)-22(x-9)=0
=>\(\orbr{\begin{cases}3x-22=0\\x-9=0\end{cases}=>\orbr{\begin{cases}x=\frac{22}{3}\\x=9\end{cases}}}\)
nhớ k mk nha
\(\dfrac{1}{2}x^3-49x=0\)
\(\dfrac{1}{2}x^2.x-49x=0\)
\(x.\left(\dfrac{1}{2}x^2-49\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x^2-49=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x^2=49\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=98\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=7\sqrt{2}\end{matrix}\right.\)
Vậy \(x\in\left\{0,7\sqrt{2}\right\}\)
\(x^3+8-x\left(x^2-9\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow9x+8=26\Leftrightarrow9x=18\)
\(\Leftrightarrow x=2\)