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Gớm Tú ơi, làm gì mà Dis nhiều thế :)) Nghiếp khiếp vậy mày:))))
1/ \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow9x^2-6x-35=0\)
\(\Leftrightarrow\left(2x-1\right)^2-36=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+6\right)=0\)
2/ \(\left(3x+5\right)^2-4x^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(5x+5\right)=0\)
3/ \(25x^2-\left(4x-3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(9x-3\right)=0\)
1) ( 9x2 - 25 ) - ( 6x - 10 ) = 0
\(\Leftrightarrow\) [ ( 3x)2 - 52 ] - 2.( 3x + 5 ) = 0
\(\Leftrightarrow\)( 3x - 5 ).( 3x + 5 ) - 2.( 3x - 5 ) = 0
\(\Leftrightarrow\) ( 3x + 5 ).( 3x + 5 - 2 ) = 0
\(\Leftrightarrow\)( 3x + 5 ).( 3x + 3 ) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+5=0\\3x+3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-5\\3x=-3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{-5}{3}\\x=-1\end{cases}}\)
Vậy x = \(\frac{-5}{3}\) , x = -1
2) ( 3x + 5 )2 - 4x2 = 0
\(\Leftrightarrow\) ( 3x + 5 - 2x ).( 3x + 5 + 2x ) = 0
\(\Leftrightarrow\)( x + 5 ).( 5x + 5 ) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+5=0\\5x+5=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-5\\x=-1\end{cases}}\)
Vậy x = -5 , x = -1
3) 25x2 - ( 4x - 3 )2 = 0
\(\Leftrightarrow\)( 5x )2 - ( 4x - 3 )2 = 0
\(\Leftrightarrow\) ( 5x - 4x + 3 ).(5x + 4x - 3 ) = 0
\(\Leftrightarrow\)( x + 3 ).( 9x - 3 ) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\9x-3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\9x=3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
Vậy x = 3 , x = \(\frac{1}{3}\)
1) \(\left(3x+7\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(3x+7-2x+3\right)\left(3x+7+2x-3\right)=0\)
\(\Leftrightarrow\left(x+10\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+10=0\\5x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-10\\x=\frac{-4}{5}\end{cases}}\)
Vạy ...
phần 2 tương tự áp dụng \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\((4x-1)^2-(5-3x)^2=0\)
\(\Leftrightarrow(4x-1-5-3x)(4x+1+5-3x)=0\)
\(\Leftrightarrow(x-6)(x+6)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Vậy : ...
4x2 - 36x + 56
= 4(x2 - 9x + 14)
= 4(x2 - 2x - 7x + 14)
= 4[x(x - 2) - 7(x - 2)]
= 4(x - 2)(x - 7)
\(4x^2\)−36x+56=04x2−36x+56
⇒4(x2−9x+14)=0⇒4(x2−9x+14)
⇒4(x2−7x−2x+14)=0⇒4(x2−7x−2x+14)
⇒4x(x−2)−7(x−2)=0⇒4x(x−2)−7(x−2)
⇒4(x−7)(x−2)=0⇒4(x−7)(x−2)
⇒(x−7)(x−2)=0⇒(x−7)(x−2)
⇒[x−7=0x−2=0⇒[x−7=0x−2=0
⇒x=7;x=2⇒x=7;x=2.
4x - 4x2 - 1 + 81z2
= 81z2 - ( 4x2 - 4x + 1 )
= ( 9z )2 - ( 2x - 1 )2
= [ 9z - ( 2x - 1 ) ][ 9z + ( 2x - 1 ) ]
= ( 9z - 2x + 1 )( 9z + 2x - 1 )
\(4x-4x^2-1+81z^2\)
\(=-\left(4x^2-4x+1\right)+81z^2=81z^2-\left(2x-1\right)^2\)
\(=\left(9z-2z+1\right)\left(9z+2z-1\right)\)