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b)(x+3)\(^2\)-x\(^2\)=9
(x+3-x)(x+3+x)=9
3(2x+3)=9
2x+3=3
2x=0
x=0
c)\(x^2+4x+4+x^2-6x+9-2x^2+2=9\)
-2x+15=9
-2x=-6
x=3
a) x(x-2) - 5x + 10 = 0
x(x - 2) - 5(x - 2)=0
(x - 2)(x - 5)=0
=> x=2 ; x=5
a) x2 - 9 = 0
=> x2 = 0 + 9
=> x2 = 9
=> x2 = 32
=> x = 3
b) 64 - x2 = 0
=> - x2 = 0 - 64
=> - x2 = - 64
=> x2 = 64
=> x2 = 82
=> x = 8
b, ( x + 2)^2 - x^2 + 4 = 0
( x + 2)^2 - ( x^2 - 2^2) = 0
( x + 2)^2 - ( x - 2)( x +2) = 0
( x+ 2)( x + 2 - x + 2) = 0
4( x + 2) = 0
=> x + 2 = 0
=> x = - 2
a, ( 2x - 1)^2 -9 = 0
( 2x - 1)^2 - 3^2 = 0
( 2x- 1 - 3)( 2x - 1 + 3 ) = 0
( 2x - 4)( 2x + 2) = 0
=> 2x - 4 = 0 hoặc 2x+ 2 = 0
=> x = 2 hoặc x = -2
a) 4x2 - 9=0
(2x)2 - 32 = 0
=》(2x - 3)(2x+3) =0
=》 2x - 3 = 0 hoặc 2x +3 = 0
=》x = 1,5 hoặc x = - 1,5
b) (x + 1)2 - 16 = 0
=》( x + 1)2 - 42 = 0
=》( x - 3 )( x + 5 ) =0
=》 x - 3 = 0 hoặc x + 5 = 0
=》 x = 3 hoặc x = -5
c) ( x + 1)2 - (2x + 3)2 = 0
=》 ( x + 1 - 2x - 3)(x+1 +2x +3 ) =0
=》 ( -x - 2 )( 3x + 4 ) = 0
=》 -x -2 =0 hoặc 3x + 4 = 0
=》 x = -2 hoặc x = -4/3
d) 4(3x +2)2 - 9( x + 1 )2 =0
=》 [ 2(3x +2) ]2 - [3 (x + 1)] 2 = 0
=> ( 6x +4 )2 - ( 3x + 3)2 = 0
=》 ( 6x +4 -3x -3 )( 6x + 4 + 3x + 3 )=0
=》 (3x +1)(9x + 7 ) =0
=》 3x + 1 =0 hoặc 9x + 7 =0
=》 x = -1/3 hoặc x = -7/9
\(4.\left(x-1\right)^2-9.\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x-2\right)^2-\left(3x+6\right)^2=0\)
\(\Rightarrow\left(2x-2+3x+3\right).\left(2x-2-3x-6\right)=0\)
\(\Rightarrow\left(5x+1\right).\left(-x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+1=0\\-x-8=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\x=-8\end{cases}}\)
Vậy \(x=-\frac{1}{5}\text{ hoặc }x=-8\)
4.(x2-2.x.1+22) -9.(x2+2.x.2+22)=0 (4.x2-2.x.1.4+22.4)-(9.x2+2.x.2.9+22.4)=0 4.x2-8.x+16-9.x2-36.x -16=0 (4.x2-9.x2)-(8.x+36.x)+(16-16)=0 (-5.x2)-44.x+0=0 x.(-5x-44)=0-0 \(\hept{\begin{cases}x=0\\-5x-44=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\-5x=0+44\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\-5x=44\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-44}{5}\end{cases}}\)
a)x2 +5x =0
=>x.x +5x =0
=> x.(5+x) =0
=>Hoặc 5+x =0 =>x= -5
Hoặc x= 0
Vậy x=-5 ; x=0
1) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right).-2=0\)
\(\Leftrightarrow-4x+10=0\)
\(\Leftrightarrow-4x=-10\)
\(\Leftrightarrow x=\frac{5}{2}.\)
Vậy \(S=\left\{\frac{5}{2}\right\}\)
2)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right).\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\)
\(\Leftrightarrow x+3=0\)hoặc \(x=0\)hoặc \(x-2=0\)
\(\Leftrightarrow x=-3\)hoặc \(x=0\)hoặc \(x=2\)
Vậy \(S=\left\{-3;0;2\right\}\)
3
\(\Leftrightarrow\left(x-3\right).\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)