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a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)
\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)
\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)
b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)
c) \(x\left(x-3\right)+4x-12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
d) \(x^2-36=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
e) \(x^2+3x+1=2\)
\(\Leftrightarrow x^2+3x+1-2=0\)
\(\Leftrightarrow x^2+3x-1=0\)
\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)
Còn lại ........... Tự lm nất nha
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
a) ( x - 1 )2 - ( x - 1 )( x + 1 ) = 0
<=> x2 - 2x + 1 - ( x2 - 1 ) = 0
<=> x2 - 2x + 1 - x2 + 1 = 0
<=> 2 - 2x = 0
<=> 2x = 2
<=> x = 1
b) ( 2x - 1 )2 - ( 2x + 1 )2 = 0
<=> [ ( 2x - 1 ) - ( 2x + 1 ) ][ ( 2x - 1 ) + ( 2x + 1 ) ] = 0
<=> ( 2x - 1 - 2x - 1 )( 2x - 1 + 2x + 1 ) = 0
<=> -2.4x = 0
<=> -8x = 0
<=> x = 0
c) 25( x + 3 )2 + ( 1 - 5x )( 1 + 5x ) = 8
<=> 52( x + 3 )2 + 12 - 25x2 = 8
<=> [ 5( x + 3 ) ]2 + 1 - 25x2 = 8
<=> ( 5x + 15 )2 + 1 - 25x2 = 8
<=> 25x2 + 150x + 225 + 1 - 25x2 = 8
<=> 150x + 226 = 8
<=> 150x = -218
<=> x = -218/150 = -109/75
d) 9( x + 1 )2 - ( 3x - 2 )( 3x + 2 ) = 10
<=> 32( x + 1 )2 - ( 9x2 - 4 ) = 10
<=> [ 3( x + 1 ) ]2 - 9x2 + 4 = 10
<=> ( 3x + 3 )2 - 9x2 + 4 = 10
<=> 9x2 + 18x + 9 - 9x2 + 4 = 10
<=> 18x + 13 = 10
<=> 18x = -3
<=> x = -3/18 = -1/6
a) (x - 1)2 - (x - 1)(x + 1) = 0
=> (x - 1)2 - (x2 - 12) = 0
=> x2 - 2.x.1 + 12 - x2 + 1 = 0
=> x2 - 2x + 1 - x2 + 1 = 0
=> -2x + 1 + 1 = 0
=> -2x + 2 = 0
=> -2x = -2 => x = 1
b) (2x - 1)2 - (2x + 1)2 = 0
=> (2x - 1 - 2x + 1)(2x - 1 + 2x + 1) = 0
=> 0 = 0(đúng)
c) 25(x + 3)2 + (1 - 5x)(1 + 5x) = 8
=> 25(x2 + 2.x.3 + 32) + (12 - (5x)2) = 8
=> 25x2 + 150x + 225 + 1 - 25x2 = 8
=> 150x +225 + 1 = 8
=> 150x = -218
=> x = -109/75
d) 9(x + 1)2 - (3x - 2)(3x + 2) = 10
=> 9(x2 + 2x + 1) - [(3x)2 - 22 ] = 10
=> 9x2 + 18x + 9 - (9x2 - 4) = 10
=> 9x2 + 18x + 9 - 9x2 + 4 = 10
=> 18x + 9 + 4 = 10
=> 18x = -3
=> x = -1/6
Bài 1.
a) x2 + 7x +12 = 0
Ta có Δ = 72 - 4.12 = 1> 0 => \(\sqrt{\Delta}=\sqrt{1}=1\)
Phương trình có 2 nghiệm phân biệt:
x1 = \(\frac{-7+1}{2}=-3\)
x2= \(\frac{-7-1}{2}=-4\)
Bài 1
b) 2x2 + 5x - 3=0
Ta có: Δ = 52 + 4.2.3 = 49 > 0 => \(\sqrt{\Delta}=\sqrt{49}=7\)
Phương tình có 2 nghiệm phân biệt:
x1 = \(\frac{-5+7}{2.2}=\frac{1}{2}\)
x2 = \(\frac{-5-7}{2.2}-3\)
c) 3x2 +10x+7 = 0
Ta có: Δ = 102 - 4.3.7= 16> 0 => \(\sqrt{\Delta}=\sqrt{16}=4\)
Phương tình có 2 nghiệm phân biệt:
x1= \(\frac{-10+4}{2.3}=-1\)
x2= \(\frac{-10-4}{2.3}=-\frac{7}{3}\)
a, \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=4\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{-\frac{2}{3};4\right\}\)
b, \(5x\left(x-3\right)+3-x=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\frac{1}{5}\right\}\)
d, \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x^3+3^3\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3-;0;2\right\}\)
a) ( 4x - 1 ) (x - 3) - ( x - 3 ) ( 5x + 2 ) = 0
<=> (x - 3)(4x - 1 - 5x - 2) = 0
<=> (x - 3)(-x - 3) = 0
<=> x = 3 hoặc x = -3
b) ( x + 3 ) ( x - 5 ) + ( x + 3 ) ( 3x - 4) = 0
<=> (x + 3)(x - 5 + 3x - 4) = 0
<=> (x + 3)(4x - 9) = 0
<=> x = -3 hoặc x = 9/4
c) ( x + 6 ) ( 3x - 1 )+ x2 - 36 = 0
<=> 3x^2 + 17x - 6 + x^2 - 36 = 0
<=> 4x^2 + 17x - 42 = 0
<=> 4x^2 + 24x - 7x - 42 = 0
<=> 4x(x + 6) - 7(x + 6) = 0
<=> (4x - 7)(x + 6) = 0
<=> x = -6 hoặc x = 7/4
d) ( x + 4 ) ( 5x + 9 ) - x2 + 16 = 0
<=> 5x^2 + 29x + 36 - x^2 + 16 = 0
<=> 4x^2 + 29x + 52 = 0
<=> 4x^2 + 16x + 13x + 42 = 0
<=> 4x(x + 4) + 13(x + 4) = 0
<=> (4x + 13)(x + 4) = 0
<=> x = -13/4 và x = -4
a)
\(\Rightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b)
\(\Rightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\3x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{2}{3}\end{array}\right.\)
c)
\(\Rightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)
\(\Rightarrow\left(3x-2\right)^2.2=0\)
\(\Rightarrow3x-2=0\)
\(\Rightarrow x=\frac{2}{3}\)