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a) \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)
⇔ \(6x^2-13x+5-6x^2-11x+2=0\)
⇔ \(24x=7\)⇔\(x=\frac{7}{24}\)
b) \(\left(3x-2\right)\left(3x+2\right)-\left(3x-1\right)^2=-5\)
⇔ \(9x^2-4-9x^2+6x-1=5\)
⇔ \(6x=10\)⇔ \(x=\frac{5}{3}\)
c) \(x^2=-6x-8\)⇔\(x^2+6x+8=0\)⇔\(\left(x+2\right)\left(x+4\right)=0\)
⇔\(\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0
=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0
=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0
=> -24x + 7 = 0
=> - 24x = -7
=> x = 7/24
b, (3x - 2)(3x + 2) - (3x - 1)^2 = -5
=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5
=> 6x - 5 = -5
=> 6x = 0
=> x = 0
c, x^2 = -6x - 8
=> x^2 + 6x + 8 = 0
=> x^2 + 2.x.3 + 9 - 1 = 0
=> (x + 3)^2 = 1
=> x + 3 = 1 hoặc x + 3 = -1
=> x = -2 hoặc x = -4
Bài 1:
a)(4x-3)(3x+2)-(6x+1)(2x-5)+1
=12x2-x-6-12x2+28x+5+1
=27x
b)(3x+4)2+(4x-1)2+(2+5x)(2-5x)
=9x2+24x+16+16x2-8x+1+4-25x2
=16x+21
c)(2x+1)(4x2-2x+1)+(2-3x)(4+6x+9x2)-9
=8x3+1+8-27x3-9
=-19x3
Bài 2:
a)3x(x-4)-x(5+3x)=-34
=>3x2-12x-3x2-5x=-34
=>-17x=-34
=>x=2
Vậy x=2
b)(3x+1)2+(5x-2)2=34(x+2)(x-2)
=>9x2+6x+1+25x2-20x+4=34(x2-4)
=>34x2-14x+5-34x2+136=0
=>-14x+141=0
=>-14x=-141
=>x=\(\frac{141}{14}\)
Vậy x=\(\frac{141}{14}\)
c)x3+3x2+3x+28=0
=>x3-x2+7x+4x2-4x+28=0
=>x(x2-x+7)+4(x2-x+7)=0
=>(x+4)(x2-x+7)=0
\(\Rightarrow\left[\begin{array}{nghiempt}x+4=0\\x^2-x+7=0\left(2\right)\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-4\\\left(2\right)\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}>0\end{array}\right.\)
=>(2) vô nghiệm
Vậy x=-4
a) ( x - 5 )( 2x + 3 ) + 2x( 1 - x )
= 2x2 - 7x - 15 + 2x - 2x2
= -5x - 15
= -5( x + 3 )
b) ( 3x - 5 )2 - ( x + 5 )( 5 - x ) - 5/2( -2x )2
= 9x2 - 30x + 25 + ( x + 5 )( x - 5 ) - 5/2.4x2
= 9x2 - 30x + 25 + x2 - 25 - 10x2
= -30x
c) ( 3x + 2 )( 4 - 6x + 9x2 ) - 3x( 3x - 2 )2 + 12( -2/3 - 3x2 )
= ( 3x )3 + 23 - 3x( 9x2 - 12x + 4 ) - 8 - 36x2
= 27x3 + 8 - 27x3 + 36x2 - 12x - 8 - 36x2
= -12x
\(\text{a) }3x^2y^2:x^2=3y^2\)
\(\text{b) }\left(x^5+4x^3-6x^2\right):4x^2\\ =\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)
\(\text{c) }\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)
\(\text{d) }\left(3x^2-6x\right):\left(2-x\right)\\ =3x\left(x-2\right):\left(2-x\right)\\ =-3x\left(2-x\right):\left(2-x\right)\\ =-3x\)
\(\text{e) }\left(x^3+2x^2-2x-1\right):\left(x^2+3x+1\right)\\ =\left(x^3+3x^2-x^2+x-3x-1\right):\left(x^2+3x+1\right)\\ =\left[\left(x^3+3x^2+x\right)-\left(x^2+3x+1\right)\right]:\left(x^2+3x+1\right)\\ =\left[x\left(x^2+3x+1\right)-\left(x^2+3x-1\right)\right]:\left(x^2+3x+1\right)\\ =\left(x-1\right)\left(x^2+3x+1\right):\left(x^2+3x+1\right)\\ =x-1\)
a) 3x2y2 : x2 = 3y2
b)( x5 + 4x3 - 6x2 ) : 4x2
=\(\dfrac{1}{4}\)x3+ x - \(\dfrac{3}{2}\)
\(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1\right)=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-2\end{cases}}\)
\(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{-4}{5}\end{cases}}\)