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Câu a:
(x + 2)(x -3) - (x -2)(x + 5) = 0
x^2 + 2x - 3x - 6 - x^2 - 5x + 2x + 10 = 0
(x^2 - x^2) + (2x - 3x - 5x + 2x) + (10 - 6) = 0
0 + (-x - 5x + 2x) + 4 = 0
-6x + 2x + 4 = 0
-4x + 4 = 0
4x = 4
x = 4 : 4
x = 1
Vậy x = 1
Câu b:
(2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
2x^2 - 8x + 3x - 12+ x^2 - 2x - 5x + 10 = 3x^2 - 12x - 5x + 20
(2x^2 + x^2) - (8x+5x-3x +2x) - (12 - 10) = 3x^2 - (12x + 5x) + 20
3x^2 - (13x - 3x + 2x) - 2 = 3x^2 - 17x + 20
3x^2 - (10x + 2x) - 2 = 3x^2 - 17x + 20
3x^2 - 12x - 2 = 3x^2 - 17x + 20
3x^2 - 12x - 2 - 3x^2 + 17x - 20 = 0
(3x^2 - 3x^2) + (-12x + 17x) - (2 + 20) = 0
0 + 5x - 22 = 0
5x = 22
x = 22/5
Vậy x = 22/5
a) 4( 18 - 5x ) - 12( 3x - 16 ) = 15( 2x - 16 ) - 6( x + 14 )
<=> 72 - 20x - 36x + 192 = 30x - 240 - 6x - 84
<=> -20x - 36x - 30x + 6x = -240 - 84 - 72 - 192
<=> -80x = -588
<=> x = -588/-80 = 147/20
b) ( x + 3 )( x + 2 ) - ( x - 2 )( x + 5 ) = 6
<=> x2 + 5x + 6 - ( x2 + 3x - 10 ) = 6
<=> x2 + 5x + 6 - x2 - 3x + 10 = 6
<=> 2x + 16 = 6
<=> 2x = -10
<=> x = -5
c) -x( x + 3 ) + 2 = ( 4x + 1 )( x - 1 ) + 2x
<=> -x2 - 3x + 2 = 4x2 - 3x - 1 + 2x
<=> -x2 - 3x - 4x2 + 3x - 2x = -1 - 2
<=> -5x2 - 2x = -3
<=> -5x2 - 2x + 3 = 0
<=> -( 5x2 + 2x - 3 ) = 0
<=> -( 5x2 + 5x - 3x - 3 ) = 0
<=> -[ 5x( x + 1 ) - 3( x + 1 ) ] = 0
<=> -( x + 1 )( 5x - 3 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{5}\end{cases}}\)
d) ( 2x + 3 )( x - 3 ) - ( x - 3 )( x + 1 ) = ( 2 - x )( 3x + 1 ) + 3
<=> 2x2 - 3x - 9 - ( x2 - 2x - 3 ) = -3x2 + 5x + 2 + 3
<=> 2x2 - 3x - 9 - x2 + 2x + 3 = -3x2 + 5x + 2 + 3
<=> 2x2 - 3x - x2 + 2x + 3x2 - 5x = 2 + 3 + 9 - 3
<=> 4x2 - 6x = 11
<=> 4x2 - 6x - 11 = 0
=> Vô nghiệm ( Lớp 8 chưa học nghiệm vô tỉ nên để vậy ) :))
vẫn làm được nha quỳnh !
\(4x^2-6x-11=0\)
\(< =>\left(4x^2-6x+\frac{9}{4}\right)-13\frac{1}{4}=0\)
\(< =>\left(2x-\frac{3}{2}\right)^2=\frac{53}{4}\)
\(< =>\orbr{\begin{cases}2x-\frac{3}{2}=\frac{\sqrt{53}}{2}\\2x-\frac{3}{2}=-\frac{\sqrt{53}}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}2x=\frac{3+\sqrt{53}}{2}\\2x=\frac{3-\sqrt{53}}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{3+\sqrt{53}}{4}\\x=\frac{3-\sqrt{53}}{4}\end{cases}}\)
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
a: =>(x-6)(x+2)=0
=>x=6 hoặc x=-2
b: \(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
=>10x=20
hay x=2
c: =>x3-1-x3+4x=5
=>4x=6
hay x=3/2
a/ \(\left(x+2\right)^2-9=0\)
<=> \(\left(x+2-3\right)\left(x+2+3\right)=0\)
<=> \(\left(x-1\right)\left(x+5\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
b/ \(x^2-2x+1=25\)
<=> \(\left(x-1\right)^2=25\)
<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
<=> \(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
a) ( x - 1 )( x2 + x + 1 ) + x( x + 2 )( 2 - x ) = 5
<=> x3 - 1 - x( x + 2 )( x - 2 ) = 5
<=> x3 - 1 - x( x2 - 4 ) = 5
<=> x3 - 1 - x3 + 4x = 5
<=> 4x - 1 = 5
<=> 4x = 6
<=> x = 6/4 = 3/2
b) 5x( x - 3 )2 - 5( x - 1 )3 + 15( x + 4 )( x - 4 ) = 5
<=> 5x( x2 - 6x + 9 ) - 5( x3 - 3x2 + 3x - 1 ) + 15( x2 - 16 ) = 5
<=> 5x3 - 30x2 + 45x - 5x3 + 15x2 - 15x + 5 + 15x2 - 240 = 5
<=> 30x - 235 = 5
<=> 30x = 240
<=> x = 8
a,\(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(< =>x^3-1+x\left(4-x^2\right)=5\)
\(< =>x^3-1+4x-x^3=5\)
\(< =>4x-1-5=0< =>4x-6=0< =>x=\frac{3}{2}\)
b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)
\(< =>5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)=5\)
\(< =>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)
\(< =>\left(5x^3-5x^3\right)+\left(15x^2+15x^2-30x^2\right)+\left(45x-15x\right)+5-240=5\)
\(< =>30x-240=5-5=0< =>x=\frac{24}{3}=8\)
a, \(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3-1+4x-x^3=5\Leftrightarrow-6+4x=0\Leftrightarrow x=\frac{3}{2}\)
b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)
\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)
\(\Leftrightarrow30x-240=5\Leftrightarrow x=8\)