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a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
a)<=>|3x+1|-|2x-5|+|x-12|=2x+3
=>x=15/2
b)<=>|x-1|+2|3x+2|-|5x-3|=-(|5x-3|-2|3x+2|-|x-1|)
=>-(|5x-3|-2|3x+2|-|x-1|)=|3x+7|
=>x=4/3 hoặc x=2/3
Bài 1:
a) -6x + 3(7 + 2x)
= -6x + 21 + 6x
= (-6x + 6x) + 21
= 21
b) 15y - 5(6x + 3y)
= 15y - 30 - 15y
= (15y - 15y) - 30
= -30
c) x(2x + 1) - x2(x + 2) + (x3 - x + 3)
= 2x2 + x - x3 - 2x2 + x3 - x + 3
= (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
= 3
d) x(5x - 4)3x2(x - 1) ??? :V
Bài 2:
a) 3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = -10
=> x = -10
b) 3x2 - 3x(-2 + x) = 36
<=> 3x2 + 2x - 3x2 = 36
<=> 6x = 36
<=> x = 6
=> x = 5
c) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
=> x = -2
a, 4-|3x+5|=7
|3x+5|=4-7
|3x+5|=-3
ta có
vì trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 mà -3<0
suy ra x thuộc rỗng
b, 3/4-|1/2:x-2/3|=1/2
|1/2:x-2/3|=3/4-1/2
|1/2:x-2/3|=1/4
th1
1/2:x-2/3=1/4
1/2:x=1/4+2/3
1/2:x=11/12
x=1/2:11/12
x=6/11
th2
1/2:x-2/3=-1/2
1/2:x=-1/2+2/3
1/2:x=1/6
x=1/2:1/6
x=3
Vậy x={3;6/11}
c,|x:-2/3+1|+3/2=3/2
|x:-2/3+1|=3/2-3/2
|x:-2/3+1|=0
x:-2/3+1=0
x:-2/3=0-1
x:-2/3=-1
x=-1.-2/3
x=2/3
Vậy x=2/3
a)
\(3,5x-2=\dfrac{1}{3}x+\dfrac{2}{7}\\ \Leftrightarrow\dfrac{7}{2}x-\dfrac{1}{3}x=\dfrac{2}{7}+2\\ \Leftrightarrow x\left(\dfrac{7}{2}-\dfrac{1}{3}\right)=\dfrac{16}{7}\\ \Leftrightarrow\dfrac{19}{6}\cdot x=\dfrac{16}{7}\\ \Leftrightarrow x=\dfrac{16}{7}:\dfrac{19}{6}\\ \Leftrightarrow x=\dfrac{96}{133}\)
b)
\(2\left[\dfrac{x-1}{40}-3\left(x-1\right)\right]=2\\ \Leftrightarrow\dfrac{x-1}{40}-3\left(x-1\right)=1\\ \Leftrightarrow\dfrac{x-1-120\left(x-1\right)}{40}=1\\ \Leftrightarrow x-1-120x+120=40\\ \Leftrightarrow-119x+119=40\\ \Leftrightarrow-119x=40-119=-79\\ \Leftrightarrow x=\dfrac{79}{119}\)
a, \(3,5x-2=\dfrac{1}{3}x+\dfrac{2}{7}\Leftrightarrow\dfrac{19}{6}x=\dfrac{2}{7}+2=\dfrac{16}{7}\Leftrightarrow x=\dfrac{16}{7}:\dfrac{19}{6}=\dfrac{96}{133}\)
b, \(2\left[x-\dfrac{1}{40}-3\left(x-1\right)\right]=2\)
\(\Leftrightarrow x-\dfrac{1}{40}-3x+3=1\Leftrightarrow-2x=1+\dfrac{1}{40}-3=-\dfrac{79}{40}\Leftrightarrow x=\dfrac{79}{80}\)