Tìm x biết \(|2x+5|+|2x+3|=8\)
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\(-4x^2+20-16x+4x^2=-3\)
\(\left(-4x^2+4x^2\right)-16x=-3-20\)
\(0-16x=-23\)
\(x=\frac{23}{16}\)
CHÚC BN HX TỐT :))
\(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)
\(\Leftrightarrow-4x^2+20x-16x+4x^2=-3\)
\(\Leftrightarrow4x=-3\Leftrightarrow x=-\frac{3}{4}\)


-4x(x-5)-2x(8-2x)=-3
có:-4x^2+20x-4x^2-16x=-3
(-4x^2+4x^2)+(20x-16x)=-3
4x=-3
suy ra:x=-3/4


-4x(x-5)-2x(8-2x)=3
-4x2-(-4)x5-2x(8-2x)=3
-4x2+20x-2x(8-2x)=3
-4x2+20x-2x8+2x2x=3
-4x2+20x-16x+4x2=3
(-4x2+4x2)+(20x-16x)=3
4x=3
x=3:4
x=3/4


\(a,5\left(3x+5\right)-4\left(2x-3\right)=5x+8\left(2x+12\right)+1\)
\(\Rightarrow5\left(3x+5\right)-4\left(2x-3\right)-5x-8\left(2x+12\right)-1=0\)
\(\Rightarrow15x+25-8x+12-5x-16x-96-1=0\)
\(\Rightarrow-14x-60=0\)
\(\Rightarrow-14x=60\) \(\Rightarrow x=-\frac{60}{14}=\frac{-30}{7}\)
\(b,\left(2x+3\right)\left(x-4\right)-\left(3x-5\right)\left(x-4\right)=\left(5-x\right)\left(x-2\right)\)
\(\Rightarrow2x^2+3x-8x-12-3x^2+5x+12x-20=5x-x^2-10+2x\)
\(\Rightarrow-x^2+12x-32=7x-x^2-10\)
\(\Rightarrow-x^2+12x-32-7x+x^2+10=0\)
\(\Rightarrow5x-22=0\)
\(\Rightarrow5x=22\Rightarrow x=\frac{22}{5}\)
a) 5(3x+5)-4(2x-3) = 5x+8(2x+12)+1
15x + 25 - 8x + 12 = 5x + 16x + 96 + 1
15x - 8x - 5x - 16x = 96 + 1 - 25 - 12
-14x = 60
x = \(\frac{60}{-14}\)
x = \(-\frac{30}{7}\)
b) (2x+3)(x-4)-(3x-5)(x-4) = (5-x).(x-2)
(x - 4)(2x + 3 - 3x +5) = 5x - 10 - x2 + 2x
(x - 4)[(2x - 3x) + (3 + 5)] = 5x - 10 - x2 + 2x
(x - 4)(-x + 8) = 5x - 10 - x2 + 2x
-x2 + 8x + 4x - 32 = 5x - 10 - x2 + 2x
(-x2 + x2) + (8x + 4x - 5x - 2x) = -10 + 32
5x = 22
x = \(\frac{22}{5}\)
Với \(x< -\frac{5}{2}\)ta có : \(-2x-5-2x-3=8\Leftrightarrow-4x-8=8\Leftrightarrow x=-4\left(tm\right)\)
Với \(-\frac{5}{2}\le x< -\frac{3}{2}\)ta có :
\(2x+5-2x-3=8\Leftrightarrow2=8\)( vô lí )
Với \(x\ge-\frac{3}{2}\)ta có : \(2x+5+2x+3=8\Leftrightarrow4x+8=8\Leftrightarrow x=0\left(tm\right)\)