Tìm x biết
-4x (x-5)-2x (8-2x) =-3
nhanh dung minh tich cho nhiu
tìm x biết -4x(x-5)-2x(8-2x)=-3
\(-4x^2+20-16x+4x^2=-3\)
\(\left(-4x^2+4x^2\right)-16x=-3-20\)
\(0-16x=-23\)
\(x=\frac{23}{16}\)
CHÚC BN HX TỐT :))
\(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)
\(\Leftrightarrow-4x^2+20x-16x+4x^2=-3\)
\(\Leftrightarrow4x=-3\Leftrightarrow x=-\frac{3}{4}\)
Tìm x biết: -4x(x-5)-2x(8-2x)=-3
Tìm x biết -4x(x-5) - 2x(8 - 2x) = -3
-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
Tìm x biết:
-4x(x-5) - 2x(8-2x) = -3
-4x(x-5)-2x(8-2x)=-3
-4x^2+20x-16x+4x^2=-3
4x=-3
x=-3/4
<=>(-4)x(x-5)-2x(8-2x)=4x
=>4x=-3
=>x=\(-\frac{3}{4}\)
-4x(x-5)-2x(8-2x)=-3
=>-4x-(-20)-16-4x=-3
=>-4x+20-4x=-3+16
=>-4x-4x=13-20
=>-8x=-7
=>x=(-7):(-8)
=>x=0,875
Tìm x biết:
-4x(x-5)-2x(8-2x)=-3
-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
ý, ghi nhầm -3 thành 3 rồi, kết
tìm x biết:
a)(2x+2)(2x-2)-4x(x+5)=8
b)(4x+5)(4x-5)-8x(2x-7)=11
c)(1/2x-3)(1/2x+3)-1/4x(x+5)=11/2
d)(3x+2)(3x-2)-4x(x+2-5x2=18
\(a.x=-0,6\)
\(c.x=-11,6\)
Pt nhju ak!!!
tìm x,y,z biết x/-3=y/-7=z/12 và -2x-4y+5z=146
bài 2
tim x, y z biet 4x/-5=2y/7=-3z/8 và x+3y-2z=-273
giai nhanh dung minh tick
3. Tìm x, biết: a) 2x + 8 ≤ 0 b) 4x-7 ≥ 2x -5 c) (2x-8)(15-3x)>0 d) (10-2x)(8+2x)≤0
a) 2x+8≤ 0
⇔2x≤-8
⇔x≤-4
b) 4x-7 ≥ 2x -5
⇔2x-12 ≥ 0
⇔2x≥12
⇔x≥6
c) (2x-8)(15-3x)>0
TH1: 2x-8>0 ⇒x>4
15-3x>0⇒x<5
TH2: 2x-8<0 ⇒x<4
15-3x<0⇒x>5 (vô lí)
vậy 4<x<5
Tìm x biết :
a) (3x³ + x² – 13x + 5) : (x² + 2x – 1) = 10
b) (x⁴ – 2x² – 8) : (x – 2) = 0
c) (x²-4x) : (x²-8x+16) = 0
\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)
b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
hay x=-2