Tìm x :
x.(3.x-1)^10=(1-3.x)^20
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Bài 1:
a: x+1/2=5/6
nên x=5/6-1/2=1/3
b: x+1/4=3/4
nên x=3/4-1/4=2/4=1/2
c: x+3/10=1/2
nên x=1/2-3/10=5/10-3/10=1/5
d: x+1/4=3/8
nên x=3/8-1/4=3/8-2/8=1/8
Bài 1:
a. $(-20)+x=-30$
$x-20=-30$
$x=-30+20=-(30-20)=-10$
b.
$(-10)-x=-20$
$x=(-10)-(-20)=-10+20=20-10=10$
c. Đề sai. Bạn xem lại.
d.
$x+(-3)=-7$
$x=-7-(-3)=-7+3=-(7-3)=-4$
e.
$x-(-5)=-9$
$x=(-9)+(-5)=-14$
f.
$x(-11)=12$
$x=\frac{12}{-11}=\frac{-12}{11}$
h.
$2x-10=20$
$2x=20+10=30$
$x=30:2=15$
l.
$4x-8=-8$
$4x=-8+8=0$
$x=0:4=0$
k.
$-12-(-2)x=-8$
$(-2)x=-12-(-8)=-12+8=-(12-8)=-4$
$x=(-4):(-2)=2$
Bài 2:
a. $-20-(10-x)=-3$
$10-x=-20-(-3)=-20+3=-(20-3)=-17$
$x=10-(-17)=10+17=27$
b.
$14+(14-x)=-2$
$14-x=-2-14=-16$
$x=14-(-16)=14+16=30$
c.
$-15-(x-3)=-7$
$x-3=-15-(-7)=-15+7=-8$
x=-8+3=-5$
d.
$(x+4)+(-20)=-8$
$x+4=-8-(-20)=-8+20=12$
$x=12-4=8$
e.
$-2x-2=-4$
$-2x=-4+2=-2$
$x=(-2):(-2)=1$
f.
$-2x+4=-4$
$-2x=-4-4=-8$
$x=(-8):(-2)=4$
l.
$-12-(-2)x=-2-4=-6$
$(-2)x=-12-(-6)=-12+6=-6$
$x=(-6):(-2)=3$
a) (125 - x) + 25 = 98
125 - x = 98 - 25
125 - x = 73
x = 125 - 73
x = 52
b) 100 : (x + 20) = 68 : 34
100 : (x + 20) = 2
x + 20 = 100 : 2
x + 20 = 50
x = 50 - 20
x = 30
c) (3x + 12) . 5 = 25 . 4 - 10
(3x + 12) . 5 = 100 - 10
(3x + 12) . 5 = 90
3x + 12 = 90 : 5
3x + 12 = 18
3x = 18 - 12
3x = 6
x = 6 : 2
x = 3
d) 210 : (2x - 3) - 20 = 10
210 : (2x - 3) = 10 + 20
210 : (2x - 3) = 30
2x - 3 = 210 : 30
2x - 3 = 70
2x = 70 + 3
2x = 73
x = 73/2
a, ( 125 - x ) + 25 = 98
⇒ 125 - x = 73
⇒ x = 52.
Vậy..
b, 100 : ( x + 20 ) = 68 : 34
⇒ 100 : ( x + 20 ) = 2
⇒ x + 20 = 50
⇒ x = 30
Vậy...
c, ( 3 . x + 12 ) . 5 = 25 . 4 - 10
⇒ ( 3 . x + 12 ) . 5 = 90
⇒ 3 .x + 12 = 18
⇒ 3x = 6
⇒ x = 2.
Vậy...
d, 210 : ( 2 .x - 3 ) - 20 = 10
⇒ ( 2 .x - 3 ) - 20 = 21
⇒ 2x - 3 = 41
⇒ 2x = 44
⇒ x = 22.
Vậy..
a: -12/x=3/5
=>x=-12x5:3=-20
b: (x-1)+(x-2)+(x-3)+...+(x-20)=10
=>20x-210=10
=>20x=220
hay x=11
\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)
\(\Rightarrow x+5=20\)
\(\Rightarrow x=20-5\)
\(\Rightarrow x=15\)
Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333