tìm x biết
15/12x + 3/7 = 6/5x - 1/2
tìm X nà :
2(x+3)=3(1-x)-2
-15^12x+3^7=6^5x-1^2
Ta có 2(x+3)=3(1-x)-2
suy ra : 2x + 6 = 3 -3x -2
2x +6 = 1-3x
2x +6 -1 +3x=0
5x+5 =0
5(x+1)=0
Vậy x bằng -1
bí rùi, nếu chỉ có 1 cái x thì mình giải được
Tìm x biết:
-15/12x+3/7=6/5x -1/2
Tìm x:
a) 4x(3x-7)-6(2x^2-5x+1)=12
b) (5x+3)(4x-1)+(10x-7)(-2x+3)=27
c) (8x-5)(3x+2)-(12x+7)(2x-1)=17
d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190
a) 4x(3x-7)-6(2x2-5x+1)=12
=>4x.3x-4x.7-6.2x2-6.(-5x)-6.1=12
=>12x2-28x-12x2+30x-6=12
=>2x-6 =12
=>2x =12+6
=>2x =18
=>x =18:2
=>x =6
b)(5x+3)(4x-1)+(10x-7)(-2x+3)=27
=>5x.4x-5x.1+3.4x+3.(-1)+10x.(-2x)+10x.3-7.-(2x)-7.3=27
=>20x2-5x+12x-3-20x2+30x+14x-21=27
=>39x-36 =27
=>39x =27+36
=>39x =63
=>x =63:39
=>x =21/13
c) (8x-5)(3x+2)-(12x+7)(2x-1)=17
=>8x.3x+8x.2-5.3x-5.2-12x.2x-12x.(-1)+7.2x+7.(-1)=17
=>24x2+16x-15x-10-24x2+12x+14x-7=17
=>27x-17 =17
=>27x =17+17
=>27x =34
=>x =34:27
=>x =34/27
d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190
=>30x2-5x+63x-9 - 30x2-2x-45x-3=-190
=>11x-12 =-190
=>11x =-190+12
=>11x =-178
=>x = -178:11
=>x =-178/11
x.(1/4+1/5)-(1/7+1/8)=0
(5x-1)(2x-1/3)=0
-15/12x+3/7=6/5x-1/2
5x+5x+2=650
Tìm x
a) 6x(5x + 3) + 3x(1 – 10x) = 7 b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
c) (x + 1)(x + 2)(x + 5) – x2(x + 8) = 27
d) 5x(12x + 7) – 3x(20x – 5) = - 100
e) 0,6x(x – 0,5) – 0,3x(2x + 1,3) = 0,138
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
d) 5x(12x + 7) – 3x(20x – 5) = - 100
⇒60x2+35x-60x2+15=-100
⇒35x+15=-100
⇒35x=-100-15
⇒35x=-115
⇒x=\(\dfrac{-115}{35}\)
⇒x=\(\dfrac{-23}{7}\)
A.5x^2y^3-25x^3y^4+10x^3y^3
B.12x^2y-18xy^2-30y^2
C.5(x-y)-y(x-y)
D.y(x-z)+7(z-x)
E.27x^2(y-1)-9x^3(1-y)
F.36-12x+x^2
G.x^2+2xy+y^2-xz-yz
H.x^4+64
I.27x^2(y-1)-9x^3(1-y)
K.36-12x+x^2
M.-4x^2+4x-1
N.x^2+5x+6
P.x^2-x-6
Q.x^4-5x^2+4
-15/12X+3/7=6/5x-1/2
bài 1: tính
a) -6x+3(7+2x)
b)15y-5(6x+3y)
c) x(2x+1)-x2(x+2)+(x3-x+3)
d)4(x-6)-x2(2+3x)+x(5x-4)3x2(x-1)
bài 2: tìm x
a)3x+2(5-x)=0
b)3x2-3x(-2+x)=36
c)5x(12x+7)-3x(20x-5)=-100
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
1, Chứng tỏ rằng các đa thức sau ko phụ thuộc vào biến:
a,x(2x+1)-x2(x+2)+(x3-x+3)
b,4(x-6)-x2(2+3x)+x(5x-4)+3x2(x-1)
2,Tìm x
a,5x(12x+7)-3x(20x-5)=-100
b,0,6x(x-0,5)-0,3x(2x+1,3)=0,138
1)
a) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\)
\(=2x^2+x-x^3-2x^2+x^3-x+3=3\)
=>đpcm
b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2=-24\)
=>đpcm
2,
a) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)
\(\Leftrightarrow50x=-100\)
\(\Leftrightarrow x=-2\)
b) \(0,6x\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)
\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)
\(\Leftrightarrow-0,69x=0,138\)
\(\Leftrightarrow x=-0,2\)
Câu 1:
a)\(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^2-x+3\right)\)
\(=2x^2+x-x^3-2x^2+x^2-x+3\)
\(=x^3+3\)(ko thể CM)
b)\(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)
\(=-24\)(đpcm)
1 a ) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\)
\(=2x^2+x-x^3-2x^2+x^3-x+3\)
\(=3\)
=> đpcm
b ) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)
\(=-24\)
=> đpcm
2 ) Tìm x :
a ) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x=100\)
\(\Leftrightarrow50x=100\)
\(\Leftrightarrow x=2\)
b ) \(0,6c\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)
\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)
\(\Leftrightarrow-0,69x=0,138\)
\(\Leftrightarrow-0,2\)