TÌM X
(2x) :5=12
12x(3X)=36
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a, 36:(x–5) = 2 2
(x–5) = 9
x = 14
b, [3.(70–x)+5]:2 = 46
[3.(70–x)+5] = 92
70–x = 29
x = 41
c, 450:[41–(2x–5)] = 3 2 .5
41–(2x–5) = 10
2x–5 = 31
2x = 36
x = 18
d, 230+[ 2 4 +(x–5)] = 315. 2018 0
16+(x–5) = 315–230
x–5 = 85–16
x = 69+5
x = 74
e, 2 x + 2 x + 1 = 48
2 x .(2+1) = 48
2 x = 16 = 2 4
x = 4
f, 3 x + 2 + 3 x = 2430
3 x . 3 2 + 1 = 2430
3 x = 2430:10 = 243 = 3 5
x = 5
a) \(2x\left(x-5\right)-x\left(2x+3\right)=36\)
\(\Leftrightarrow\)\(2x^2-10x-2x^2-3x=36\)
\(\Leftrightarrow\)\(-13x=36\)
\(\Leftrightarrow\)\(x=2\)
Vậy..
b) \(\left(3x-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
\(\Leftrightarrow\)\(2x^2-x-1+4x^2-12x^3=\frac{5}{2}\)
\(\Leftrightarrow\)\(-12x^3+6x^2-x-\frac{7}{2}=0\)
\(\Leftrightarrow\)\(24x^3-12x^2+2x+7=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(12x^2-12x+7\right)=0\)
\(\Leftrightarrow\)\(2x+1=0\) ( do \(12x^2-12x+7=12\left(x-\frac{1}{2}\right)^2+4>0\))
\(\Leftrightarrow\)\(x=-\frac{1}{2}\)
Vậy...
b)3x2 - 3x(x - 2)=36 c) (3x2 - x + 1)(x - 1)+ x2(4 - 3x) = 5/2
3x2 - 3x2 + 6x= 36 3x3 - 3x2 - x2 + x + x - 1 + 4x2 - 3x3= 5/2
6x=36 =>x=36 : 6= 6 (3x3 - 3x3) + (-3x2 - x2 + 4x2) + (x + x) - 1= 5/2
2x - 1= 5/2 =>2x= 5/2 + 1= 7/2
x= 7/2 : 2 =7/4
a: =>3x+10-2x=0
hay x=-10
c: \(\Leftrightarrow3x^2-3x^2+6x=36\)
=>6x=36
hay x=6
Bài 1:
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^3-10x^2-6x\)
Bài 4:
a: =>3x+10-2x=0
=>x=-10
c: =>3x2-3x2+6x=36
=>6x=36
hay x=6
Bài 1:
\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)
Bài 4:
\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)
Bài 1:
\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)
(2x) : 5 = 12
=> 2x = 60
=> x = 30
12.(3x) = 36
=> 3x = 3
=> x = 1