The solution of 0.4x-9 - X-5 of 2 = 0.03 of 0.03 + 0.02x

The solution of 0.4x-9 - X-5 of 2 = 0.03 of 0.03 + 0.02x

Original question = {20 (0.4x-9) - 5 (X-5)} / 10 = {0.03 + 0.02x} / 0.03
0.3+0.2x=0.6(0.4x-9)-0.15(x-5)
0.19x=6.45
x=33.947
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0.5 (0.4x + 0.9) - 0.03 (0.03 + 0.02x) = 2 (X-5)

0.4x/0.5+0.9-0.03/0.03+0.02x=x/2-5
0.8x+0.9-1+0.02x=0.5x-5
0.8x+0.02x-0.5x=-0.9+1-5
0.32x=-4.9
x=15.3125

0.5x + 2-x of 0.03 = 0.3 (0.5 + 2) - 10 of 0.2 and 11 of 12
solve equations

　　0.5x+2)/0.03-x=0.3(0.5x+2)/0.2-(10+11/12)　　100*(0.5x+2)/100*0.03-x=10*0.3(0.5x+2)/10*0.2-(10+11/12)　　100*(0.5x+2)/3-x=3(0.5x+2)/2-(10+11/12)　　(50x+200)/3-x=3*1/2(x+4)/2-(10+11/12)　　(50x+20...

(1) 6 / 5x = 1 / 3 (2) x ／ 8 / 3 = 4 / 3 (3) 4 / 3-x = 0.65 '(4) x ／ 6 = 11 / 12
(5) 1 / 2 + x = 3 / 4
(6) 1 out of 5 x = 3 out of 10
(7) X-7 / 12 = 5 / 6

(1) 5x of 6 = 1 of 3 x = 1 / 6 of 3 x = 1 of 3 * 6x of 5 = 2 of 5 (2) x △ 3 of 8 = 3 of 4 x = 3 of 4 * 3x of 8 = 9 of 32 (3) 3 of 4-x = 0.65 'x = 3 of 4-0.65x = 0.75-0.65x = 0.1X = 1 of 10 (4) x △ 6 = 11 of 12 x = 11 of 12 * 1 of 6

5X/1*2+5X/2*3+…… 5X/2004*2005

5x/1*2+5x/2*3+…… 5x/2004*2005
=5x(1/1*2+1/2*3+…… 1/2004*2005)
=5x[(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+…… +(1/2003-1/2004)+(1/2004-1/2005)]
=5x(1-1/2005)
=2004x/401

Solution equation (0.5x-3) / 0.04-x = [0.4 (0.5x-3)] / 0.5

（0.5x-3）/0.04-x=[0.4(0.5x-3)]/0.5
Multiply both sides of the equation by 0.02 to get
0.5（0.5x-3）-0.02x=0.04[0.4(0.5x-3)]
0.25x-1.5-0.02x=0.008x-0.048
0.222x=1.452
x=242/37

First simplify, in the evaluation: (1) given x ^ 2-5x = 14, find the value of (x-1) (2x-1) - (x + 1) ^ 2 + 1

x^2-5x-14=0
（x+2)(x-7)=0 x1=-2,x2=7
（x-1）（2x-1）-（x+1）^2+1
=2x^2-3x+1-x^2-2x-1+1
=x^2-5x+1
X1 = - 2, the original formula = 4-20 + 1 = - 15
X2 = 7, original formula = 49-35 + 1 = 15

Solving inequality system: 4x-10 ＜ 05x + 2 ＞ 3x11-2x ≥ 1 + 3x

If we solve inequality 1, we get x ≤ 2.5, if we solve inequality 2, we get x ＞ - 1, if we solve inequality 3, we get x ≤ 2, so the solution set of this inequality system is - 1 < x ≤ 2

2X ^ 2-5x + 3x-5 = 0 (x ^ 2 is the square of x)

2x^2-5x+3x-5=0
2x^2-2x-5=0
△＝4＋4×2×5＝44
x=（2±√44）／4＝（1±√11）／2

When m is a value, the polynomial X & # 178; + MXY + 2Y & # 178; - 5x-13y-24 can be decomposed into the product of two first-order factors?

Solution ∵ x ^ 2-5x-24 = (X-8) (x + 3)
2y^2-13y-24=(y-8)(2y+3)
Double cross multiplication
x 2y 3
x y -8
∴m=1•1+2•1=3