h) x/y = 9/10 ⇒  y/10 = x/9

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

y/10 = x/9 = (y - x)/(10 - 9) = 120/1 = 120

*) x/9 = 120 ⇒ x = 120.9 = 1080

*) y/10 = 120 ⇒ y = 120.10 = 1200

Vậy x = 1080; y = 1200

k) x/y = 3/4

⇒ x/3 = y/4

⇒ 5y/20 = 3x/9

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

5y/20 = 3x/9 = (5y - 3x)/(20 - 9) = 33/11 = 3

*) 3x/9 = 3 ⇒ x = 3.9:3 = 9

*) 5y/20 = 3 ⇒ y = 3.20:5 = 12

Vậy x = 9; y = 12

\(\Leftrightarrow\dfrac{3x-1}{4}-\dfrac{3x-6}{8}-\dfrac{5-3x}{2}>1\)

\(\Leftrightarrow\dfrac{\left(3x-1\right).2-\left(3x-6\right)-\left(5-3x\right).4}{8}>1\)

\(\Leftrightarrow\dfrac{6x-2-3x+6-20+12x}{8}>1\)

<=> 15x - 16 > 8

<=> 15x > 24

<=> x > 8/5

Ta có: \(\dfrac{3x-1}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{5-3x}{2}\)

\(\Leftrightarrow2\left(3x-1\right)-3\left(x-2\right)-8>4\left(5-3x\right)\)

\(\Leftrightarrow6x-2-3x+6-8>20-12x\)

\(\Leftrightarrow3x-4-20+12x>0\)

\(\Leftrightarrow15x>24\)

hay \(x>\dfrac{8}{5}\)

= (6x-2-3x+6-8)/8=(20-12x)/8

= 12x-3x+6x=20+8-6

<=>21x=22

<=>x=22/21

(Ủa đúng không ha)

\(\dfrac{3x-1}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{5-3x}{2}\)

MTC : 8

\(\Rightarrow\dfrac{2\left(3x-1\right)}{8}-\dfrac{3\left(x-2\right)}{8}-\dfrac{8}{8}>\dfrac{4\left(5-3x\right)}{8}\)

Suy ra : 2(3x - 1) - 3(x - 2) - 8 > 4(5 - 3x)

          \(\Leftrightarrow\) 6x - 2 - 3x + 6 - 8 > 20 - 12x

          \(\Leftrightarrow\) 6x - 3x + 12x > 20 + 2 - 6 + 8

          \(\Leftrightarrow\) 15x > 24

          \(\Leftrightarrow\) x > \(\dfrac{24}{15}=\dfrac{8}{5}\)

 Vay x >\(\dfrac{8}{5}\)

Chuc ban hoc tot

a: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)

\(\Leftrightarrow20x^2-12x+15x-5< 20x^2+30x-100\)

=>3x-5<=30x-100

=>30x-100>3x-5

=>27x>95

hay x>95/27

b: \(\Leftrightarrow4\left(5x-2\right)-6\left(2x^2-x\right)< 4x\left(1-3x\right)-15x\)

\(\Leftrightarrow20x-8-12x^2+6x< 4x-12x^2-15x\)

=>26x-8<-11x

=>37x<8

hay x<8/37

a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)

⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\) 

⇔ 2x + 2 - 5 - 2x = 12 -16x

⇔ 16x = 15 

⇔ x = 15/16

b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)

⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)

⇔ 8 - 6x - 4 + x = 5x + 10

⇔ 10x = -6

⇔ x = -6/10

Câu 1:

x + 1/4 - 5 + 2x/8 = 3 - 4x/2

<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8

<=> 2x + 2 - 5 - 2x = 12 - 16x

<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16

Câu 2:

4 - 3x/5 - 4 - x/10 = x + 2/2

<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10

<=> 8 - 6x - 4 + x = 5x + 10

<=> 4 - 5x = 5x + 10

<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5

\(\Leftrightarrow\dfrac{20\left(5x+6\right)-35\left(3x+1\right)}{140}=\dfrac{28\left(x+16\right)}{140}\)

\(\Leftrightarrow20\left(5x+6\right)-35\left(3x+1\right)=28\left(x+16\right)\)

\(\Leftrightarrow100x+120-105x-35=28x+448\)

\(\Leftrightarrow-33x=363\)

\(\Leftrightarrow x=-11\)

a. có vấn đề

b.

\(\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-30\)

\(\Leftrightarrow-22x+5x< -30-5\)

\(\Leftrightarrow-17x< -35\)

\(\Leftrightarrow x>\dfrac{35}{17}\)