\(b.\dfrac{3x+1}{-4}=\dfrac{5x-4}{3}\)
giúp mk với ạ TT
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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 \(x>\dfrac{8}{5}\)
\(\dfrac{3x-1}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{5-3x}{2}\)
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
\(\Leftrightarrow\dfrac{20\left(5x+6\right)-35\left(3x+1\right)}{140}=\dfrac{28\left(x+16\right)}{140}\)
\(\Leftrightarrow20\left(5x+6\right)-35\left(3x+1\right)=28\left(x+16\right)\)
\(\Leftrightarrow100x+120-105x-35=28x+448\)
\(\Leftrightarrow-33x=363\)
\(\Leftrightarrow x=-11\)
\(\Leftrightarrow9x+3=-20x+16\)
=>29x=13
hay x=13/29