\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}< \frac{7x^2-14x-5}{15}\)
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13 tháng 3 2020
<=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)
<=> \(\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}-\frac{7x^2-14x-5}{15}=0\)
<=> \(\frac{12x^2+12x+3-5x^2+10x-5-7x^2+14x+5}{15}=0\)
=> 36x + 3 = 0
<=> 36x = -3
<=> x = -1/12
Vậy S = { -1/12 }
A
11 tháng 8 2020
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{4x^2+4x+1}{5}-\frac{x^2-2x+1}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\Leftrightarrow x=-\frac{3}{36}=-\frac{1}{12}\)
11 tháng 3 2020
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
11 tháng 3 2020
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}
3 tháng 2 2019
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\frac{1}{12}\)
4 tháng 2 2019
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2-2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow\)12x2+12x+3-5x2+10x-5-7x2+14x+5=0
\(\Leftrightarrow\)36x+3=0
\(\Leftrightarrow\)36x=-3
\(\Leftrightarrow\)x=\(-\frac{1}{12}\)
Vậy pt có tập nghiêm S={\(-\frac{1}{12}\)}
8 tháng 1 2020
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}< \frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}< \frac{7x^2-14x-5}{15}\)
\(\Rightarrow3\left(2x+1\right)^2-5\left(x-1\right)^2< 7x^2-14x-5\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)< 7x^2-14x-5\)
\(\Leftrightarrow\left(12x^2+12x+3\right)-\left(5x^2-10x+5\right)< 7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5< 7x^2-14x-5\)
\(\Leftrightarrow7x^2+22x-2< 7x^2-14x-5\)
\(\Leftrightarrow7x^2+22x-2-7x^2+14x+5< 0\)
\(\Leftrightarrow36x+3< 0\)
\(\Leftrightarrow36x< -3\)
\(\Leftrightarrow x< -\frac{3}{36}\)
\(\Leftrightarrow x< -\frac{1}{12}\)
bn ơi kl phải là <=>x >\(-\frac{1}{12}\)