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Đặt 3154=a; 6517=b
Theo đề, ta có:
\(B=2\dfrac{1}{a}\cdot\dfrac{3}{b}-\dfrac{3}{a}\cdot\dfrac{b-1}{b}+\dfrac{6}{\dfrac{1}{2}a}-\dfrac{6}{a\cdot b}\)
\(=\dfrac{2a+1}{a}\cdot\dfrac{3}{b}-\dfrac{3b-3}{ab}+\dfrac{12}{a}-\dfrac{6}{a\cdot b}\)
\(=\dfrac{6a+3-3b+3-6}{ab}+\dfrac{12b}{ab}\)
\(=\dfrac{6a-3b+12b}{ab}=\dfrac{6a+9b}{ab}=\dfrac{3\left(2a+3b\right)}{ab}\)
\(=\dfrac{3\left(2\cdot3154+3\cdot6517\right)}{3154\cdot6517}\)
a)\(\frac{\frac{x-3}{5}+1}{4}=\frac{\frac{2x}{3}-\frac{1}{2}}{6}\)
\(\Rightarrow\frac{\frac{x-3+5}{5}}{4}=\frac{\frac{4x}{6}-\frac{3}{6}}{6}\)
\(\Rightarrow\frac{\frac{x+2}{5}}{4}=\frac{\frac{4x-3}{6}}{6}\)
\(\Rightarrow\frac{x+2}{5}:4=\frac{4x-3}{6}:6\)
\(\Rightarrow\frac{x+2}{5}.\frac{1}{4}=\frac{4x-3}{6}.\frac{1}{6}\)
\(\Rightarrow\frac{x+2}{20}=\frac{4x-3}{36}\)
\(\Rightarrow36.\left(x+2\right)=20.\left(4x-3\right)\)
\(\Rightarrow36x+72=80x-60\)
\(\Rightarrow72+60=80x-36x\)
\(\Rightarrow132=44x\)
\(\Rightarrow x=\frac{132}{44}=3\)
Vậy x=3
b)Bn làm tương tự phần a nha
Chúc bn học tốt
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\(\Leftrightarrow\frac{3x-3+5}{20}=|\frac{4x-3}{36}|\)
\(\Leftrightarrow\frac{3x+2}{5}=|\frac{4x-3}{9}|\)
\(\Leftrightarrow\left(3x+2\right)9=5|4x-3|\)
ĐK: \(3x+2\ge0\Rightarrow x\ge\frac{-2}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}5.\left(4x-3\right)=9.\left(3x+2\right)\\-5.\left(4x-3\right)=9.\left(3x+2\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}20x-15=27x+18\\-20x+15=27x+18\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}7x=-33\\-47x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-33}{7}\left(L\right)\\x=\frac{-3}{47}\end{cases}}}\)
Vậy x= - 3/47
b) \(\Leftrightarrow\frac{10x+x+2}{18}=\frac{x+3+75}{60}-2\)
\(\Leftrightarrow\frac{11x+2}{18}=\frac{x-42}{60}\)
\(\Leftrightarrow10.\left(11x+2\right)=3.\left(x-42\right)\)
\(\Leftrightarrow110x+20=3x-126\)
\(\Leftrightarrow107x=-146\Leftrightarrow x=\frac{-146}{107}\)
a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)
⇔ 3(x - 3) = 90 - 5(1 - 2x)
⇔ 3x - 9 = 90 - 5 + 10x
⇔ 3x - 10x = 90 - 5 + 9
⇔ -7x = 94
⇔ x = \(\frac{-94}{7}\)
S = { \(\frac{-94}{7}\) }
b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)
⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)
⇔ 6x - 4 - 60 = 9 - 6x - 42
⇔ 6x + 6x = 9 - 42 + 60 + 4
⇔ 12x = 31
⇔ x = \(\frac{31}{12}\)
S = { \(\frac{31}{12}\) }
c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7
⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210
⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210
⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40
⇔ 13x = 150
⇔ x = \(\frac{150}{13}\)
S = { \(\frac{150}{13}\) }
d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)
⇔ 21x - 120(x - 9) = 4(2x + 1,5)
⇔ 21x - 120x + 1080 = 8x + 6
⇔ 21x - 120x - 8x = 6 - 1080
⇔ -107x = -1074
⇔ x = \(\frac{1074}{107}\)
S = { \(\frac{1074}{107}\) }
e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5
⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840
⇔ 140x -140+56 -294x+42= 96x+48 -840
⇔ 140x -294x -96x = 48 -840 -42 -56+140
⇔ -250x = -750
⇔ x = 3
S = { 3 }
f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)
⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x
⇔ 4x+4+18x+9 = 4x+6x+6+7+12x
⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4
⇔ 0x = 0
S = R
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