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`(x+19)/3 +(x+13)/5 = (x+7)/7 + (x+1)/9`
`<=> x/3 + 19/3 +x/5 +13/5 = x/7 +1 +x/9 +1/9`
`<=> x/3 +x/5 -x/7 -x/9 = 1+1/9 -19/3 -13/5`
`<=> x (1/3 +1/5 -1/7 -1/9) = -118/45`
`<=> x * 88/315 = -352/45`
`<=> x = -28`
Vậy `S={-28}`
\(\begin{array}{l}a)\dfrac{{{x^2} - 49}}{{{x^2} + 5}}.\left( {\dfrac{{{x^2} + 5}}{{x - 7}} - \dfrac{{{x^2} + 5}}{{x + 7}}} \right)\\ = \dfrac{{\left( {x - 7} \right)\left( {x + 7} \right)}}{{{x^2} + 5}}.\dfrac{{{x^2} + 5}}{{x - 7}} - \dfrac{{\left( {x - 7} \right)\left( {x + 7} \right)}}{{{x^2} + 5}}.\dfrac{{{x^2} + 5}}{{x + 7}}\\ = x + 7 - \left( {x - 7} \right) = 14\end{array}\)
\(\begin{array}{l}b)\dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2000 - x}}{{x + 1945}} + \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2{\rm{x}} - 25}}{{x + 1945}}\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\left( {\dfrac{{2000 - x}}{{x + 1945}} + \dfrac{{2{\rm{x}} - 25}}{{x + 1945}}} \right)\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2000 - x + 2{\rm{x}} - 25}}{{x + 1945}}\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{x + 1975}}{{x + 1945}} = \dfrac{{19{\rm{x}} + 8}}{{x + 1945}}\end{array}\)
dễ mà
\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)thay x = 1;x=3 vào ra kq thui
\(\Leftrightarrow\)\(\dfrac{x}{7}-\dfrac{20}{7}+\dfrac{x}{33}-\dfrac{46}{33}+\dfrac{x}{8}-\dfrac{5}{8}+\dfrac{x}{19}+\dfrac{6}{19}=0\)
\(\dfrac{5016x}{35112}+\dfrac{1064x}{35112}+\dfrac{4389x}{35112}+\dfrac{1848x}{35112}=\dfrac{100320}{35112}+\dfrac{48994}{35112}-\dfrac{21945}{35112}+\dfrac{11088}{35112}\)
\(\dfrac{12317x}{35112}=\dfrac{160121}{35112}\)
\(12317x=160121\)
\(x=13\)
Làm cách này nhưng tớ thấy nó vẫn không tiến triển gì, thôi, qua cái này:
\(\dfrac{x-20}{7}+\dfrac{x-46}{33}+\dfrac{x-5}{8}+\dfrac{x+6}{19}=0\)
\(\dfrac{5016\left(x-20\right)+1064\left(x-46\right)+4389\left(x-5\right)+1848\left(x+6\right)}{35112}=0\)
\(5016x+1064x+4389x+1848x=100320+48944+21945-11088\)
\(12317x=160121\)
\(x=13\)
\(\Leftrightarrow\frac{x-20}{7}+1+\frac{x-46}{33}+1+\frac{x-5}{8}-1+\frac{x+6}{19}-1=0\)
\(\Leftrightarrow\frac{x-13}{7}+\frac{x-13}{33}+\frac{x-13}{8}+\frac{x-13}{19}=0\)
\(\Leftrightarrow\left(x-13\right)\left(\frac{1}{7}+\frac{1}{33}+\frac{1}{8}+\frac{1}{19}\right)=0\)
\(\Leftrightarrow x-13=0\)
\(\Leftrightarrow x=13\)
a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)
<=> \(25x+10-80x+10=24x+12-30\)
<=> \(25x-80x-24x=12-30-10-10\)
<=> \(-79x=-38\)
<=> \(x=\dfrac{-38}{-79}\)
\(x=\dfrac{38}{79}\)
b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)
<=> \(30x-12x+30+5x+40=210+10x-10\)
<=> \(30x-12x+5x-10x=210-10-30-40\)
<=> \(13x=130\)
<=> \(x=\dfrac{130}{13}\)
\(x=10\)
c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)
<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)
<=> \(28x+28+60x+120+105x+420+2520=0\)
<=> \(28x+60x+105x=-28-120-420-2520\)
<=> \(193x=-3088\)
<=> \(x=\dfrac{-3088}{193}\)
\(x=-16\)
d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)
<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)
<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)
<=> \(22968x=8199576\)
<=> \(x=\dfrac{8199576}{22968}\)
\(x=357\)
\(B=\dfrac{11}{17}+\left(-\dfrac{8}{19}\right)+\left(-\dfrac{3}{4}\right)+\dfrac{6}{17}-\dfrac{30}{19}\\ \Rightarrow B=\left(\dfrac{11}{17}+\dfrac{6}{17}\right)+\left(-\dfrac{8}{19}-\dfrac{30}{19}\right)-\dfrac{3}{4}\\ \Rightarrow B=1-2-\dfrac{3}{4}\\ \Rightarrow B=-\dfrac{7}{4}\)
\(\dfrac{7}{13}\cdot\dfrac{5}{19}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\cdot\dfrac{7}{19}\\ =\dfrac{7}{19}\cdot\dfrac{5}{13}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\cdot\dfrac{7}{19}\\ =\dfrac{7}{19}\cdot\left(\dfrac{5}{13}+\dfrac{8}{13}-3\right)=\dfrac{7}{19}\cdot\left(-2\right)=-\dfrac{14}{19}\)
Ta có: \(\dfrac{7}{13}\cdot\dfrac{5}{19}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\cdot\dfrac{7}{19}\)
\(=\dfrac{7}{19}\cdot\dfrac{5}{13}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\cdot\dfrac{7}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-3\right)\)
\(=\dfrac{7}{19}\cdot\left(-2\right)\)
\(=\dfrac{-14}{19}\)