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a) \(\frac{5x-2}{3}=\frac{5-3x}{2}\)\(\Leftrightarrow2\left(5x-2\right)=3\left(5-3x\right)\)\(\Leftrightarrow10x-4=15-9x\)
\(\Leftrightarrow10x+9x=15+4\)\(\Leftrightarrow19x=19\)\(\Rightarrow x=1\)
Vậy tập nghiệm của phương trình là: \(S=\left\{1\right\}\)
b) \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)\(\Leftrightarrow\frac{3\left(10x+3\right)}{36}=\frac{36}{36}+\frac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)\(\Leftrightarrow30x+9=36+24+32x\)\(\Leftrightarrow32x-30x=9-36-24\)\(\Leftrightarrow2x=-51\)\(\Leftrightarrow x=\frac{-51}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{-51}{2}\right\}\)
c) \(2\left(x+\frac{3}{5}\right)=5\left(\frac{13}{5}+x\right)\)\(\Leftrightarrow2\left(\frac{5x}{5}+\frac{3}{5}\right)=5\left(\frac{13}{5}+\frac{5x}{5}\right)\)\(\Leftrightarrow\frac{2\left(5x+3\right)}{5}=\frac{5\left(13+5x\right)}{5}\)
\(\Leftrightarrow2\left(5x+3\right)=5\left(13+5x\right)\)\(\Leftrightarrow10x+6=65+25x\)\(\Leftrightarrow25x-10x=6-65\)\(\Leftrightarrow15x=-59\)\(\Leftrightarrow x=\frac{-59}{15}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{-59}{15}\right\}\)
\(a,\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right)2}{3.2}=\frac{\left(5-3x\right)3}{2.3}\)
\(< =>\frac{10x-4}{6}=\frac{15-9x}{6}\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4=19\)
\(< =>19x=19< =>x=1\)
Bài 1:
\(B=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left(0,625-0,5+\frac{5}{11}+\frac{5}{12}\right)}+\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left[5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)\right]}+\frac{3}{5}\)
\(=\frac{-3}{5}+\frac{3}{5}\)
\(=0\)
Bài 2:
b) Giải:
Ta có: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{c^6}{d^6}=\frac{3a^6+c^6}{3b^6+d^6}\) (1)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{b+d}\)
\(\Rightarrow\left(\frac{a}{b}\right)^6=\left(\frac{a+c}{b+d}\right)^6=\frac{a^6}{b^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\left(đpcm\right)\)
a) \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow-21x=3x-60\)
\(\Leftrightarrow24x=60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{\left(8x-3\right)-2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)+\left(x+3\right)}{4}\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow x=-16\)
Vậy tập nghiệm của phương trình là \(S=\left\{-16\right\}\)
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{9\left(3-x\right)+16\left(5-x\right)}{24}=\frac{12\left(1-x\right)-48}{24}\)
\(\Leftrightarrow27-9x+80-16x=12-12x-48\)
\(\Leftrightarrow-25x+107=-12x-36\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của phương trình là \(S=\left\{11\right\}\)
\( 2\left( {x + \dfrac{3}{5}} \right) = 5 - \left( {\dfrac{{13}}{5} + x} \right)\\ \Leftrightarrow 2x + \dfrac{6}{5} = \dfrac{{12}}{5} - x\\ \Leftrightarrow 3x = \dfrac{6}{5}\\ \Leftrightarrow x = \dfrac{2}{5} \)
\( \dfrac{{10x + 3}}{{12}} = 1 + \dfrac{{6 + 8x}}{9}\\ \Leftrightarrow 3\left( {10x + 3} \right) = 36 + 4\left( {6 + 8x} \right)\\ \Leftrightarrow 30x + 9 = 36 + 24 + 32x\\ \Leftrightarrow - 2x = 51\\ \Leftrightarrow x = - \dfrac{{51}}{2} \)
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)