giải phương trình:
a/ \(\frac{x-15}{65}+\frac{x-2}{78}+\frac{x-60}{20}+\frac{x-68}{12}=4\)
b/\(\frac{x-1}{6}+\frac{x-2}{3}+\frac{x-3}{2}=\frac{11}{3}\)
Giải các phương trình:
a) \(\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{64}\)
b) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
g) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
h) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
i) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2\frac{9-3x}{5}}{12}+\frac{2}{3}\)
Giải hộ mik cái mn!!
g) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+98}{98}\right)+\left(\frac{x+4+96}{96}\right)=\left(\frac{x+6+94}{94}\right)+\left(\frac{x+8+92}{92}\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0.\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=0-100\)
\(\Leftrightarrow x=-100.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-100\right\}.\)
h) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)
\(\Leftrightarrow\left(\frac{x-12-77}{77}\right)+\left(\frac{x-11-78}{78}\right)=\left(\frac{x-74-15}{15}\right)+\left(\frac{x-73-16}{16}\right)\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Vì \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0.\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=0+89\)
\(\Leftrightarrow x=89.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{89\right\}.\)
Chúc bạn học tốt!
Câu g) bạn cộng 1 vào mỗi hạng tử của 2 vế
Câu h) bạn trừ một vào mỗi hạng tử ở hai vế
Quy đồng mẫu thì được tử giống nhau sau đó đặt nhân tử chung là xong
16, giải phương trình.
1, \(\frac{x+5}{65}+\frac{x+10}{60}=\frac{x+15}{55}+\frac{x+20}{50}\)
2, \(\frac{x+91}{81}+\frac{x+92}{82}+\frac{x+93}{83}=3\)
3, \(\frac{59-x}{19}+\frac{58-x}{18}=\frac{57-x}{17}+\frac{56-x}{16}\)
4, \(\frac{x}{15}+\frac{x+1}{16}+\frac{x+2}{17}+\frac{x+3}{18}+\frac{x+4}{19}=5\)
Giải các phương trình:
\(a,\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(b,\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
a) \(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}+2=\frac{x+6}{94}+\frac{x+8}{92}+2\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
b)\(\Leftrightarrow\frac{x-12}{77}+\frac{x-11}{78}-2=\frac{x-74}{15}+\frac{x-73}{16}-2\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-89=0\Leftrightarrow x=89\)
Giải các phương trình:
\(a,\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(b,\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
Lời giải:
a)
PT \(\Leftrightarrow \frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Dễ thấy \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}<0\) nên $x+100=0$
$\Rightarrow x=-100$
b)
PT \(\Leftrightarrow \frac{x-12}{77}-1+\frac{x-11}{78}-1=\frac{x-74}{15}-1+\frac{x-73}{16}-1\)
\(\Leftrightarrow \frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow (x-89)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Dễ thấy \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}< 0\)
\(\Rightarrow x-89=0\Rightarrow x=89\)
16,giải phương trình sau.
1, \(\frac{\frac{x-3}{5}+1}{4}=\frac{\frac{2x}{3}-\frac{1}{2}}{6}\)
2, \(\frac{5x+\frac{x+2}{2}}{9}-x=\frac{\frac{x+3}{5}+15}{12}-2\)
3, 5x+\(\frac{\frac{x+2}{2}}{9}-x=\frac{\frac{x+3}{5}+15}{12}-2\)
\(\Leftrightarrow\frac{\frac{x-3+5}{5}}{4}=\frac{\frac{4x-3}{6}}{6}\Leftrightarrow\frac{x+2}{20}=\frac{4x-3}{36}\Leftrightarrow36x+72=80x-60\Leftrightarrow44x=132\Rightarrow x=2\)
\(\Leftrightarrow\frac{\frac{10x+x+2}{2}}{9}-\frac{\frac{x+3+75}{5}}{12}=x-2\)\(\Leftrightarrow\frac{11x+2}{18}-\frac{x+78}{60}=x-2\)\(\Leftrightarrow\left(\frac{11}{18}-\frac{1}{60}-1\right)x+\left(\frac{2}{18}-\frac{78}{60}+2\right)=0\).Giải típ nha, ko có Casio nên mk ko bấm
\(\Leftrightarrow5x+\frac{x+2}{18}-x-\frac{x+3+75}{60}+2=0\Leftrightarrow\left(5+\frac{1}{18}-1-\frac{1}{60}\right)x+\left(\frac{2}{18}-\frac{78}{60}+2\right)=0\).Giải típ nha
1. giải pt
a. 5(x-3)-4=2(x-1)+7
b. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
c.\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
d. \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}\)\(=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
e. \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
a, 5(x-3)-4=2(x-1)+7
<=>\(5x-15-4=2x-2+7\)
\(\Leftrightarrow5x-2x=15+4-2+7\)
\(\Leftrightarrow3x=24\)
\(\Leftrightarrow x=8\)
b, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)}{4}+\frac{x+3}{4}\)
\(\Rightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow8x-6x-4x-x=3+4-2+3\)
\(\Leftrightarrow-3x=8\)
\(\Leftrightarrow x=\frac{-8}{3}\)
c,\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
<=>\(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
\(\Leftrightarrow\frac{4x+20}{6}+\frac{3x+36}{6}-\frac{5x-10}{6}=\frac{2x}{6}+\frac{66}{6}\)
\(\Rightarrow4x+20+3x+36-5x+10=2x+66\)
\(\Leftrightarrow4x+3x-5x-2x=66-20-36-10\)
\(\Leftrightarrow0=0\)
1,GIẢI BẤT PHƯƠNG TRÌNH và PHƯƠNG TRÌNH
a,\(\frac{2x-3}{2}>\frac{8x-11}{6}\)
b,\(\frac{2x+1}{6}-\frac{x-2}{9}>x-3\)
c,\(x-\frac{x-3}{4}>3-\frac{x-3}{12}\)
d,\(\frac{-12}{x^2-9}+\frac{2}{x-3}+\frac{3}{x+3}=1\)
Câu 1 tính hợp lí
a,A=\(\frac{-9}{11}.\frac{3}{8}-\frac{9}{11}.\frac{5}{8}+\frac{17}{11}\)
b,B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{53.55}\)
Câu2 tìm x
a,\(x+5-\frac{1}{2}=3\frac{1}{2}\)
b,\(3^{x+1}=27\)
c,\(2016.\left[25-\left(3x+2\right)\right]=3^2.7\)
d,\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{60}+\frac{x}{78}=\frac{220}{39}\)
Ai giải nhanh tớ tick cho
Thank you
\(-\frac{9}{11}\cdot\frac{3}{8}-\frac{9}{11}\cdot\frac{5}{8}+\frac{17}{11}=-\frac{9}{11}\left(\frac{3}{8}+\frac{5}{8}\right)+\frac{17}{11}=-\frac{9}{11}\cdot1+\frac{17}{11}=1\)
\(\frac{2}{1.3}+....+\frac{2}{53.55}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{53}-\frac{1}{55}=1-\frac{1}{55}=\frac{54}{55}\)
\(x+5-\frac{1}{2}=3\frac{1}{2}\)
\(x+5=3.5+0.5=4\)
\(x=4-5=-1\)
\(3^{x+1}=27=3^3\)
\(x+1=3\)
vậy x=2
bây h bn mún tớ iair hộ bn trừ câu cuối ko
\(2016\left\{25-\left(3x+2\right)\right\}=63\)
\(3x+2=24.96875\)
tự tính