Tìm x:
1, \(\left(1,5.x-\frac{4}{5}\right).\left(\frac{1}{2019}-\frac{1}{2018}\right)=0\\\)
\(2,\frac{2x}{3}+\frac{1}{3}=\left|-\frac{2}{5}\right|\)
Những câu hỏi liên quan
Bài 1: Tìm x:
a) x - \(\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)
b)\(2\frac{1}{3}-x=\frac{-5}{9}+2x\)
c) \(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\)
d) \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
Phần a vs phần b tính toán thông thường thôi mà bạn, vs 1 h/s lớp 7 thì ít nhất phải làm được chứ?? :((
a) \(x-\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)
\(\Leftrightarrow x-\frac{4}{5}=\frac{-1}{20}\)
\(\Leftrightarrow x=\frac{-1}{20}+\frac{4}{5}=\frac{15}{20}=\frac{3}{4}\)
b) \(2\frac{1}{3}-x=\frac{-5}{9}+2x\)
\(\Leftrightarrow2\frac{1}{3}-\frac{-5}{9}=2x+x\)
\(\Leftrightarrow3x=\frac{7}{3}+\frac{5}{9}\)
\(\Leftrightarrow3x=\frac{26}{9}\)
\(\Leftrightarrow x=\frac{26}{9}:3=\frac{26}{27}\)
d) .............................. ( Đề bài)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}\)\(-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Leftrightarrow-\frac{1}{x+3}=\frac{1}{2010}\)
\(\Leftrightarrow\frac{1}{-\left(x+3\right)}=\frac{1}{2010}\)\(\Leftrightarrow-\left(x+3\right)=2010\)
\(\Leftrightarrow-x-3=2010\) \(\Leftrightarrow-x=2010+3=2013\)
\(\Leftrightarrow x=-2013\)
Bạn tự kết luận nha!
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c)
\(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\\ \Leftrightarrow\frac{x+3}{2016}+1+\frac{x+2}{2017}+1=\frac{x+1}{2018}+1+\frac{x}{2019}+1\\ \Leftrightarrow\frac{x+2019}{2016}+\frac{x+2019}{2017}-\frac{x+2019}{2018}-\frac{x+2019}{2019}=0\\ \Leftrightarrow\left(x+2019\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\\ \Rightarrow x-2019=0\\ \Rightarrow x=2019\)
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Bài 1:
a) x - \(\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)
=> x - \(\frac{4}{5}=-\frac{1}{20}\)
x = \(\left(-\frac{1}{20}\right)+\frac{4}{5}\)
x = \(\frac{3}{4}\)
Vậy x = \(\frac{3}{4}\).
b) \(2\frac{1}{3}-x=-\frac{5}{9}+2x\)
=> \(2\frac{1}{3}-\left(-\frac{5}{9}\right)=2x+x\)
=> 3x = \(\frac{7}{3}+\frac{5}{9}\)
=> 3x = \(\frac{26}{9}\)
x = \(\frac{26}{9}:3\)
x = \(\frac{26}{27}\)
Vậy x = \(\frac{26}{27}\).
Chúc bạn học tốt!
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tìm x
a) \(\frac{3}{4}+\frac{1}{4}.x=\frac{1}{2}+\frac{1}{2}.x\)
b) \(\frac{1}{3}.x-\frac{2}{5}.\left(x+1\right)=0\)
c)\(\left(\frac{1}{4}.x-1,5\right)+\left(\frac{5}{6}.x-3\right)-\left(\frac{5}{8}.x-0,5\right)=-4,5\)
a) \(\frac{3}{4}+\frac{1}{4}.x=\frac{1}{2}+\frac{1}{2}x\)
\(\Rightarrow3.\frac{1}{4}+\frac{1}{4}.x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{1}{4}.\left(x+3\right)=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{x+1}{x+3}=\frac{1}{4}:\frac{1}{2}=\frac{1}{2}\)\(\Rightarrow\left(x+1\right).2=x+3\Rightarrow2x+2=x+3\)
\(\Rightarrow2x-x=3-2\Rightarrow x=1\)
vay x=1
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ta có
frac{left(2018-xright)^2+left(2018-xright)left(x-2019right)+left(x-2019right)^2}{left(2018-xright)^2-left(2018-xright)left(x-2019right)-left(x-2019right)^2}frac{19}{49} ( điều kiện : x khác : 2018;2019 )
đặt a x - 2019 ( a khác 0 )
ta có hệ thức :
frac{left(a+1right)^2-left(a+1right)a+a^2}{left(a+1right)^2+left(a+1right)a+a^2}frac{19}{49}
Leftrightarrowfrac{a^2+a+1}{3a^2+3a+1}frac{19}{49}
Leftrightarrow49left(a^2+a+1right)19left(3a^2+3a+1right)
Leftrightarrow49a^2+49a+4957a^2+57a+19...
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ta có
\(\frac{\left(2018-x\right)^2+\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}{\left(2018-x\right)^2-\left(2018-x\right)\left(x-2019\right)-\left(x-2019\right)^2}=\frac{19}{49}\) ( điều kiện : x khác : 2018;2019 )
đặt a = x - 2019 ( a khác 0 )
ta có hệ thức :
\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\\ \Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\\ \left(2a+1\right)^2-4^2=0\\ \Leftrightarrow\left(2a+3\right)\left(2a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4041}{2}\\x=\frac{4033}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
vậy \(x\in\left\{\frac{4041}{2};\frac{4033}{2}\right\}\)
Bài 1 : Tìm x biết :a, left|frac{3}{2}x+frac{1}{2}right|left|4x-1right|b, left|frac{5}{4}x-frac{7}{2}right|-left|frac{5}{8}x+frac{3}{5}right|0c,left|frac{7}{5}x+frac{2}{3}right|left|frac{4}{3}x-frac{1}{4}right|Bài 2 : Tìm x biết :a, | 2x - 5 | x +1 b, | 3x - 2 | -1 x c, | 3x - 7 | 2x + 1d, | 2x-1 | +1 x
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Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
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a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
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TÌM X
a,\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=6\)
b,\(\left(x^2-4\right).\left(2x+\frac{4}{3}\right)=0\)
Giải các phương trình sau:
a) left(frac{x-2}{x-1}right)^2-5left(frac{x+2}{x+1}right)^2+4left(frac{x^2-4}{x^2-1}right)1
b) left(frac{x-1}{x}right)^2+left(frac{x-1}{x-2}right)^2frac{40}{9}
c) x.frac{4-x}{x+2}.left(frac{8-2x}{x+2}right)3
d) frac{1}{3x-2020}+frac{1}{4x-2018}+frac{1}{5x-2017}frac{1}{12x-2019}
Đọc tiếp
Giải các phương trình sau:
a) \(\left(\frac{x-2}{x-1}\right)^2-5\left(\frac{x+2}{x+1}\right)^2+4\left(\frac{x^2-4}{x^2-1}\right)=1\)
b) \(\left(\frac{x-1}{x}\right)^2+\left(\frac{x-1}{x-2}\right)^2=\frac{40}{9}\)
c) \(x.\frac{4-x}{x+2}.\left(\frac{8-2x}{x+2}\right)=3\)
d) \(\frac{1}{3x-2020}+\frac{1}{4x-2018}+\frac{1}{5x-2017}=\frac{1}{12x-2019}\)
1) frac{2}{5}xfrac{1}{3}-frac{2}{15}:frac{1}{5}+frac{3}{5}xfrac{1}{3}2)left(3frac{1}{3}+2,5right):left(3frac{1}{6}-4frac{1}{5}right)-frac{11}{31}3)left[6+left(frac{1}{2}right)^3-left|-frac{1}{2}right|right]:frac{3}{12}4)frac{18}{37}+frac{8}{24}+frac{19}{37}-1frac{23}{24}+frac{2}{3}5)left(-2right)^3xleft(frac{3}{4}-0,25right):left(2frac{1}{4}-1frac{1}{6}right)6)left(frac{2}{5}right)^2+5frac{1}{2}xleft(45-2right)+frac{23}{4}7)frac{4}{9}-19frac{1}{3}-frac{4}{9}x39frac{1}{3}8)left(-frac{1}{2}right)^...
Đọc tiếp
1) \(\frac{2}{5}x\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}x\frac{1}{3}\)
2)\(\left(3\frac{1}{3}+2,5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)
3)\(\left[6+\left(\frac{1}{2}\right)^3-\left|-\frac{1}{2}\right|\right]:\frac{3}{12}\)
4)\(\frac{18}{37}+\frac{8}{24}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)
5)\(\left(-2\right)^3x\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
6)\(\left(\frac{2}{5}\right)^2+5\frac{1}{2}x\left(45-2\right)+\frac{23}{4}\)
7)\(\frac{4}{9}-19\frac{1}{3}-\frac{4}{9}x39\frac{1}{3}\)
8)\(\left(-\frac{1}{2}\right)^2:\frac{1}{4}-2x\left(-\frac{1}{2}\right)^2\)
9)\(125\%x\left(-\frac{1}{2}\right)^3:\left(1\frac{5}{16}-1,5\right)+2008^0\)
giúp mình nha mai mình học rùi
Giải phương trình:
1. frac{2x+3}{4}-frac{5x+3}{6}frac{3-4x}{12}
2. frac{3.left(2x+1right)}{4}-1frac{15x-1}{10}
3. frac{2x-1}{5}-frac{x-2}{3}frac{x+7}{15}
4. frac{x+3}{2}-frac{x-1}{3}frac{x+5}{6}+1
5. frac{x-4}{5}-frac{3x-2}{10}-xfrac{2x-5}{3}-frac{7x+2}{6}
6. frac{left(x+2right)left(x+10right)}{3}-frac{left(x+4right)left(x+10right)}{12}frac{left(x-2right)left(x+4right)}{4}
7. frac{left(x+2right)^2}{8}-2left(2x-1right)25+frac{left(x-2right)^2}{8}
8.frac{7x^2-14x-5}{5}frac{left(2x+1right)^...
Đọc tiếp
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
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 :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
1. \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
2 . \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
3 . \(4\left(3x-2\right)-3\left(x-4\right)=7x+10\)
4. \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
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}
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}