\(\frac{1}{2}x+2\frac{1}{2}=3\frac{1}{2}x-\frac{3}{4}\)
\(\frac{1}{3}x+\frac{2}{5}\left\{x+1\right\}=0\)
\(\frac{2}{3}-\frac{1}{3}\left\{x-\frac{3}{2}\right\}-\frac{1}{2}\left\{2x+1\right\}=5\)
Những câu hỏi liên quan
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)2x(25x-4)-(5x-2)(5x+1)8 / 5)2left(x-2right)-3left(3x-1right)left(x-3right)
2)x(4x-3)-(2x-2)(2x-1)5 / 6)frac{2}{x+1}-frac{1}{x-2}frac{3x-11}{left(x+1right)left(x-2right)}
3)frac{5}{2x+3}+frac{3}{9-x^2}frac{8}{7left(x3right)} / 7)frac{5x-2}{6}+frac{3-4x}{2}2-frac{x+7}{3}
4)frac{2}{3left(x-2right)}+frac{5}{12-3x^2}frac{3}{4left(x+2right)} /...
Đọc tiếp
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
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hệ phương trình
1, left{{}begin{matrix}frac{1}{x+y}+frac{1}{x-y}frac{5}{8}frac{1}{x+y}-frac{1}{x-y}-frac{3}{8}end{matrix}right.
2, left{{}begin{matrix}frac{4}{2x-3y}+frac{5}{3x+y}2frac{3}{3x+y}-frac{5}{2x-3y}21end{matrix}right.
3, left{{}begin{matrix}frac{7}{x-y+2}+frac{5}{x+y-1}frac{9}{2}frac{3}{x-y+2}+frac{2}{x+y-1}4end{matrix}right.
4, left{{}begin{matrix}frac{3}{x}+frac{5}{y}-frac{3}{2}frac{5}{x}-frac{2}{y}frac{8}{3}end{matrix}right.
5 , left{{}begin{matrix}frac{2}{x+y-1}-frac{4}{x-y+1...
Đọc tiếp
hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Giải các phương trình sau
a) frac{1}{x+1}-frac{5}{x-2}frac{15}{left(x+1right)left(2-xright)}
b) frac{1}{2x-3}-frac{3}{xleft(2x-3right)}frac{5}{x}
c) frac{6}{x-1}-frac{4}{x-3}frac{8}{2x-6}
d) frac{3}{left(x-1right)left(x-2right)}+frac{2}{left(x-3right)left(x-1right)}frac{1}{left(x-2right)left(x-3right)}
e) frac{1}{x-2}+frac{5}{x+1}frac{3}{2-x}
f) frac{5x}{2x+2}+1-frac{6}{x+1}
g) frac{x+1}{x-1}-frac{x-1}{x+1}frac{4}{x^2-1}
h) frac{3x}{x-2}-frac{x}{x-5}frac{3x}{left(x-2right)left(5-xright)}
Đọc tiếp
Giải các phương trình sau
a) \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
b) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
c) \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
d) \(\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\)
e) \(\frac{1}{x-2}+\frac{5}{x+1}=\frac{3}{2-x}\)
f) \(\frac{5x}{2x+2}+1=-\frac{6}{x+1}\)
g) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
h) \(\frac{3x}{x-2}-\frac{x}{x-5}=\frac{3x}{\left(x-2\right)\left(5-x\right)}\)
a) ĐKXĐ: \(x\ne-1;x\ne2\)
Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
⇔\(\frac{1}{x+1}-\frac{5}{x-2}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(x-2-5x-5+15=0\)
⇔\(-4x+8=0\)
⇔\(-4x=-8\)
⇔\(x=\frac{-8}{-4}=2\)(loại)
Vậy: x không có giá trị
b) ĐKXĐ: \(x\ne0;x\ne\frac{3}{2}\)
Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
⇔\(\frac{x}{\left(2x-3\right)\cdot x}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)
⇔\(x-3-10x+15=0\)
⇔\(-9x+12=0\)
⇔\(-9x=-12\)
⇔\(x=\frac{-12}{-9}=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
c) ĐKXĐ:\(x\ne3;x\ne1\)
Ta có: \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2\left(x-3\right)}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{4}{x-3}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}-\frac{4}{x-3}=0\)
⇔\(\frac{6}{x-1}-\frac{8}{x-3}=0\)
⇔\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=0\)
⇔\(6\left(x-3\right)-8\left(x-1\right)=0\)
⇔6x-18-8x+8=0
⇔-2x-10=0
⇔-2(x+5)=0
Vì 2≠0 nên x+5=0
hay x=-5
Vậy: x=-5
tìm xa) frac{x-1}{2}+frac{x-2}{5}frac{1}{4}+frac{x-7}{10}b) 3-frac{2}{2x-3}frac{2}{5}+frac{1}{2x-3}-frac{3}{2}c)7cdotleft(x-1right)+2xcdotleft(1-xright)0d) frac{x+1}{2008}+frac{x+2}{2017}+frac{x+3}{2016}frac{x+10}{2009}+frac{x+11}{2008}+frac{x+12}{2007}e) frac{2}{left(x-1right)cdotleft(x-3right)}+frac{5}{left(x-3right)cdotleft(x-8right)}+frac{12}{left(x-8right)cdotleft(x-20right)}-frac{1}{x-20}frac{-3}{4}
Đọc tiếp
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
frac{12}{16}frac{-x}{4}frac{21}{y}frac{z}{80} (-0,6x-frac{1}{2}).frac{3}{4}-left(-1right)frac{1}{3}
frac{1}{3}x+frac{2}{5}left(x-1right)0
left(2x-3right).left(6-2xright)0
frac{-2}{3}-frac{1}{3}left(2x-5right)frac{3}{2}
2|frac{1}{2}x-frac{1}{3}|-frac{3}{2}frac{1}{4}
frac{3}{4}-2|2x-frac{2}{3}|2
Đọc tiếp
\(\frac{12}{16}=\frac{-x}{4}=\frac{21}{y}=\frac{z}{80}\) \((-0,6x-\frac{1}{2}).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\left(2x-3\right).\left(6-2x\right)=0\)
\(\frac{-2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
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Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
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a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
b)\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
c)\(\frac{x +1}{x-2}+\frac{x-1}{x +2}=\frac{2\left(x^{2^{ }}+2\right)}{x^2-4}\)
d)(2x+3)\(\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)
Bài 1: Thu gọn
a) frac{1}{5}x^4y^3-3x^4y^3
b) 5x^2y^5-frac{1}{4}x^2y^5
c) frac{1}{7}x^2y^3.left(-frac{14}{3}xy^2right)-frac{1}{2}xy.left(x^2y^{text{4}}right)
d) left(3xyright)^2.left(-frac{1}{2}x^3y^2right)
e) -frac{1}{4}xy^2+frac{2}{5}x^2y+frac{1}{2}xy^2-x^2y
f) frac{1}{2}x^4y.left(-frac{2}{3}x^3y^2right)-frac{1}{3}x^7y^3
g) frac{1}{2}x^2y.left(-10x^3yz^2right).frac{1}{4}x^5y^3z
h) 4.left(-frac{1}{2}xright)^2-frac{3}{2}x.left(-xright)+frac{1}{3}x^2
i) 1frac{2}{3}x^3y.left(frac{-1}{2}xy...
Đọc tiếp
Bài 1: Thu gọn
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)
d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)
e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)
f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)
g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)
h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)
k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)
n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)
m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)
p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
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