1. x : (\(\frac{2}{9}-\frac{1}{5}\))=\(\frac{8}{16}\)
2. a)\(\frac{x-3}{-2}\)>0
b) \(x^2\)+x >0
c) \(\frac{x+3}{x-5}\)<0
Những câu hỏi liên quan
Tìm x biết
a) (8-5x).(x+2)+4.(x-2).(x+1)+2.(x-2).(x+2)=0
b)\(\left(-\frac{2}{5}+x\right):\frac{7}{9}+\left(-\frac{3}{5}+\frac{5}{6}\right):\frac{7}{9}=0\)
c)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2004}\)
Ai giúp vs !!!
\(a.\frac{3x-7}{5}=\frac{2x-1}{3}\\ b.\frac{4x-7}{12}-x=\frac{3x}{8}\\ c.\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\\ d.\frac{5x-8}{3}=\frac{1-3x}{2}\\ e.\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\\ f.\frac{x-1}{\frac{2}{5}}-3-\frac{3x-2}{\frac{5}{4}}-2=1\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)
\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)
\(\Rightarrow-16x-272=120x+312\)
\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)
Xem thêm câu trả lời
a, \(\frac{3}{5}.x-\frac{1}{2}=\frac{1}{7}\)
b, \(\frac{1}{4}+\frac{1}{3}:3x=-5\)
c, \(\frac{1}{3}.x+\frac{2}{5}\left(x+1\right)=0\)
d, \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
tìm x, biết :
a, \(1-(5\frac{3}{8}+x-7\frac{5}{24}):(-16\frac{2}{3})=0\)
b, \((\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+x:\frac{1}{3}=-4\)
c, \(1\frac{3}{5}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}.x=\frac{16}{5}\)
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)0
b) \(\left(\frac{x}{3}-5\frac{1}{4}\right)^2-\frac{-2}{5}=1\frac{1}{25}\)
c) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
Các bạn giúp mk với, nhanh nhé, mk cần gấp
Câu a có 1 số 0 to thôi nhé!
Dạng 1: Phương trình bậc nhất
Bài 1: Giải các phương trình sau :
a) 0,5x (2x - 9) 1,5x (x - 5)
b) 28 (x - 1) - 9 (x - 2) 14x
c) 8 (3x - 2) - 14x 2 (4 - 7x) + 18x
d) 2 (x - 5) - 6 (1 - 2x) 3x + 2
e) frac{x+7}{2}-frac{x-3}{5}frac{x}{6}
f) frac{2x-3}{3}-frac{5x+2}{12}frac{x-3}{4}+1
g) frac{x+6}{2}+frac{2left(x+17right)}{2}+frac{5left(x-10right)}{6}2x+6
h) frac{3x+2}{5}-frac{4x-3}{7}4+frac{x-2}{35}
i) frac{x-1}{2}+frac{x+3}{3}frac{5x+3}{6}
j) frac{x-3}{5}-1frac{4x+1}{4}
Dạng 2: Phươn...
Đọc tiếp
Dạng 1: Phương trình bậc nhất
Bài 1: Giải các phương trình sau :
a) 0,5x (2x - 9) = 1,5x (x - 5)
b) 28 (x - 1) - 9 (x - 2) = 14x
c) 8 (3x - 2) - 14x = 2 (4 - 7x) + 18x
d) 2 (x - 5) - 6 (1 - 2x) = 3x + 2
e) \(\frac{x+7}{2}-\frac{x-3}{5}=\frac{x}{6}\)
f) \(\frac{2x-3}{3}-\frac{5x+2}{12}=\frac{x-3}{4}+1\)
g) \(\frac{x+6}{2}+\frac{2\left(x+17\right)}{2}+\frac{5\left(x-10\right)}{6}=2x+6\)
h) \(\frac{3x+2}{5}-\frac{4x-3}{7}=4+\frac{x-2}{35}\)
i) \(\frac{x-1}{2}+\frac{x+3}{3}=\frac{5x+3}{6}\)
j) \(\frac{x-3}{5}-1=\frac{4x+1}{4}\)
Dạng 2: Phương trình tích
Bài 2: Giải phương trình sau :
a) (x + 1) (5x + 3) = (3x - 8) (x - 1)
b) (x - 1) (2x - 1) = x(1 - x)
c) (2x - 3) (4 - x) (x - 3) = 0
d) (x + 1)2 - 4x2 = 0
e) (2x + 5)2 = (x + 3)2
f) (2x - 7) (x + 3) = x2 - 9
g) (3x + 4) (x - 4) = (x - 4)2
h) x2 - 6x + 8 = 0
i) x2 + 3x + 2 = 0
j) 2x2 - 5x + 3 = 0
k) x (2x - 7) - 4x + 14 = 9
l) (x - 2)2 - x + 2 = 0
Dạng 3: Phương trình chứa ẩn ở mẫu
Bài 3: Giải phương trình sau :
| \(\frac{90}{x}-\frac{36}{x-6}=2\) | \(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\) |
| \(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\) | \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) |
| \(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\) | \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\) |
| \(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\) | \(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\) |
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
b) \(\left(\frac{x}{3}-5\frac{1}{4}\right)^2-\frac{-2}{5}=1\frac{1}{25}\)
c) \(\frac{17-x}{12}=\frac{3}{17-x}\)
d) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
Giái phương trình :
a,\(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
b,\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
c,\(\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)
d,\(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\)
a) ĐKXĐ: x∉{2;5}
Ta có: \(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(\Leftrightarrow\frac{6x+1}{\left(x-2\right)\left(x-5\right)}+\frac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{3\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=0\)
\(\Leftrightarrow6x+1+5x-25-3\left(x-2\right)=0\)
\(\Leftrightarrow11x-24-3x+6=0\)
\(\Leftrightarrow8x-18=0\)
\(\Leftrightarrow8x=18\)
hay \(x=\frac{9}{4}\)(tm)
Vậy: \(x=\frac{9}{4}\)
b) ĐKXĐ: x∉{0;2;-2}
Ta có: \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow2x-\left(x^2+x-2\right)+x^2-6x+8=0\)
\(\Leftrightarrow2x-x^2-x+2+x^2-6x+8=0\)
\(\Leftrightarrow-5x+10=0\)
\(\Leftrightarrow-5x=-10\)
hay x=2(ktm)
Vậy: x∈∅
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Bài 1: Tìm điều kiện xác định của phương trình:a.frac{5-x}{x^2+6x+9}frac{3x+2}{x^2+6x+8}b.frac{x-7}{x^2+1}frac{x+6}{x^2+x+1}Bài 2: Giải phương trình:a.frac{15x-10}{x^2+3}0b.frac{x^2-4x-5}{x-5}0c.frac{3x-1}{x-1}-frac{2x+5}{x+3}-frac{8}{x^2+2x-3}0
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Bài 1: Tìm điều kiện xác định của phương trình:
\(a.\frac{5-x}{x^2+6x+9}=\frac{3x+2}{x^2+6x+8}\)
\(b.\frac{x-7}{x^2+1}=\frac{x+6}{x^2+x+1}\)
Bài 2: Giải phương trình:
\(a.\frac{15x-10}{x^2+3}=0\)
\(b.\frac{x^2-4x-5}{x-5}=0\)
\(c.\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{x^2+2x-3}=0\)