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a) Ta có: \(\frac{3x-2}{6}-\frac{4-3x}{18}=\frac{4-x}{9}\)

\(\Leftrightarrow\frac{3\left(3x-2\right)}{18}-\frac{4-3x}{18}-\frac{2\left(4-x\right)}{18}=0\)

\(\Leftrightarrow9x-6-4+3x-\left(8-2x\right)=0\)

\(\Leftrightarrow12x-10-8+2x=0\)

\(\Leftrightarrow10x-18=0\)

\(\Leftrightarrow10x=18\)

hay \(x=\frac{9}{5}\)

Vậy: \(x=\frac{9}{5}\)

b) Ta có: \(\frac{2+3x}{6}-x+2=\frac{x-7}{9}\)

\(\Leftrightarrow\frac{3\left(2+3x\right)}{18}-\frac{18x}{18}+\frac{36}{18}-\frac{2\left(x-7\right)}{18}=0\)

\(\Leftrightarrow6+9x-18x+36-\left(2x-14\right)=0\)

\(\Leftrightarrow42-9x-2x+14=0\)

\(\Leftrightarrow56-11x=0\)

\(\Leftrightarrow11x=56\)

hay \(x=\frac{56}{11}\)

Vậy: \(x=\frac{56}{11}\)

c) ĐKXĐ: x∉{3;-3}

Ta có: \(\frac{6-x}{x^2-9}+\frac{2}{x+3}=\frac{-5}{x-3}\)

\(\Leftrightarrow\frac{6-x}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow6-x+2x-6=-5x-15\)

\(\Leftrightarrow x+5x+15=0\)

\(\Leftrightarrow6x=-15\)

hay \(x=\frac{-5}{2}\)(tm)

Vậy: \(x=\frac{-5}{2}\)

d) Ta có: \(\left(5x+2\right)\left(x^2-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x^2-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-2\\x^2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{5}\\x=\pm\sqrt{7}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-2}{5};\sqrt{7};-\sqrt{7}\right\}\)

e) ĐKXĐ: x∉{4;-4}

Ta có: \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)

\(\Leftrightarrow\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{5x-2}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)

\(\Leftrightarrow3x+12+5x-2-\left(4x-16\right)=0\)

\(\Leftrightarrow8x+10-4x+16=0\)

\(\Leftrightarrow4x+26=0\)

\(\Leftrightarrow4x=-26\)

hay \(x=\frac{-13}{2}\)(tm)

Vậy: \(x=\frac{-13}{2}\)

5 tháng 3 2020

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

5 tháng 3 2020

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

24 tháng 3 2020

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

19 tháng 5 2021

có bị viết nhầm thì thông cảm nha!

Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0 1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\) e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\) g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\) i,...
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Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0

1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\)

g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)

i, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\); k, \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)

m, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\); n, \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right).\frac{1}{3}\left(x+2\right)\)

p, \(\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x\); q, \(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)

r, \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\); s, \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)

t, \(\frac{2x-8}{6}.\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\); u, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{3}+\frac{2x-1}{12}\)

v, \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\); w, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)

17

Đây là những bài cơ bản mà bạn!

29 tháng 3 2020

bạn ấy muốn thách xem bạn nào đủ kiên nhẫn đánh hết chỗ này

26 tháng 3 2020

a)

\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)

\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)

\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)

\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)

\(\Leftrightarrow\frac{49-13x}{12}=0\)

\(\Rightarrow49-13x=0\)

\(\Rightarrow x=\frac{-49}{13}\)

26 tháng 3 2020

b)

\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)

\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)

\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)

\(\Leftrightarrow\frac{-3x}{4}=0\)

\(\Rightarrow-3x=0\)

\(\Rightarrow x=0\)

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)...
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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 :

\(\frac{90}{x}-\frac{36}{x-6}=2\) \(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
\(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\) \(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)

0
15 tháng 4 2020

\(a\text{) }7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow x=7\)

\(b\text{) }\frac{3x-1}{3}=\frac{2-x}{2}\)

\(\Leftrightarrow2\left(3x-1\right)=3\left(2-x\right)\)

\(\Leftrightarrow6x-2=6-3x\)

\(\Leftrightarrow9x=8\Leftrightarrow x=\frac{8}{9}\)

\(c\text{) }\frac{2\left(3x+5\right)}{3}-\frac{x}{2}=5-\frac{3\left(x+1\right)}{4}\)

\(\Leftrightarrow8\left(3x+5\right)-6x=60-9\left(x+1\right)\)

\(\Leftrightarrow24x+40-6x=60-9x-9\)

\(\Leftrightarrow27x=11\Leftrightarrow x=\frac{11}{27}\)

\(d\text{) }x^2-4x+4=9\)

\(\Leftrightarrow\left(x-2\right)^2=3^2\)

\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)

\(e\text{) }\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-8}{x^2-4}\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)-x\left(x+2\right)=5x-8\)

\(\Leftrightarrow x^2-x-2x+3-x^2-2x=5x-8\)

\(\Leftrightarrow11-10x=0\Leftrightarrow x=\frac{11}{10}\)

15 tháng 4 2020

Bổ sung: e) ĐKXĐ: x ≠ \(\pm\) 2