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KN
5 tháng 3 2020
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
KN
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
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
16 tháng 2 2017
b,(2x -3 )(3x - 1) =(2x+3)(x-2)
<=> 6x2 - 11x + 3 = 2x2 - x - 6
<=.> 4x2 - 10x + 9 = 0
<=> (2x - \(\frac{5}{2}\))2 +\(\frac{11}{4}\)= 0 ( vô lí )
( Vì (2x - \(\frac{5}{2}\))2 \(\ge\) 0 => (2x - \(\frac{5}{2}\))2 + \(\frac{11}{4}\)\(\ge\)\(\frac{11}{4}\))
Vậy pt vô nghiệm
CP
17 tháng 2 2017
câu C : (4-3x)(2x+3)=(5-2x)(3x-4)
<=> (4-3x)(2x+3)-(5-2x)(4-3x)=0
<=>(4-3x)(2x+3-5+2x)=0
<=>(4-3x)(4x-2)=0
<=>\(\left[\begin{matrix}3x=4\\4x=2\end{matrix}\right.\)
<=>\(\left[\begin{matrix}x=\frac{4}{3}\\x=\frac{1}{2}\end{matrix}\right.\)
XU
8 tháng 2 2020
a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)
⇔ 3(x - 3) = 90 - 5(1 - 2x)
⇔ 3x - 9 = 90 - 5 + 10x
⇔ 3x - 10x = 90 - 5 + 9
⇔ -7x = 94
⇔ x = \(\frac{-94}{7}\)
S = { \(\frac{-94}{7}\) }
b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)
⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)
⇔ 6x - 4 - 60 = 9 - 6x - 42
⇔ 6x + 6x = 9 - 42 + 60 + 4
⇔ 12x = 31
⇔ x = \(\frac{31}{12}\)
S = { \(\frac{31}{12}\) }
c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7
⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210
⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210
⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40
⇔ 13x = 150
⇔ x = \(\frac{150}{13}\)
S = { \(\frac{150}{13}\) }
d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)
⇔ 21x - 120(x - 9) = 4(2x + 1,5)
⇔ 21x - 120x + 1080 = 8x + 6
⇔ 21x - 120x - 8x = 6 - 1080
⇔ -107x = -1074
⇔ x = \(\frac{1074}{107}\)
S = { \(\frac{1074}{107}\) }
e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5
⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840
⇔ 140x -140+56 -294x+42= 96x+48 -840
⇔ 140x -294x -96x = 48 -840 -42 -56+140
⇔ -250x = -750
⇔ x = 3
S = { 3 }
f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)
⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x
⇔ 4x+4+18x+9 = 4x+6x+6+7+12x
⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4
⇔ 0x = 0
S = R
Chúc bạn học tốt !
22 tháng 4 2020
Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html
Mình cảm ơn trước nhaa
HP
27 tháng 2 2020
1. \(\frac{3x-7}{5}=\frac{2x-1}{3}\)
<=> 3(3x-7)=5(2x-1)
<=> 9x-21=10x-5
<=> -21+5=10x-9x
<=> x=-16
2. \(\frac{3x-7}{2}+\frac{2x-1}{3}=-16\)
<=> \(\frac{3\left(3x-7\right)}{6}+\frac{2\left(2x-1\right)}{6}=\frac{-96}{6}\)
=> 9x-21+4x-2=-96
<=> 13x-23=-96
<=> 13x=-73
<=> x=\(\frac{-73}{13}\)
3. \(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
<=> \(\frac{15x}{15}-\frac{5\left(x+1\right)}{15}=\frac{3\left(2x+1\right)}{15}\)
=> 15x-5x-5=6x+3
<=> 15x-5x-6x=3+5
<=> 4x=8
<=> x=2
4. \(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5-\left(5-2x\right)}{6}\)
<=>\(\frac{7-3x}{12}+\frac{9}{12}=\frac{24\left(x-2\right)}{12}+\frac{2\left[5-\left(5-2x\right)\right]}{12}\)
=> 7-3x+9=24x-48+4x
<=> -3x-24x-4x=-48-7
<=> -31x=-55
<=> x= \(\frac{55}{31}\)
5. \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
<=> \(\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21\left(x+13\right)}{21}\)
=> 14x-7-15x-6=21x+273
<=> 14x-15x-21x=273+7+6
<=> -22x=286
<=> x= -13
27 tháng 2 2020
a/\(\Leftrightarrow3\left(3x-7\right)=5\left(2x-1\right)\Leftrightarrow9x-21=10x-5\Leftrightarrow x=-16\)
b/\(\Leftrightarrow\frac{9x-21+4x-2}{6}=-16\)\(\Leftrightarrow13x-23=-96\Leftrightarrow x=x=-\frac{73}{13}\)
c/\(\Leftrightarrow\frac{3x-x+1}{3}-\frac{2x+1}{5}=0\Leftrightarrow\left(2x+1\right)\left(\frac{1}{3}-\frac{1}{5}\right)=0\Leftrightarrow x=-\frac{1}{2}\)
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)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
1 tháng 4 2020
f, \(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)
\(\Leftrightarrow\) \(\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{2\left(10x-4\right)\left(11x-4\right)}{18\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\)
\(\Leftrightarrow\) 18(12x + 1) + 2(10x - 4)(11x - 4) = (20x + 17)(11x - 4)
\(\Leftrightarrow\) 216x + 18 + 220x2 − 168x + 32 = 220x2 + 107x − 68
\(\Leftrightarrow\) 216x + 18 + 220x2 − 168x + 32 - 220x2 - 107x + 68 = 0
\(\Leftrightarrow\) −59x + 118 = 0
\(\Leftrightarrow\) -59x = -118
\(\Leftrightarrow\) x = 2
Vậy S = {2}
Chúc bạn học tốt!
AH
Akai Haruma
Giáo viên
12 tháng 8 2020
f)
$\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{(2-3x)^3}$
$=\frac{2x-3x^2}{x^2-1}.\frac{x^4-1}{(2-3x)^3}=\frac{x(2-3x)(x^2-1)(x^2+1)}{(x^2-1)(2-3x)^3}$
$=\frac{x(x^2+1)}{(2-3x)^2}$
g)
$\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15xy^3}$
$=\frac{5xy}{2x-3}.\frac{-4(2x-3)}{15xy^3}=\frac{-4}{3y^2}$
h)
$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$
$=\frac{x(x+2)}{3(x-1)^2}.\frac{5(x-1)}{2(x+2)}$
$=\frac{5x}{6(x-1)}$
AH
Akai Haruma
Giáo viên
12 tháng 8 2020
d)
$\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}=\frac{x+8}{(x-4)(x+4)}-\frac{2}{x(x+4)}$
$=\frac{x(x+8)}{x(x-4)(x+4)}-\frac{2(x-4)}{x(x+4)(x-4)}$
$=\frac{x^2+8x-2(x-4)}{x(x+4)(x-4)}=\frac{x^2+6x+8}{x(x+4)(x-4)}$
$=\frac{(x+2)(x+4)}{x(x+4)(x-4)}=\frac{x+2}{x(x-4)}$
e)
$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$
$=\frac{-3(x+7)}{2x+1}$
c/ \(\frac{2x-5}{7}=\frac{5-2x}{5}\)
=> 5 (2x - 5) = 7 (5 - 2x)
=> 10x - 25 = 35 - 14x
=> 10x + 14x = 35 + 25
=> 24x = 60
=> x = 2,5
a)-4
b)6
c)5/2