(1 - \(\frac{1}{5}\)) x (1 - \(\frac{1}{6}\)) x (1 - \(\frac{1}{7}\)) x (1 - \(\frac{1}{8}\))
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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
XO
18 tháng 2 2021
a) ĐKXĐ : \(x\ne-2;x\ne5\)
\(\frac{7}{x+2}=\frac{3}{x-5}\)
<=> 3(x + 2) = 7(x - 5)
<=> 3x + 6 = 7x - 35
<=> 4x = 41
<=>x = 41/4 (tm)
Vậy x = 41/4 là ngiệm phương trình
b) ĐKXĐ \(x\ne\pm3\)
\(\frac{2x-1}{x+3}=\frac{2x}{x-3}\)
<=> \(\frac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
<=> (2x - 1)(x - 3) = 2x(x + 3)
<=> 2x2 - 7x + 3 = 2x2 + 6x
<=> 13x = 3
<=> x = 3/13 (tm)
Vậy x = 3/13 là nghiệm phương trình
c) ĐKXĐ : \(x\ne-7;x\ne1,5\)
Khi đó \(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\)
<=> \(\frac{\left(3x-2\right)\left(2x-3\right)}{\left(x+7\right)\left(2x-3\right)}=\frac{\left(6x+1\right)\left(x+7\right)}{\left(x+7\right)\left(2x-3\right)}\)
<=> (3x - 2)(2x - 3) = (6x + 1)(x + 7)
<=> 6x2 - 13x + 6 = 6x2 + 43x + 7
<=> 56x = -1
<=> x = -1/56 (tm)
Vậy x = -1/56 là nghiệm phương trình
d) ĐKXĐ : \(x\ne\pm1\)
Khi đó \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
<=> \(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{5\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
<=> (2x + 1)(x + 1) = 5(x - 1)2
<=> 2x2 + 3x + 1 = 5x2 - 10x + 5
<=> 3x2 - 13x + 4 = 0
<=> 3x2 - 12x - x + 4 = 0
<=> 3x(x - 4) - (x - 4) = 0
<=> (3x - 1)(x - 4) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)
Vậy x \(\in\left\{\frac{1}{3};4\right\}\)là nghiệm phương trình
XO
18 tháng 2 2021
e) ĐKXĐ : \(x\ne1\)
Khi đó \(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\)
<=> \(\frac{3x-5}{x-1}=2\)
<=> 3x - 5 = 2(x - 1)
<=> 3x - 5 = 2x - 2
<=> x = 3 (tm)
Vậy x = 3 là nghiệm phương trình
f) ĐKXĐ : \(x\ne-1\)
\(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> \(\frac{3x+2}{x+1}=3\)
<=> 3x + 2 = 3(x + 1)
<=> 3x + 2 = 3x + 3
<=> 0x = 1
<=> \(x\in\varnothing\)
Vậy tập nghiệm phương trình S = \(\varnothing\)
g) ĐKXĐ : \(x\ne2\)
Khi đó \(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)
<=>\(\frac{x-2}{x-2}=3\)
<=> (x - 2) = 3(x - 2)
<=> x - 2 = 3x - 6
<=> -2x = -4
<=> x = 2 (loại)
Vậy tập nghiệm phương trình S = \(\varnothing\)
h) ĐKXĐ : \(x\ne7\)
Khi đó \(\frac{1}{7-x}=\frac{x-8}{x-7}-8\)
<=> \(\frac{x-7}{x-7}=8\)
<=> x - 7 = 8(x - 7)
<=> x - 7 = 8x - 56
<=> 7x = 49
<=> x = 7 (loại)
Vậy tập nghiệm phương trình S = \(\varnothing\)
i) ĐKXĐ : \(x\ne0;x\ne6\)
Ta có : \(\frac{x+6}{x}=\frac{1}{2}+\frac{15}{2\left(x-6\right)}\)
<=> \(\frac{x+6}{x}-\frac{15}{2\left(x-6\right)}=\frac{1}{2}\)
<=> \(\frac{2\left(x+6\right)\left(x-6\right)}{2x\left(x-6\right)}-\frac{15x}{2x\left(x-6\right)}=\frac{1}{2}\)
<=> \(\frac{2x^2-72-15x}{2x\left(x-6\right)}=\frac{1}{2}\)
<=> 4x2 - 144 - 30x = 2x(x - 6)
<=> 2x2 - 18x - 144 = 0
<=> x2 - 9x - 72 = 0
<=> x2 - 9x + 81/4 - 72- 81/4 = 0
<=> \(\left(x-\frac{9}{2}\right)^2-\frac{369}{4}=0\)
<=> \(\left(x-\frac{9}{2}+\sqrt{\frac{369}{4}}\right)\left(x-\frac{9}{2}-\sqrt{\frac{369}{4}}\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{9}{2}-\sqrt{\frac{369}{4}}\\x=\frac{9}{2}+\sqrt{\frac{369}{4}}\end{cases}}\)(tm)
Vậy x \(\in\left\{\frac{9}{2}-\sqrt{\frac{369}{4}};\frac{9}{2}+\sqrt{\frac{369}{4}}\right\}\)
28 tháng 8 2019
a,\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\) (1)
<=> \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
<=> \(\left(x+1\right)\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
=> x+1=0 (vì \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\ne0\))
<=> x=-1
Vậy pt (1) có tập nghiệm S\(=\left\{-1\right\}\)
b, \(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)(2)
<=> \(\frac{x+6}{2015}+1+\frac{x+5}{2016}+1+\frac{x+4}{2017}+1=\frac{x+3}{2018}+1+\frac{x+2}{2019}+1+\frac{x+1}{2020}+1\)
<=> \(\frac{x+2021}{2015}+\frac{x+2021}{2016}+\frac{x+2021}{2017}-\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
<=> \(\left(x+2021\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x+2021=0(vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=-2021
Vậy pt (2) có tập nghiệm S=\(\left\{-2021\right\}\)
c,\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\) (3)
<=> \(\frac{x+6}{2016}-1+\frac{x+7}{2017}-1+\frac{x+8}{2018}-1=\frac{x+9}{2019}-1+\frac{x+10}{2020}-1+1-1\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}=\frac{x-2010}{2019}+\frac{x-2010}{2020}\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}-\frac{x-2010}{2019}-\frac{x-2010}{2020}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x-2010=0 (vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=2010
Vậy pt (3) có tập nghiệm S=\(\left\{2010\right\}\)
d, \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\) (4)
<=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=15-1-2-3-4-5\)
<=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
<=> (x-100)(\(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\))=0
=> x -100=0(vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\))
<=> x=100
Vậy pt (4) có tập nghiệm S=\(\left\{100\right\}\)
VM
28 tháng 8 2019
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
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
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
(1 - 1/5) x (1 - 1/6) x (1 - 1/7) x (1 - 1/8) =
= 4/5 x 5/6 x 6/7 x 7/8 =
= (2/1 x 1/3) x (3/1 x 1/4) =
=2/3 x 3/4 =
= 6/12 = 1/2
(1-1/5 ) *(1- 1/7 ) * (1- 1/7) *(1- 1/8)
=4/5* 5/6* 6/7* 7/8
=1/2