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27 tháng 7 2018

\(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}\)

⇒ (x-18) . (x+16) = (x-17) . (x+4)

x(x+16) -18(x+16) = x(x+4) - 17(x+4)

\(x^2+16x-18x-288=x^2+4x-17x-68\)

\(x^2+16-18x-x^2-4x+17x=-68+288\)

11x=220

x= 220 : 11

x = 20

27 tháng 7 2018

\(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}\)

\(\Leftrightarrow\left(x-18\right)\left(x+16\right)=\left(x-17\right)\left(x+4\right)\)

\(\Leftrightarrow x^2-2x-288=x^2-13x-68\)

\(\Leftrightarrow220=11x\)

\(\Leftrightarrow x=20\)

\(\)

30 tháng 7 2018

Đặt \(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}=k\)

\(\Rightarrow x-18=k.\left(x+4\right)\Rightarrow x=\dfrac{4k+18}{1-k}\left(1\right)\)

\(x-17=k.\left(x+16\right)\Rightarrow x=\dfrac{16k+17}{1-k}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow4k+18=16k+17\Rightarrow-12k=-1\Rightarrow k=\dfrac{1}{12}\)

\(\Rightarrow x=\dfrac{4.\dfrac{1}{12}+18}{1-\dfrac{1}{12}}=\dfrac{\dfrac{55}{3}}{\dfrac{11}{12}}=20\)

Vậy x = 20

30 tháng 7 2018

\(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}\)

(x\(-\)18).(x+16)=(x-17).(x+4)

x\(^2\)-18x+16x-18.16=x\(^2\)-17x+4x-4.17

x\(^2\)-18x+16x-288=x\(^2\)-17x+4x-68

x\(^2\)-18x+16x-x\(^2\)+17x-4x=-68+288

11x=220

x=220/11

x=20

21 tháng 12 2022

`B17:`

`a)` Với `x \ne +-3` có:

`A=[x+15]/[x^2-9]+2/[x+3]`

`A=[x+15+2(x-3)]/[(x-3)(x+3)]`

`A=[x+15+2x-6]/[(x-3)(x+3)]`

`A=[3x+9]/[(x-3)(x+3)]=3/[x-3]`

`b)A=[-1]/2<=>3/[x-3]=-1/2<=>-x+3=6<=>x=-3` (ko t/m)

   `=>` Ko có gtr nào của `x` t/m

`c)A in ZZ<=>3/[x-3] in ZZ`

   `=>x-3 in Ư_3`

 Mà `Ư_3={+-1;+-3}`

`@x-3=1=>x=4`

`@x-3=-1=>x=2`

`@x-3=3=>x=6`

`@x-3=-3=>x=0`

________________________________

`B18:`

`a)M=1/3`             `ĐK: x  \ne +-4`

`<=>(4/[x-4]-4/[x+4]).[x^2+8x+16]/32=1/3`

`<=>[4(x+4)-4(x-4)]/[(x-4)(x+4)].[(x+4)^2]/32=1/3`

`<=>32/[x-4].[x+4]/32=1/3`

`<=>3x+12=x-4`

`<=>x=-8` (t/m)

d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7

\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

=>x^2+11x+28=54

=>x^2+11x-26=0

=>(x+13)(x-2)=0

=>x=2 hoặc x=-13

e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)

=>x-258=0

=>x=258

19 tháng 4 2023

a/\(\left(\dfrac{7}{9}\times\dfrac{9}{7}\right)\times\dfrac{25}{28}\)
\(=1\times\dfrac{25}{28}\)
\(=\dfrac{25}{28}\)
b/\(\dfrac{4}{7}\times\dfrac{17}{18}\times\dfrac{7}{4}\times\dfrac{18}{17}\)
\(=\left(\dfrac{4}{7}\times\dfrac{7}{4}\right)\times\left(\dfrac{17}{18}\times\dfrac{18}{17}\right)\)
\(=1\times1\)
\(=1\)

a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)

\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)

nên x-17=0

hay x=17

Vậy: x=17

b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)

\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)

\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)

nên x+20=0

hay x=-20

Vậy: x=-20

Ta có: \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)

\(\Rightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)

\(\Rightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)

\(\Rightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)

\(\Rightarrow\left(x+18\right).\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\) (1)

\(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\ne0\) (2)

Từ (1) và (2) => x+18=0 => x=-18

Vậy x=-18

1 tháng 7 2017

Đăng trùng.

15 tháng 1 2022

Bài 2:

\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)

\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)

\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)

1 tháng 7 2017

\(\left(1+\dfrac{x+1}{17}\right)+\left(1+\dfrac{x+2}{16}\right)=\left(1+\dfrac{x+3}{15}\right)+\left(1+\dfrac{x+4}{14}\right)\)

\(\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)

\(\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)

\(\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)

Vì : \(\dfrac{1}{17}< \dfrac{1}{15};\dfrac{1}{16}< \dfrac{1}{14}\Rightarrow\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}< 0\)

\(\Rightarrow x+18=0\Rightarrow x=0-18=-18\)

1 tháng 7 2017

Từ đầu đến đến dòng thứ tư thì mình đồng ý, nhưng mình nghĩ không nhất thiết phải so sánh, chỉ cần làm tiếp như sau:

\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+18=0\\\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}=0\end{matrix}\right.\)

rồi tính tiếp là OK rồi. Dù gì thì cũng xin cảm ơn nha ;)

17 tháng 4 2018

15

\(\dfrac{7}{x-2}\)+\(\dfrac{8}{x-5}\)=3 (x khác 2 khác 5)

\(\Leftrightarrow\)7*(x-5)+8(x-2)=3(x-2)(x-5)

\(\Leftrightarrow\)15x-51=3x^2-21x+30\(\Leftrightarrow\)3x^2-36x+81=0

\(\Leftrightarrow\)\(\begin{matrix}&\end{matrix}\)\(\left[{}\begin{matrix}9\\3\end{matrix}\right.\) tmđk

16\(\dfrac{x^2-3x+6}{x^2-9}\)=\(\dfrac{1}{x-3}\)(x khác +_3)

\(\Leftrightarrow\)x^2-3x+6=x+3

\(\Leftrightarrow\)x^2-4x+3=0\(\Leftrightarrow\)\(\left[{}\begin{matrix}3loại\\1\end{matrix}\right.\)

vậy x=1 là nghiệm của pt

25 tháng 4 2018

17 \(\dfrac{3}{x^2-4}\) = \(\dfrac{1}{x-2}+\dfrac{1}{x+2}\)

<=> x + 2 + x - 2 = 3

<=> 2x = 3

<=> x = \(\dfrac{3}{2}\)