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Ta có :
\(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
\(\Leftrightarrow\left(\frac{392-x}{32}+1\right)+\left(\frac{390-x}{34}+1\right)+\left(\frac{388-x}{36}+1\right)+\left(\frac{386-x}{38}+1\right)+\left(\frac{384-x}{40}\right)=0\)
\(\Leftrightarrow\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
\(\Leftrightarrow\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
Mà : \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\ne0\)
\(\Leftrightarrow424-x=0\)
\(\Leftrightarrow x=424\)
Vậy x = 424
a, Ta có : \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}+1=-5+5=0\)
=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
=> \(424-x=0\)
=> \(x=424\)
Vậy phương trình có nghiệm là x = 424 .
b, Ta có : \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
=> \(\frac{x+1}{2014}+1+\frac{x+3}{2012}+1=\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}-\frac{x+2015}{2010}-\frac{x+2015}{2009}=0\)
=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> \(x+2015=0\)
=> \(x=-2015\)
Vậy phương trình có nghiệm là x = -2015 .
a) \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
<=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}=0\)
<=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{40}=0\)
<=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{40}\right)=0\)
<=> 424 - x = 0
<=> x = 424
Vậy S = {424}
b) \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
<=> x + 2015 = 0
<=> x= -2015
Vậy S = {-2015}
a) \(\dfrac{x+43}{57}+\dfrac{x+46}{54}=\dfrac{x+49}{51}+\dfrac{x+52}{48}\)
\(\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}\right)\)
\(\dfrac{x+43+57}{57}+\dfrac{x+46+54}{54}-\dfrac{x+49+51}{51}-\dfrac{x+52+48}{48}=0\)
\(\dfrac{x+100}{57}+\dfrac{x+100}{54}-\dfrac{x+100}{51}-\dfrac{x+100}{48}=0\)
\(\left(x+100\right)\left(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\right)=0\)
Mà \(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\ne0\)
Nên: \(x+100=0\)
\(x=-100\)
\(\Leftrightarrow\dfrac{x-5}{95}-1+\dfrac{x-132}{32}+1=\dfrac{x-131}{31}+1+\dfrac{x-10}{90}-1\)
=>x-100=0
hay x=100
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
Lời giải:
PT $\Leftrightarrow \frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1$
$\Leftrightarrow \frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}$
$\Leftrightarrow (x+100)(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60})=0$
Dễ thấy $\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}<0$
$\Rightarrow x+100=0$
$\Leftrightarrow x=-100$ (tm)
1: Ta có: \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)
Suy ra: \(x^2+4x+4+2x-4=x^2\)
\(\Leftrightarrow6x=0\)
hay \(x=0\left(nhận\right)\)
2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)
Suy ra: \(x+6-2x+12=3x+6\)
\(\Leftrightarrow-x-3x=6-18=-12\)
hay \(x=3\left(nhận\right)\)
Lời giải:
1. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)
\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)
2. ĐKXĐ: $x\neq \pm 6$
PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)
\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)
\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)
Câu x ) là bằng - 5 nhé mấy bạn. Làm giúp mình tất cả nhé ! Mình cảm ơn nhiều lắm !
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
ta có : \(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}\)+\(\dfrac{384-x}{40}=-5\)
\(\Leftrightarrow\)\(\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1\)+\(\dfrac{384-x}{40}+1=0\)
\(\Leftrightarrow\)\(\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\)\(\Leftrightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\)
\(\Leftrightarrow x=424\)(vì \(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\ne0\))
Vậy tập nghiệm của phương trình là s=\(\left\{424\right\}\)