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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}
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=\(\left\{424\right\}\)
Câu x ) là bằng - 5 nhé mấy bạn. Làm giúp mình tất cả nhé ! Mình cảm ơn nhiều lắm !
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
a, \(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
\(\Leftrightarrow1+\frac{x+16}{49}+1+\frac{x+18}{47}=\frac{x+20}{45}-1+2\)
\(\Leftrightarrow\frac{x+16+49}{49}+\frac{x+18+47}{47}=\frac{x+20+45}{45}\)
\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\)
\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)
Ta có: \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\)>0
\(\Rightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
Vậy x = -65
b, \(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}+\frac{x-59}{40}\)
\(\Leftrightarrow\frac{x-69}{30}-1+\frac{x-67}{32}-1+\frac{x-65}{34}-1+\frac{x-63}{36}-1+\frac{x-61}{38}-1+\frac{x-59}{40}-1\)
\(\Leftrightarrow\frac{x-99}{30}+\frac{x-99}{32}+\frac{x-99}{34}-\frac{x-99}{36}-\frac{x-99}{38}-\frac{x-99}{40}=0\)
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\right)=0\)
Vì \(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\)>0
\(\Rightarrow x-99=0\)
\(\Leftrightarrow x=99\)
Vậy x =99
\(ĐKXĐ:x\ne\pm5\)
\(\frac{5}{x+5}-\frac{x-3}{5-x}=\frac{2x-40}{x^2-25}\)
\(\Leftrightarrow\frac{5}{x+5}+\frac{x-3}{x-5}=\frac{2x-40}{x^2-25}\)
\(\Leftrightarrow\frac{5x-25+5x-15}{x^2-25}=\frac{2x-40}{x^2-25}\)
\(\Rightarrow10x-40=2x-40\)
\(\Leftrightarrow x=0\left(TMĐKXĐ\right)\)
Vậy x=0
\(\frac{5}{x+5}-\frac{x-3}{5-x}=\frac{2x-40}{x^2-25}\) ( đkxđ : \(x\ne\pm5\))
( 5 - x ) = -( 5 - x ) = -5 + x = x - 5
<=> \(\frac{5}{x+5}-\frac{x-3}{x-5}=\frac{2x-40}{\left(x+5\right)\left(x-5\right)}\)
<=> \(\frac{5\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}-\frac{\left(x+5\right)\left(x-3\right)}{\left(x+5\right)\left(x-5\right)}=\frac{2x-40}{\left(x+5\right)\left(x-5\right)}\)
<=> \(5x-25-x^2+2x-15=2x-40\)
<=> \(5x-x^2+2x-2x=-40+25+15\)
<=> \(5x-x^2=0\)
<=> \(x^2-5x=0\)
<=> \(x\left(x-5\right)=0\)
<=> x = 0 ( nhận ) hoặc x = 5 ( loại do đkxđ )
Vậy nghiệm của phương trình là x = 0
<=>
Gợi ý :
Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)
Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)
Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)
bài 3
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)
=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x=100
\(\frac{x+34}{33}+\frac{x+33}{29}+\frac{x}{25}=8\)
\(\frac{725\left(x+34\right)}{23925}+\frac{825\left(x+33\right)}{23925}+\frac{957x}{23925}=\frac{191400}{23925}\)( ko hiểu thì inbox riêng nha )
\(725x+24650+825x+27225+957x=191400\)
\(2507x+51875=191400\)
\(2507x+51875-191400=0\)
\(2507x-139525=0\)
\(2507x=139525\)
\(x=\frac{139525}{2507}\)
pt <=> \(\left(\frac{392-x}{32}+1\right)+\left(\frac{390-x}{34}+1\right)+\left(\frac{388-x}{36}+1\right)+\left(\frac{386-x}{38}+1\right)+\left(\frac{384-x}{40}+1\right)=0\)
<=> \(\frac{392-x+32}{32}+\frac{390-x+34}{34}+\frac{388-x+36}{36}+\frac{386-x+38}{38}+\frac{384-x+40}{40}=0\)
<=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
<=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
Vì \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}>0\)( quá rõ ràng rồi =)) )
=> 424 - x = 0 <=> x = 424
Vậy S = { 424 }
Trả lời:
\(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}=-5\)
\(\Leftrightarrow\left(\frac{392-x}{32}+1\right)+\left(\frac{390-x}{34}+1\right)+\left(\frac{388-x}{36}+1\right)+\left(\frac{386-x}{38}+1\right)=0\)
\(\Leftrightarrow\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
\(\Leftrightarrow\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
\(\Leftrightarrow424-x=0\) ( vì \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\ne0\))
\(\Leftrightarrow x=424\)
Vậy \(S=\left\{424\right\}\)