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\(\frac{43-x}{57}+\frac{46-x}{54}=\frac{49-x}{51}+\frac{52-x}{48}\)
\(\Leftrightarrow\left(\frac{43-x}{57}+1\right)+\left(\frac{46-x}{54}+1\right)=\left(\frac{49-x}{51}+1\right)+\left(\frac{52-x}{48}+1\right)\)
\(\Leftrightarrow\frac{43-x+57}{57}+\frac{46-x+54}{54}=\frac{49-x+51}{51}+\frac{52-x+48}{48}\)
\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}=\frac{100-x}{51}+\frac{100-x}{48}\)
\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}-\left(\frac{100-x}{51}+\frac{100-x}{48}\right)=0\)
\(\Leftrightarrow\left(100-x\right)\left[\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)\right]=0\) (*)
Vì\(\frac{1}{57}< \frac{1}{51},\frac{1}{54}< \frac{1}{48}\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)< \left(\frac{1}{51}+\frac{1}{48}\right)\)
\(\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)< 0\)
Phương trình (*) xảy ra khi: \(100-x=0\Leftrightarrow x=100\)
Vậy phương trình có nghiệm duy nhất là x = 100
Phương trình đầu bài tương đương với
\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)\(\Leftrightarrow\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
\(\Leftrightarrow\orbr{\begin{cases}x+100=0\\\frac{1}{57}+\frac{1}{54}=\frac{1}{51}+\frac{1}{48}\left(sai\right)\end{cases}\Leftrightarrow x+100=0\Leftrightarrow x=-100}\)
Vậy phương trình có nghiệm duy nhất là x=-100
<=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
<=> \(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=> \(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
vi \(\frac{1}{57}< \frac{1}{51};\frac{1}{54}< \frac{1}{48}\Rightarrow\frac{1}{57}-\frac{1}{51}+\frac{1}{54}-\frac{1}{48}< 0\)
=> x+100=0 => x= -100
vay pt co nghiem \(x=-100\)
\(\frac{x+43}{57}+\frac{x+46}{54}+\frac{x+49}{51}+\frac{x+235}{45}=0\)
\(\Leftrightarrow\text{}\text{}\)\(\frac{x+43}{57}+1+\frac{x+46}{54}+1+\frac{x+49}{51}+1+\frac{x+235}{45}-3=0\)
\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}+\frac{x+100}{51}+\frac{x+100}{45}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}+\frac{1}{51}+\frac{1}{45}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy x = -100
<=>\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
<=>\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)
=>x+100=0
<=>x=-100
k nha bạn
\(\Leftrightarrow\frac{37x+1648}{1026}=\frac{11x+556}{272}\Rightarrow\left(37x+1648\right)272=1026\left(11x+556\right)\)
<=>(37x+1648)272=272(37x+1648)
=>272(37x+1648)=1026(11x+556)
=>10064x+448256=11286x+570456
<=>-1222x=122200
=>x=122200:-1222
=>x=-100 ( dễ hiểu chưa hả )
a: \(\Leftrightarrow\dfrac{x-51}{9}-1+\dfrac{x-52}{8}-1=\dfrac{x-53}{7}-1+\dfrac{x-54}{6}-1\)
=>x-60=0
hay x=60
b: \(\Leftrightarrow\left(x-2\right)^2-3\left(x+2\right)=x-14\)
\(\Leftrightarrow x^2-4x+4-3x-6-x+14=0\)
\(\Leftrightarrow x^2-8x+12=0\)
=>(x-2)(x-6)=0
=>x=2(loại) hoặc x=6(nhận)
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: \(x+100=0\)
\(x=-100\)
1: =>x^3-5x^2+x^2-5x+3x-15=0
=>(x-5)(x^2+x+3)=0
=>x-5=0
=>x=5
2: =>x^3+6x^2+12x+35=0
=>x^3+5x^2+x^2+5x+7x+35=0
=>(x+5)(x^2+x+7)=0
=>x+5=0
=>x=-5
3: \(\Leftrightarrow\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}+1\right)\)
=>x+100=0
=>x=-100
ta có: (59-x)/41 +(57-x)/43 +(55-x)/45 +(53-x)/47 +(51-x)/49 =-5
<=>[(59-x)/41 +1 ] +[(57-x)/43 +1] +[(55-x)/45 +1] +[(53-x)/47 +1] +[(51-x)/49 +1] =0
<=>(59-x-41)/41 + (57-x-43)/43 +(55-x-45)/45 +(53-x-47)/47 +(51-x-49)/49 =0
<=>(100-x)/41 + (100-x)/43 + (100-x)/45 +(100-x)/47 + (100-x)/49 =0
<=>(100-x).( 1/41 + 1/43 + 1/45 + 1/47 + 1/49 ) =0
mà (1/41 + 1/43 + 1/45 + 1/47 + 1/49) khác 0 nên 100-x =0 <=>x=100
vậy nghiệm của pt là x=100
a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
Vậy x = -10
b) \(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
\(\Rightarrow\left(\frac{x+43}{57}+1\right)+\left(\frac{x+46}{54}+1\right)=\left(\frac{x+49}{51}+1\right)+\left(\frac{x+52}{48}+1\right)\)
\(\Rightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
\(\Rightarrow\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Mà \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
Vậy x = -100
a.\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
=>\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
<=> \(\frac{x+1+9}{9}+\frac{x+2+8}{8}=\frac{x+3+7}{7}+\frac{x+4+6}{6}\)
<=>\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
<=> \(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
<=> \(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
<=> x+10=0
<=> x=-10
Vậy tập nghiệm của phương trình trên là S=\(\left\{-10\right\}\)
b. \(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)<=>\(\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
<=>(x+100)\(\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)\)=0
<=>x+100=0
<=>x= -100
Vậy tập nghiệm của phương trình trên là S=\(\left\{-100\right\}\)