\(\frac{43-x}{57}+\frac{46-x}{54}=\frac{49-x}{51}+\frac{52-x}{48}\)

\(\Leftrightarrow\left(\frac{43-x}{57}+1\right)+\left(\frac{46-x}{54}+1\right)=\left(\frac{49-x}{51}+1\right)+\left(\frac{52-x}{48}+1\right)\)

\(\Leftrightarrow\frac{43-x+57}{57}+\frac{46-x+54}{54}=\frac{49-x+51}{51}+\frac{52-x+48}{48}\)

\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}=\frac{100-x}{51}+\frac{100-x}{48}\)

\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}-\left(\frac{100-x}{51}+\frac{100-x}{48}\right)=0\)

\(\Leftrightarrow\left(100-x\right)\left[\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)\right]=0\) (*)

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 \(x=-100\)

Ta thấy:\(\frac{x+43}{57}\)\(+\)\(\frac{x+46}{54}\)\(+\)\(2\)\(=\)\(\frac{x+49}{51}\)\(+\)\(\frac{x+52}{48}\)\(+\)\(2\)

\(\Rightarrow\)\(\frac{x+43}{57}\)\(+\)\(\frac{57}{57}\)\(+\)\(\frac{x+46}{54}\)\(+\)\(\frac{54}{54}\)\(=\)\(\frac{x+49}{51}\)\(+\)\(\frac{51}{51}\)\(+\)\(\frac{x+48}{52}\)\(+\)\(\frac{52}{52}\)

\(\Leftrightarrow\)\(\frac{x+100}{57}\)\(+\)\(\frac{x+100}{54}\)\(=\)\(\frac{x+100}{51}\)\(+\)\(\frac{x+100}{52}\)

\(\Leftrightarrow\)\((\)\(x+100)\)\((\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\()\)\(=\)\((x+100)\)\((\frac{1}{52}\)\(+\)\(\frac{1}{51})\)

\(\Leftrightarrow\)\((x+100)\)\((\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\(-\)\(\frac{1}{52}\)\(-\)\(\frac{1}{51}\)\()\)\(=\)\(0\)\((1)\)

Ta thấy: \(\frac{1}{57}\)< \(\frac{1}{52}\)

          \(\frac{1}{54}\)<\(\frac{1}{51}\)

\(\Rightarrow\)\(\frac{1}{57}\)\(+\)\(\frac{1}{54}\)< \(\frac{1}{52}\)\(+\)\(\frac{1}{51}\)

\(\Rightarrow\)\(\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\(-\)\(\frac{1}{52}\)\(-\)\(\frac{1}{51}\)< 0 \((2)\)

Từ \((1)\)và \(\left(2\right)\)\(\Rightarrow\)\(x+100\)\(=0\)

\(\Leftrightarrow x=-100\)

Vậy phương trình có tập nghiệm \(x=-100\)

\(\frac{x+43}{57}+\frac{x+46}{54}+\frac{x+49}{51}+\frac{x+235}{45}=0\)

\(\Leftrightarrow\text{​​}\text{​​}\)\(\frac{x+43}{57}+1+\frac{x+46}{54}+1+\frac{x+49}{51}+1+\frac{x+235}{45}-3=0\)

\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}+\frac{x+100}{51}+\frac{x+100}{45}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}+\frac{1}{51}+\frac{1}{45}\right)=0\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

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 )

\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)

\(\Leftrightarrow\)\(\frac{x+43}{57}+1+\frac{x+46}{54}=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

\(\Leftrightarrow\)\(\frac{x+43}{57}+\frac{57}{57}+\frac{x+46}{54}+\frac{54}{54}=\frac{x+49}{51}+\frac{51}{51}+\frac{x+52}{48}+\frac{48}{48}\)

\(\Leftrightarrow\)\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)nên \(x+100=0\Leftrightarrow x=-100\)

Vậy \(x=-100\)

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

Bài 1:

Đặt \(\left\{\begin{matrix} 5x+3=a\\ 2x+4=b\end{matrix}\right.\) \(\Rightarrow 3x-1=a-b\)

PT trở thành:

\(a^3-b^3=(a-b)^3\)

\(\Leftrightarrow (a-b)(a^2+ab+b^2)=(a-b)^3\)

\(\Leftrightarrow (a-b)[a^2+ab+b^2-(a^2-2ab+b^2)]=0\)

\(\Leftrightarrow 3ab(a-b)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a=b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{5}\\x=-2\\5x+3=2x+4\Leftrightarrow x=\dfrac{1}{3}\end{matrix}\right.\)

Thử lại thấy đều thỏa mãn

Vậy \(x\in\left\{\frac{-3}{5};-2;\frac{1}{3}\right\}\)

Bài 2:

\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}=\frac{x-4}{2010}\)

\(\Leftrightarrow \frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\left(\frac{x-3}{2011}-1\right)=\frac{x-4}{2010}-1\)

\(\Leftrightarrow \frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}=\frac{x-2014}{2010}\)

\(\Leftrightarrow (x-2014)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\) (1)

Thấy rằng \(2013> 2011; 2012> 2010\Rightarrow \frac{1}{2013}< \frac{1}{2011}; \frac{1}{2012}< \frac{1}{2010}\)

\(\Rightarrow \frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}< 0\) (2)

Từ (1),(2) suy ra \(x-2014=0\Leftrightarrow x=2014\)

Bài 3:

Đặt \(\left\{\begin{matrix} 2x-5=a\\ x-2=b\end{matrix}\right.\Rightarrow x-3=a-b\)

PT trở thành: \(a^3-b^3=(a-b)^3\)

\(\Leftrightarrow (a-b)(a^2+ab+b^2)-(a-b)(a^2-2ab+b^2)=0\)

\(\Leftrightarrow 3ab(a-b)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\\x-3=0\Leftrightarrow x=3\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{5}{2}; 2; 3\right\}\)

Bài 4:

\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)

\(\Leftrightarrow \frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

\(\Leftrightarrow \frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

\(\Leftrightarrow (x+100)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\) (1)

Dễ thấy: \(\frac{1}{57}< \frac{1}{51}; \frac{1}{54}< \frac{1}{48}\Rightarrow \frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}< 0\) (2)

Từ \((1);(2)\Rightarrow x+100=0\Leftrightarrow x=-100\)

Bài 5:

\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Leftrightarrow \frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)

\(\Leftrightarrow \frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Leftrightarrow (x-50)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\neq 0\) suy ra \(x-50=0\Leftrightarrow x=50\)