trong quá trình bạn xem bài mk thấy chỗ nào sai dấu thì sửa giùm mk nha trong quá trình làm mk cx có thể sai sót nhầm lẫn nha

\( m){x^4} + 2{x^3} - 13{x^2} - 14x + 24 = 0\\ \Leftrightarrow {x^4} - {x^3} + 3{x^3} - 3{x^2} - 10{x^2} + 10x - 24x + 24 = 0\\ \Leftrightarrow {x^3}\left( {x - 1} \right) + 3{x^2}\left( {x - 1} \right) - 10x\left( {x - 1} \right) - 24\left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^3} + 3{x^2} - 10x - 24} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^3} + 2{x^2} + {x^2} + 2x - 12x - 24} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left[ {{x^2}\left( {x + 2} \right) + x\left( {x + 2} \right) - 12\left( {x + 2} \right)} \right] = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {{x^2} + x - 12} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {{x^2} + 4x - 3x - 12} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left[ {x\left( {x + 4} \right) - 3\left( {x + 4} \right)} \right] = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)\left( {x + 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 1 = 0\\ x + 2 = 0\\ x - 3 = 0\\ x + 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - 2\\ x = 3\\ x = - 4 \end{array} \right. \)

\(n)\dfrac{x+1}{99}+\dfrac{x+2}{98}=\dfrac{x+3}{97}+\dfrac{x+4}{96}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right) = \left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}=\dfrac{x+100}{97}+\dfrac{x+100}{96}\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}-\dfrac{x+100}{97}-\dfrac{x+100}{96}=0\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{97}-\dfrac{1}{96}\right)=0 \)

Mà \(\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{97}-\dfrac{1}{96}\ne0 \\ \)

\(\Rightarrow x+100=0\\ \Rightarrow x=-100\\ \)

Vậy \(x=-100\)

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)

\(\Leftrightarrow\frac{x+10}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)\(\Rightarrow x+100=0\Leftrightarrow x=-100\)

Ta có:\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\Rightarrow x+100=0\)

\(\Rightarrow x=-100\)

x=100

Ta sẽ có: 1-1+1+1-1+1-1+1=0

\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)

\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)

\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)

Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)

\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(x=100\)

x thuoc R

\(\dfrac{x-1}{99}-\dfrac{x+1}{101}+\dfrac{x-2}{98}-\dfrac{x+2}{102}+\dfrac{x-3}{97}-\dfrac{x+3}{103}+\dfrac{x-4}{96}-\dfrac{x+4}{104}=0\)
<=> \(\dfrac{x-1}{99}-1-\dfrac{x+1}{101}-1+\dfrac{x-2}{98}-1-\dfrac{x-2}{102}-1+\dfrac{x-3}{97}-1-\dfrac{x+3}{103}-1+\dfrac{x-4}{96}-1-\dfrac{x+4}{104}=0\)

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}>0\)

\(\Rightarrow x+100=0\)

\(\Rightarrow x=0-100\)

\(\Rightarrow x=-100\)

Chúc bạn học tốt !!!! 

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\) 

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\) 

\(\Leftrightarrow x+100=0\) 

\(\Leftrightarrow x=-100\) 

Vậy...

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+1+1+1+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì 1/99+1/98+1/97+1/96>0

\(\Rightarrow x+100=0\Rightarrow x=-100\)

Ta có :

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=-4+4\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)nên \(x+100=0\)

\(\Rightarrow x=0-100=-100\)

sai đề

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yk nha

(x+1)/99+(x+2)/98+(x+3)/97=(x+4)/96+(x+5)/95+(x+6)/94

[(x+1)/99 +1]+[(x+2)/98 +1]+[(x+3)/97 +1]-3=[(x+4)/96 +1]+[(x+5)/95 +1]+[(x+6)/94 +1]-3

[(x+1+99)/99+(x+2+98)/98+(x+3+97)/97]-3=[(x+4+96)/96+(x+5+95)/95+(x+6+94)/94]-3

(x+100)/99+(x+100)/98+(x+100)/97=(x+100)/96+(x+100)/95+(x+100)/94

(x+100)(1/99+1/98+1/97)=(x+100)(1/96+1/95+1/94)

(x+100)(1/99+1/98+1/97)-(x+100)(1/96+1/95+1/94)=0

(x+100)(1/99+1/98+1/97-1/96-1/95-1/94)=0

Ma : 1/99+1/98+1/97-1/96-1/95-1/94 \(\ne\)0

=>x+100=0

=>x=-100

k mk nha khong hieu noi mk nha.

1/3x-1/2=(3/5-4x)15/7

1/3x-1/2=9/7-60/7x

1/3x+60/7x=1/2+9/7

187/21x=25/14

x=75/374

k mk nha ban.

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