\(\Leftrightarrow\frac{108-x}{92}+1+\frac{107-x}{93}+1+\frac{106-x}{94}+1+\frac{105-x}{95}=0.\)

\(\Leftrightarrow\frac{108+92-x}{92}+\frac{107+93-x}{93}+\frac{106+94-x}{94}+\frac{105+95-x}{95}=0\)

\(\Leftrightarrow\frac{200-x}{92}+\frac{200-x}{93}+\frac{200-x}{94}+\frac{200-x}{95}=0\)

\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}200-x=0\Leftrightarrow x=200\\\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\ne0\end{cases}}\)

Vậy nghiệm của phương trình là 200

Ta có : 

\(\frac{108-x}{92}+\frac{107-x}{93}+\frac{106-x}{94}+\frac{105-x}{95}+4=0\)

\(\Leftrightarrow\)\(\left(\frac{108-x}{92}+1\right)+\left(\frac{107-x}{93}+1\right)+\left(\frac{106-x}{94}+1\right)+\left(\frac{105-x}{95}+1\right)+\left(4-4\right)=0\)

\(\Leftrightarrow\)\(\frac{200-x}{92}+\frac{200-x}{93}+\frac{200-x}{94}+\frac{200-x}{95}=0\)

\(\Leftrightarrow\)\(\left(200-x\right)\left(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\right)=0\)

Vì \(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\ne0\)

\(\Rightarrow\)\(200-x=0\)

\(\Rightarrow\)\(x=200\)

Vậy \(x=200\)

Chúc bạn học tốt ~

a, 96,75 + x - 47,8 = 107,54

=> 96,75 + x  = 107,54 + 47,8

=> 96,75 + x  = 155,34

=> x = 155,34 - 96,75

=> x = 58,59

b, x : 2,5 : 3,4 = 95

=> x : 2,5 = 95 x 3,4

=> x : 2,5 = 98,4

=> x = 98,4 x 2,5

=> x = 246

a, 96,75 + x - 47,8 = 107,54

=> 96,75 + x = 107,54 + 47,8

=> 96,75 + x = 155,34

=> x = 155,34 - 96,75

=> x = 58,59

b, dấu " : " hay dấu " ; " vầy vậy mà sao mình tìm được

Sửa lại đề b, nha

Chúc bạn học tốt :3

\(96,75+x-47,8=107,54\)

\(\Rightarrow48,85+x=107,54\)

\(\Rightarrow x=107,54-48,85\)

\(\Rightarrow x=58,69\)

Ta có: \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)

\(\Leftrightarrow\frac{99-x}{101}-1+\frac{97-x}{103}-1+\frac{95-x}{105}-1+\frac{93-x}{107}-1=-4+4\)

\(\Leftrightarrow\frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)

\(\Leftrightarrow\left(200-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)

=> 200 - x = 0

=> x          = 200

Vậy x = 200

\(x=200\)

\(\left(2x-6\right)\left(x^2+2\right)=\left(2x-6\right)\left(8x-10\right)\)

\(\Leftrightarrow\left(2x-6\right)\left(x^2+2\right)-\left(2x-6\right)\left(8x-10\right)=0\)

\(\Leftrightarrow\left(2x-6\right)\left(x^2+2-8x+10\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x^2-6x-2x-12\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x-6\right)\left(x-2\right)=0\)

\(\Rightarrow x\in\left\{3;6;2\right\}\)

\(\left(5x-1\right)^2=\left(3x+5\right)^2\)

\(\Leftrightarrow\left(5x-1\right)^2-\left(3x+5\right)^2=0\)

\(\Leftrightarrow\left(5x-1-3x-5\right)\left(5x-1+3x+5\right)=0\)

\(\Leftrightarrow\left(2x-6\right)\left(8x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\8x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}=-4\)

\(\Leftrightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

\(\Leftrightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)

\(\Rightarrow200-x=0\)

\(\Leftrightarrow x=200\)

Vậy....

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}=-2\)

\(\Leftrightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=2\)

\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=2\)

\(\Rightarrow x=200-\frac{2}{\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}}\)

Bạn tự bấm máy, kết quả chẳng đẹp gì

\(A=2B\) thì còn có lý

(109-x)/91+(107-x)/93+(105-x)/95+(103-x)/97=-4

[(109-x)/91 +1]+[(107-x)/93 +1]+[(105-x)/95 +1]+[(103-x)/97 +1]-4=-4

(109+91-x)/91+(107+93-x)/93+(105+95-x)/95+(103+97-x)/97=-4+4

(200-x)/91+(200-x)/93+(200-x)/95+(200-x)/97=0

(200-x)(1/91+1/93+1/95+1/97)=0

Ma : 1/91+1/93+1/95+1/97\(\ne\)0

=>200-x=0

=>x=200

6: =>x=9/10+1/5=9/10+2/10=11/10

7: =>x=3/8-5/12=36/96-40/96=-1/24

8: =>x=7/6-5/4=14/12-15/12=-1/12

9: =>x=1/35+2/7=1/35+10/35=11/35

10: =>x=2/7-7/10=20/70-49/70=-29/70

1, \(x=-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{10}{6}-\dfrac{3}{6}=-\dfrac{13}{6}\)

2, \(x=\dfrac{1}{3}-\dfrac{3}{5}=\dfrac{5-9}{15}=-\dfrac{4}{15}\)

4, \(x=-\dfrac{7}{9}+\dfrac{4}{3}=\dfrac{-7+12}{9}=\dfrac{5}{9}\)

5, \(x=\dfrac{5}{6}-\dfrac{7}{3}=\dfrac{5-14}{6}=-\dfrac{9}{6}=-\dfrac{3}{2}\)

6, \(x=\dfrac{9}{10}+\dfrac{1}{5}=\dfrac{9+2}{10}=\dfrac{11}{10}\)

7, \(x=\dfrac{3}{8}-\dfrac{5}{12}=\dfrac{9-10}{12}=-\dfrac{1}{12}\)

8, \(x=\dfrac{7}{6}-\dfrac{5}{4}=\dfrac{14-15}{12}=-\dfrac{1}{12}\)

9, \(x=\dfrac{1}{35}+\dfrac{2}{7}=\dfrac{1+10}{35}=\dfrac{11}{35}\)

10, \(x=-\dfrac{7}{10}+\dfrac{2}{7}=\dfrac{-49+20}{70}=\dfrac{-29}{70}\)