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

     \(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)

   \(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)

\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)

==> x+200=0

<=>x=-200

Vậy nghiệm của phương trình là x=-200

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

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

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

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

==>200-x=0

<=>x=200

vậy nghiệm của pt là x=200

a, \(2\left(\frac{11x}{12}+\frac{1}{3}\right)=2-\frac{x}{6}\)

\(2\left(\frac{11x+4}{12}\right)-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-\frac{24}{12}+\frac{2x}{12}=0\)

\(\frac{44x+8-24+2x}{12}=\frac{46x-16}{12}=0\)

\(\Leftrightarrow46x-16=0\)

\(\Leftrightarrow46x=16\Rightarrow x=\frac{8}{23}\)

Vậy nghiệm của pt là x=8/23

k mk

x=-100

\(\frac{x+5}{95}+\frac{x+6}{94}+\frac{x+7}{93}+\frac{x+8}{92}+\frac{x+9}{91}=-5\)

\(\left(\frac{x+5}{95}+1\right)+\left(\frac{x+6}{94}+1\right)+\left(\frac{x+7}{93}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+9}{91}+1\right)=-5+5=0\)

x=100

k hộ với

     30% . x - x - \(\dfrac{5}{6}\) = \(\dfrac{1}{3}\)

\(\Rightarrow\) (\(\dfrac{3}{10}\) - 1) . x = \(\dfrac{1}{3}+\dfrac{5}{6}\)

\(\Rightarrow\) \(-\dfrac{7}{10}\) . x = \(\dfrac{7}{6}\)

\(\Rightarrow\) x = \(-\dfrac{3}{5}\)

KO ghi đề nhé

\(\dfrac{3}{10}.x-x=\dfrac{1}{3}+\dfrac{5}{6}\)

\(x\left(\dfrac{3}{10}-1\right)=\dfrac{2}{6}+\dfrac{5}{6}\)

\(x.\dfrac{-7}{10}=\dfrac{7}{6}\)

\(x=\dfrac{7}{10}:\dfrac{-7}{6}\)

\(x=\dfrac{7}{10}.\dfrac{6}{-7}\)

\(x=\dfrac{42}{-70}\)

\(x=\dfrac{6}{-10}\)

\(x=\dfrac{3}{-5}\)

Vậy \(x=\dfrac{3}{-5}\)

\(30\%.x-x-\dfrac{5}{6}=\dfrac{1}{3}\) 

     \(x.\left(30\%-1\right)=\dfrac{1}{3}+\dfrac{5}{6}\) 

               \(x.\dfrac{-7}{10}=\dfrac{7}{6}\)  

                       \(x=\dfrac{7}{6}:\dfrac{-7}{10}\) 

                       \(x=\dfrac{-5}{3}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow200-x=0\)

\(\Leftrightarrow x=200\)

Vậy....

\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)

\(\Leftrightarrow\frac{x+1}{94}+1+\frac{x+2}{93}+1+\frac{x+3}{92}+1=\frac{x+4}{91}+1+\frac{x+5}{90}+1+\frac{x+6}{89}+1\)

\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)

\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)

\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)

\(\Leftrightarrow x+95=0\).Do \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)

\(\Leftrightarrow x=-95\)

(x+1)/94 + ( x+2)/93 + ( x+3)/92.......

= ................ + ( x+6)/89 
<=> (x+1)/94 + 1 + ( x+2)/93 +1 .........

=.............. cộng 1 nhá 
<=> (x+95)/94 + ( x+96) / 93 + ( x+95)/92

= ( x+95)/91 + ( x+95)/90 + ( x+95)/89 
<=> ( x+95) ( 1/94 +1/93 +1/92 )

= ( x+95) ( 1/91 +1/90 +1/89) 
<=> ( x+95) ( 1/94 +1/93 +1/92 - 1/91 - 1/90 - 1/89 ) 
<=> x+95 =0 
<=>x = -95 
Vậy :x = -95

91 x 157 -  29 x 91 - 91 x 28

=9100

127 x 63 + 19 x 127 - 82 x 26

=8282

2 x 9 x 41 + 36 + 3 x 57 x 6

=1800

29 x 17 + 31 x 29 + 48 x 93 - 48 x 22

=4800

1125 - 25 x 30 + 0,06 : 0,2

=375.3

Bn k cho mk nha!

91 x 157 -  29 x 91 - 91 x 28

= 19 (157 - 29 - 28)

=19 * 100

=1900

127 x 63 + 19 x 127 - 82 x 26

= 127 (63 + 19) - 82 * 26

= 127 * 82 - 82 * 26

= 82 (127 - 26)

= 82 * 101

= 8282

2 x 9 x 41 + 36 + 3 x 57 x 6

= 738 + 36 + 1026

= 1800 

29 x 17 + 31 x 29 + 48 x 93 - 48 x 22

= 29 (17 + 31) + 48 (93 - 22)

= 29 * 48 + 48 * 71 

= 48 (29 + 71) 

= 48 * 100

= 4800

1125 - 25 x 30 + 0,06 : 0,2

= 1125 - 750 + 0,3

= 375 + 0,3

= 375,3

GIẢI

  91 x 157 - 29 x 91 - 91 x 28

=91 x (157 + 29 + 91)

=91 x 277

=25207

  127 x 63 + 19 x 127 - 82 x 26

= 127 x (63 +19) - 2132

= 127 x   92        - 2132

= 11684              - 2132

=  9552

  29 x 17 + 31 x 29 + 48 x 93 - 48 x 22

=29 x(17 +31)        + 48 x(93 - 22)

=29 x 48                + 48 x71

=48 x (29 + 71)

=48 x 100

=4800

  1125 - 25 x 30 + 0,06 : 0,2

= 1100       x 30 + 0,3

= 33000             +0,3

= 33000,3

 K HỘ MIK NGHEN