Đề : Tìm x???

\(\frac{2}{3}x+\frac{1}{4}x=\frac{-22}{27}\)

\(x\left(\frac{2}{3}+\frac{1}{4}\right)=\frac{-22}{27}\)

\(x\left(\frac{8}{12}+\frac{3}{12}\right)=\frac{-22}{27}\)

\(x.\frac{11}{12}=\frac{-22}{27}\)

\(x=\frac{-22}{27}\div\frac{11}{12}\)

\(x=\frac{-22}{27}.\frac{12}{11}\)

\(x=\frac{-8}{9}\)

Vậy \(x=-\frac{8}{9}\).

a)-x=-5=>x=5

b)x=-16

c)x=4

a, \(|-x|=-1+\left(-4\right)\)

<=>\(|-x|=-5\) (vô lý)

vậy pt vô nghiệm 

b, \(|x+22|=6\)

<=>\(\orbr{\begin{cases}x+22=6\\x+22=-6\end{cases}}\) <=>\(\orbr{\begin{cases}x=-16\\x=-28\end{cases}}\)

c, \(|x-1|=2\) <=>\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\) <=>\(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

3x-5 là bội của x+4=>3x-5\(⋮\)x+4=>\(\frac{3x-5}{x+4}\)=\(\frac{3x-5+12-12}{x+4}=3+\frac{-17}{x+4}\)=>\(x+4\inƯ\left(-17\right)=\left(\pm1;\pm17\right)\)

Ta có Bảng

\(3x-5\in B(x+4)\)

\(\Rightarrow(3x-5)⋮(x+4)\)\(\Rightarrow(3x+3\times4-12-5)⋮(x+4)\)

\(\Rightarrow[3\times(x+4)-17]⋮(x+4)\)mà \([3\times(x+4)]⋮(x+4)\)

\(\Rightarrow17⋮(x+4)\)\(\Rightarrow x+4\in\text{Ư}(17)\)

\(\Rightarrow x+4\in-1;1;-17;17\)

\(\Rightarrow x\in-5;-3;-21;13\)

a, x:x+1

b; x-2 ; x-1 ;x

     Bài làm :

a) x ; x+1 ; x+2

b)x-2 ; x-1 ; x

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\Rightarrow4\left|x-7\right|+8+2\left|x-7\right|+10=0\)

\(\Rightarrow6\left|x-7\right|+18=0\) vì \(\left|x-7\right|\ge0\) với mọi x \(\Rightarrow6\left|x-7\right|\ge0\)với mọi x\(\Rightarrow6\left|x-7\right|+18\ge18>0\)

Mà 6|x-7| +18=0 nên \(x\in\varnothing\)

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

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

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

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

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

Vậy x = 1

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

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

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

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

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

=> x + 100 = 0

=> x           = -100

c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)

=> x - 100 = 0

=> x           = 100

Chúc bạn học tốt

có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai

tớ cũng học bồi  dưỡng ,,k minh nha

        giải

ta có:B=1/1x2x3+1/2x3x4+... +1/18x19x220

=>2B=2/1x2x3+2/2x3x4+...2/18x19x20

=(1/1x2-1/2x3)+(1/2x3-1/3x4)+..+(1/18x19-1/19x20)

=1/1x2-1/19x20=189/380

=>B=189/760<1/4

kb đi có gì tớ giải cho

23 . 163-63.23+180     

= 23.( 163 - 63 + 180 ) 

= 23. 280

= 6440

    23. 163 - 63.23 + 180

=  23 . ( 163 - 63 ) + 180

= 23 . 100             + 180

= 2300                  + 180

= 2480

   280 - ( 4 . 5³ - 3² . 23 )

=  280 - ( 4. 125 - 9 . 23 )

= 280 - (   500 - 207 )

= 280 -  293

  133 - 3( x + 5 ) = 46

   3( x + 5 )        = 133 - 46

    3 ( x + 5 )       = 87

    x + 5            = 87 : 3

   x + 5            = 29

hok tốt ##