x=291

x=21

x.84=252.97

x=24444:84

x=291

x.24=9.56

x=504:24

x=21

k nha

1/ x=291

2/ x=21

CHÚC BẠN HỌC GIỎI

TK MÌNH NHÉ

a) Ta có:

252/x = 84/97

==> x.84=252.97

==> x.84= 24444

==> x=24444:84

==> x=291

b) x/17 + 19/34=1/2

X/17=1/2–19/34

X/17=17/34–19/34

X/17=—2/34

==> x.34=17.(—2)

==> x.34=-34

==> x=(—34):34

==> x=—1

+ Tìm x

\(\frac{3}{x}\)= \(\frac{-36}{84}\)=> x . ( - 36 ) = 3 . 84

                            => x = \(\frac{3.84}{-36}\)=> x = - 7

+ Tìm y

\(\frac{y}{35}\)= \(\frac{-36}{84}\)=> y . 84 = ( - 36 ) . 35

                               => y = \(\frac{\left(-36\right).35}{84}\)=> y = - 15

X= 252x97:84

=> X= 291

x = 291

\(\frac{252}{x}=\frac{84}{97}\)

\(\Leftrightarrow x.84=252.97\)

\(\Leftrightarrow x.84=24444\)

\(\Leftrightarrow x=\frac{24444}{84}=291\)

   Vậy \(x=291\)

Chú ý: Dấu \(.\) là dấu nhân nhé

252.97=x.84

24444=x.84

x=291

\(\frac{252}{x}=\frac{84}{97}\)

\(\Rightarrow252:x=\frac{84}{97}\)

              \(x=252:\frac{84}{97}\)

              \(x=291\)

x=291

\(\frac{252}{x}\)=\(\frac{84}{97}\)

Vì 252:84=3

=> x:97=3

     x=3*97

     x= 291

=> \(\frac{252}{291}\)=\(\frac{84}{97}\)

Câu a bn cứ phân tích 3^x-2 + 3^x-5 ra 3^x : 3² + 3^x : 3⁵ là ra nhé!

b, \(\left(\frac{3}{4}-x\right)^2=\frac{25}{36}\)

Ta thấy: \(\frac{25}{36}=\left(\frac{5}{6}\right)^2\Rightarrow\left(\frac{3}{4}-x\right)^2=\left(\frac{5}{6}\right)^2\Rightarrow\frac{3}{4}-x=\frac{5}{6}\)

\(\Rightarrow x=\frac{3}{4}-\frac{5}{6}=\frac{9}{12}-\frac{10}{12}=\frac{-1}{12}\)

c, \(|2015x-1|-|-2015|=-2014\)

\(\Rightarrow|2015x-1|-2015=-2014\)

\(\Rightarrow|2015x-1|=-2014+2015=1\)

Bn cứ đưa ra 2 trường hợp \(2015x-1=1\)và \(2015x-1=-1\) là được nhé!

Ban co the lam chi tiet phan a giup mik dc ko?

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

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

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

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

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

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

\(\frac{x-3}{2013}+\frac{x-4}{2012}=\frac{x-5}{2011}+\frac{x-6}{2010}\)

\(\Leftrightarrow\frac{x-3-2013}{2013}+\frac{x-2-2012}{2012}=\frac{x-5-2011}{2011}+\frac{x-6-2010}{2010}\)(mỗi vế trừ đi 2)

\(\Leftrightarrow\frac{x-2016}{2013}+\frac{x-2016}{2012}-\frac{x-2016}{2011}-\frac{x-2016}{2010}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)

Mà \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)

\(\Rightarrow x-2016=0\Leftrightarrow x=2016\)

Cộng mỗi vế cho 1 

Ta có: \(\frac{x-3-2013}{2013}+\frac{x-4-2012}{2012}=\frac{x-5-2011}{2011}+\frac{x-6-2010}{2010}\)

\(=>\left(\frac{x-2016}{2013}+\frac{x-2016}{2012}\right)-\left(\frac{x-2016}{2011}+\frac{x-2016}{2010}\right)=0\)

\(=>\left(x-2016\right).\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)\)

Mà \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\ne0\)

\(=>x-2016=0\\ =>x=2016\)