bằng âm nha

-90 nha bn

310 + x = 20.11 

310 + x  = 220

         x  = 220  - 310

        x   = - 90

Ủng hộ nha! 

310 + x = 20 x 11

310 + x = 220

          x  = 220 - 310

         x   =   220 + ( -310)

        x    = -90

ĐK : 4x \(\ge\)0 => x\(\ge\)0

=> |x + 19| \(\ge\)0 ; |x + 5| \(\ge\)0 ; |x + 20.11| \(\ge\)0

Khi đó : 

 x + 19 + x + 5 + x + 20.11 = 4x

=> 3x + 19 + 5 + 220          = 4x

=> 3x + 244                         = 4x

=> 244                                 = 4x - 3x

=> 224                                  = x

=> x                                      = 224 (t/m Đk)

cmn lỗi đánh máy : x = 244 nha >:

mình cảm ơn nhiều nhé

chúc bạn học tốt

\(20,11\times96+20,11\times0,25=20,11\times\left(96+0,25\right)=20,11\times96,25=1935,5875\)

Tham khảo

\(\Leftrightarrow2011.\left|x-2011\right|+\left(x-2011\right)^2=2013.\left|x-2011\right|\)

\(\Leftrightarrow\left(x-2011\right)^2=2.\left|x-2011\right|\Rightarrow\orbr{\begin{cases}x-2011=0\\2=\left|x-2011\right|\end{cases}}\Rightarrow\orbr{\begin{cases}x=2013\\x=2009\end{cases}}\text{hoặc }x=2011\)

Vậy ....

X x 2011-X=2011x2009+2011

Xx(2011-1)=2011x(2009+1)

Xx2010    = 2011x2010

=> X=2011

x=2011

x * 2011 - x = 2011 x 2009 + 2011

x * ( 2011 - 1 ) = 2011 x ( 2009 + 1 )

x * 2010 = 2011 x 2010

x = 2011

\(2011-\left|x-2011\right|=x\)

\(\Rightarrow\left|x-2011\right|=2011-x\)

\(\Rightarrow x-2011=\pm\left(2011-x\right)\)

+) \(x-2011=2011-x\)

\(\Rightarrow2x=2011+2011\)

\(\Rightarrow x=2011\) ( chọn )

+) \(x-2011=-\left(2011-x\right)\)

\(\Rightarrow x-2011=-2011+x\)

\(\Rightarrow0=0\) ( vô lí )

Vậy \(x=2011\)

$2011 - |x-2011| = x$

$\iff |x-2011| = 2011 - x$

$\iff x -2011 \leqslant 0$

$\iff x \leqslant 2011$

Đặt (x-2011)^x ra ngoài . Làm tương tự như bài lúc nãy nhé. 

x=2011

\(\left(x-2011\right)^{x+1}-\left(x-2011\right)^{x+2011}=0\)

\(\Rightarrow\left(x-2011\right)^{x+1}.\left(1-x^{2010}\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-2011\right)^{x+1}=0\\1-x^{2010}=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=2011\\x=\pm1\end{array}\right.\)

Vậy x = 2011 ; x = 1 ; x = - 1