167,23 - 34,89 - 67,23

= ( 167,23 - 67,23 ) - 34,89

= 100 - 34,89

= 65,11

167,23-34,89-67,23

=(167,23-67,23)-34,89

=100                -34,89

=65,11

167x23-34x89-67x23

=23x(167-67)-34x89

=23x100-34x89

= 2300-3026

=-726

109,5-67,23-9,5

=(109,5-9,5)-67,23

=100-67,23

=32,77

= (109,5-9,5)-67,23

109,5 - 67,23 - 9,5

= ( 109,5 - 9,5 ) - 67,23

= 100 - 67,23

= 32,77

Chúc các bạn buổi sáng tốt lành !
Tk mk mk tk lại 

Bài 1:

\(x+11,2=12,6\times2\\ x+11,2=25,2\\ x=25,2-11,2=14\)

\(2,5\times x=24,4:2\\ 2,5\times x=12,2\\x=\dfrac{12,2}{2,5}=4,88 \)

\(12,76-x=9,5+1,8\\ 12,76-x=11,3\\ x=12,76-11,3=1,46\)

`A=(x^2-2)(x^2+x-1)-x(x^3+x^2-3x-2)`

`=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x`

`=(x^4-x^4)+(x^3-x^3)+(3x^2-x^2-2x^2)+(2x-2x)+2`

`=2`

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\)

\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)

\(\Leftrightarrow\left(x-3\right).\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right).\left(x-4\right)\)

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow3x^2-9x-8x+24=0\)

\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\left(\text{thỏa}\right)\)

\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne\left\{2;4\right\}\right)\\ =>\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\\ =>x^2-3x-4x+12+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\\ =>2x^2-11x+16=-x^2+6x-8\\ =>3x^2-17x+24=0\\ =>\left(x-3\right)\left(3x-8\right)=0\\ =>\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\left(TMDK\right)\)

\(\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\dfrac{-\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)

\(\dfrac{x^2-7x+12}{x^2-6x+8}+\dfrac{x^2-4x+2}{x^2-6x+8}=\dfrac{-x^2+6x-8}{x^2-6x+8}\)

\(\dfrac{x^2-7x+12+x^2-4x+2+x^2-6x+8}{x^2-6x+8}=0\)

\(\dfrac{3x^2-17x+22}{x^2-6x+8}=0\)

\(\dfrac{\left(3x-11\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)

\(\dfrac{3x-11}{x-4}=0\)

Với x khác 4 

=> 3x -11 = 0 

=> \(x=\dfrac{11}{3}\)

rút gọn à banj

đúng rồi á

a, ĐKXĐ:\(x\ne-3\)

\(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+3}{x+3}\\ \Leftrightarrow x+1=1\\ \Leftrightarrow x=0\left(tm\right)\)

b, ĐKXĐ:\(x>2\)

\(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\\ \Leftrightarrow x^2-4x-2=x-2\\ \Leftrightarrow x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

\(a,=\left(x+8-x+2\right)^2=10^2=100\\ b,=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ c,=x^3+1-x^3+1=2\)