\(A=\dfrac{404\times25+75\times404}{2038-9-9}\)

\(=\dfrac{404\times\left(25+75\right)}{2029-9}\)

\(=\dfrac{404\times100}{2020}\)

\(=\dfrac{40400}{2020}\)

\(=20\)

#\(Toru\)

chịu

căn bậc hai của 404

\(460+85\times4=\left(x+200\right)\times4\)

\(\left(x+200\right)\times4=460+340\)

\(\left(x+200\right)\times4=800\)

\(x+200=800:4\)

\(x+200=200\)

\(x=200-200\)

\(x=0\)

~~~

\(\left(x-7\right)\left(2x-8\right)=0\)

\(+, TH1: x - 7 = 0\)

\(x=0+7\)

\(x=7\)

\(+, TH2 : 2x - 8 = 0 \)

\(2x=0+8\)

\(2x=8\)

\(x=8:2\)

\(x=4\)

~~~

\(x-280:35=5\times54\)

\(x-8=270\)

\(x=270+8\)

\(x=278\)

~~~

\(324+16\times\left(2x+3\right)=404\)

\(16\times\left(2x+3\right)=404-324\)

\(16\times\left(2x+3\right)=80\)

\(2x+3=80:16\)

\(2x+3=5\)

\(2x=5-3\)

\(2x=2\)

\(x=2:2\)

\(x=1\)

#\(Toru\)

`460 + 85 xx 4 = ( x + 200) xx 4`

`460 + 340 = (x+200)xx4`

` 800= (x+200)xx4`

`x+200=800:4`

`x+200=200`

`x=200-200`

`x=0`

__

`(x-7)(2x-8)=0`

`@ TH1`

`x-7=0`

`x=0+7`

`x=7`

`@ TH2`

`2x-8=0`

`2x=0+8`

`2x=8`

`x=8:2`

`x=4`

__

`x -280 : 35=5xx54`

`x -280 : 35=270`

`x-8=270`

`x=270+8`

`x=278`

__

`324 + 16xx(2x+3)=404`

`16xx(2x+3)=404 -324`

`16xx(2x+3)=80`

`2x+3=80:16`

`2x+3=5`

`2x=5-3`

`2x=2`

`x=2:2`

`x=1`

Ta có: \(x-5⋮x-1\)

=> \(\left(x-1\right)-4⋮x-1\)

=> \(-4⋮x-1\)

Vì \(x\in Z\Rightarrow x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng sau:

Vậy \(x\in\left\{2;0;3;-1;5;-3\right\}\)

X x ( 4,2 + 0,8 ) = 2,64
X x 5                   = 2,64
X                         = 2,64 : 5
X                         = 0,528

X x ( 4,2 + 0,8 ) = 2,64
X x 5                   = 2,64
X                         = 2,64 : 5
X                         = 0,528 ok

1794:x+358=404

1794:x=404-358

1794:x=46

x=1794:46

x=39

nhớ tk m nhé

39 bn nhé