Phân số chỉ tổng số tấn hàng lấy ra :

`50% + 9/14  8/7`

Phân số chỉ số tấn hàng còn lại :

`1-8/9=1/9`

Lúc đầu kho đó có :

`145:1/9=1305(tấn)`

Đ/s...

`#LeMichael`

`2/3 . x = (-5)/17`

`x= (-5)/17 : 2/3`

`x= (-5)/17 xx 3/2`

`x= (-5.3)/(17.2)`

`x=(-15)/34`

\(\dfrac{2}{3}.x=-\dfrac{5}{17}\)

\(x=-\dfrac{5}{17}:\dfrac{2}{3}\)

\(x=-\dfrac{5}{17}.\dfrac{3}{2}\)

\(x=-\dfrac{15}{34}\)

#Hphong2010

`a xx 12 +a + 25 xx a = 3268`

`a xx 12 + a xx 1 + 25 xx a = 3268`

`a xx (1+ 12 + 25) = 3268`

`a xx 38=3268`

`a=3268:38`

`a=86`

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`29xxa-a=896`

`29xxa-axx1=896`

`(29-1)xxa=896`

`28xxa=896`

`a=896:28`

`a=32`

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`4 xx a - 7 = 13`

`4 xx a = 13 + 7`

`4 xx a  =20`

`a=20:4`

`a=5`

____________________________

`1/6 = a/48`

`=> a = 1/6 xx 48`

`=>a = 8`

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`a/5 = 72/45`

`=> a = 72/45xx5`

`=> a =8`

__________________________

`a/9 = 34/51`

`a = 34/51 xx 9`

`a=6`

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`(a+4)/16 = 75/100`

`(a+4)/16 = 3/4`

`a+4=3/4xx16`

`a+4=12`

`a=12-4`

`a=8`

giả sử: A = n(n+1) , có 2 trường hợp:

nếu n chẵn thì n chia hết cho 2 do đó A chia hết chia 2

nếu n lẻ thì n+1 chẵn do đó n+1 chia hết cho 2 nên A chia hết cho 2

\(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)

\(\Leftrightarrow\left(\sqrt{x^2-4x+5}-1\right)+\left(\sqrt{x^2-4x+8}-2\right)+\left(\sqrt{x^2-4x+9}-\sqrt{5}\right)=0\)

\(\Leftrightarrow\dfrac{x^2-4x+5-1}{\sqrt{x^2-4x+5}+1}+\dfrac{x^2-4x+8-4}{\sqrt{x^2-4x+8}+2}+\dfrac{x^2-4x+9-5}{\sqrt{x^2-4x+9}+\sqrt{5}}=0\)\(\left(Vì:\sqrt{x^2-4x+5}+1>0;\sqrt{x^2-4x+8}+2>0;\sqrt{x^2-4x+9}+\sqrt{5}>0\right)\)

\(\Leftrightarrow\left(x^2-4x+4\right)\left(\dfrac{1}{\sqrt{x^2-4x+5}+1}+\dfrac{1}{\sqrt{x^2-4x+8}+2}+\dfrac{1}{\sqrt{x^2-4x+9}+\sqrt{5}}\right)=0\)

\(\Leftrightarrow x^2-4x+4=0\left(Vì:\dfrac{1}{\sqrt{x^2-4x+5}+1}+\dfrac{1}{\sqrt{x^2-4x+8}+2}+\dfrac{1}{\sqrt{x^2-4x+9}+\sqrt{5}}>0\right)\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x=2\)

Ta có: 

\(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}\)

\(=\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}\ge1+2+\sqrt{5}=3+\sqrt{5}\)

Dấu ''='' xảy ra khi và chỉ khi \(x-2=0\)\(\Leftrightarrow x=2\)

Vậy x=2

`70-5(x-3)=45`

`=>5(x-3)=70-45=25`

`=>x-3=25:5=5`

`=>x=5+3=8`

`70-5.(x-3)=45`

`5.(x-3)=70-45`

`5.(x-3)=25`

`x-3=25:5`

`x-3=5`

`x=5+3`

`x=8`

`136:x+34:x=10`

`(136+34):x=10`

`170:x=10`

`x=170:10`

`x=17`

Vậy `x=17`

`136:x+34:x=10`

`(136+34):x=10`

`170:x=10`

`x=170:10=17`