Ta có :

\(\sqrt{2}+\sqrt{3}=\left(\sqrt{2}+\sqrt{3}\right)^2\\ =\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2\\ =2+2\sqrt{6}+3\\ =5+2\sqrt{6}\left(5+2\sqrt{6}>5+2\sqrt{4}\right)\)

Mà \(5+2\sqrt{4}=5+2\cdot2=5+4=9\)

\(\Rightarrow5+2\sqrt{6}>9\)

Nên \(\sqrt{2}+\sqrt{3}>3\)

`(-2 1/3 ) .( - (-6)/4 )`

`= - (2xx3+1)/3 . 6/4`

`= - 7/3 . 6/4`

`= -42/12`

`= -7/2`

`32: ( 3xx x -2)=8`

`3xx x-2=32:8`

`3xx x-2=4`

`3 xx x=4+2`

`3xx x=6`

`x=6:3`

`x=2`

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`75 : (x-18) =25`

`x-18=75:25`

`x-18= 3`

`x=3+18`

`x=21`

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`(15-6 xx x ) xx 243 =729`

`15-6 xx x = 729 : 243`

`15-6 xx x = 3`

`6 xx x=15-3`

`6 xx x=12`

`x=12:6`

`x=2`

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`4 xx (x-12)+9=17`

`4 xx (x-12)=17-9`

`4 xx (x-12)= 8`

`x-12=8:4`

`x-12=2`

`x=2+12`

`x=14`

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`20-2xx(x+4)=4`

`2xx(x+4)=20-4`

`2xx(x+4)=16`

`x+4=16:2`

`x+4=8`

`x=8-4`

`x=4`

\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)

\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)

`b,` Tớ tính mãi ko ra, xl cậu nha=')

`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`

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`(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`

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`x -280 : 35=5xx54`

`x -280 : 35=270`

`x-8=270`

`x=270+8`

`x=278`

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`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`

hình như có 3z đkhong e nếu 3z thì 36 nhe