`(2x+5)^4=3^4*5^4`

`(2x+5)^4=(3*5)^4`

`(2x+5)^4=15^4`

`=>2x+5=15`

`x=5`

`6^2+8^2=36+64=100`

`13^2-12^2=169-144=25`

`1^3+2^3+3^3+4^3=1+8+27+64=100`

`3.1D`

`3,2B`

`3,3A`

`3,4A`

`n*n*n+p*p=n^3+p^2`

ủa cái từ ''bạn xem lại đề ạ'' phải ở cuối chứ nhỉ?=)))

có \(VT=\dfrac{\sqrt{1+\dfrac{2\sqrt{2}}{3}}+\sqrt{1-\dfrac{2\sqrt{2}}{3}}}{\sqrt{1+\dfrac{2\sqrt{2}}{3}}-\sqrt{1-\dfrac{2\sqrt{2}}{3}}}\)

\(=\dfrac{\sqrt{\dfrac{3+2\sqrt{2}}{3}}+\sqrt{\dfrac{3-2\sqrt{2}}{3}}}{\sqrt{\dfrac{3+2\sqrt{2}}{3}}-\sqrt{\dfrac{3-2\sqrt{2}}{3}}}\)

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

bạn xem lại đề ạ

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

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

\(=\dfrac{\sqrt{2}+1+\sqrt{2}-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{2}+1-\sqrt{2}+1}\) (vì \(\sqrt{2}+1>0;\sqrt{2}-1>0\))

\(=\dfrac{2\sqrt{2}}{2}\\ =\sqrt{2}\)

`x xx4/5 :2=6/7`

`x xx 4/5xx1/2=6/7`

`x xx2/5=6/7`

`x=6/7:2/5`

`x=6/7xx5/2`

`x=15/7`

b)

`6/5:x:5/4=10/15`

`6/5:x xx4/5=2/3`

`6/5:x=2/3:4/5`

`6/5:x=2/3xx5/4`

`6/5:x=5/6`

`x=6/5:5/6`

`x=6/5xx6/5`

`x=36/25`

a)

\(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{5+2\sqrt{15}+3}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}\\ =\left(\sqrt{9}-\sqrt{15}\right)\cdot\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}\)

\(=\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\) (vì \(\sqrt{5}+\sqrt{3}>0\))

\(=\sqrt{3}\cdot\dfrac{3-5}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-2}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-\sqrt{4}}{\sqrt{2}}\\ =-\sqrt{6}\)

b)

\(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\\ =\sqrt{20-2\cdot3\cdot2\sqrt{5}+9}-\sqrt{20-2\cdot2\cdot2\sqrt{5}+4}\\ =\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}-3\right|-\left|2\sqrt{5}-2\right|\)

\(=2\sqrt{5}-3-\left(2\sqrt{5}-2\right)\) (vì \(2\sqrt{5}-3>0;2\sqrt{5}-2>0\))

\(=2\sqrt{5}-3-2\sqrt{5}+2\\ =-1\)

đổi `1` tạ `=100kg`

1 kg hạt tươi phơi khô thì hạt đó giảm số kg là

`15:100=0,15(kg)`

200 kg hạt tươi đó đem phơi khô thì giảm số kg là

`200xx0,15=30(kg)`

nếu có 200kg hạt tươi đó đem phơi khô thì còn số kg là

`200-30=170(kg)`

a)

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

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

b)

\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)