Chiều cao khu đất là:
`80 xx 3/4 =60(dm)`

Diện tích khu đất là:
`80xx60=4800`(dm²)

Đáp số: 4800dm²

Chiều cao của hình thang là:
`54:18=3(cm)`

Đáp số: 3cm

a,`9/5+2/5xx4/6=9/5+4/15=27/15+4/15=31/15`

b,`3/8xx2-6/7xx1/3=3/4-2/7=21/18-8/28=13/28`

a, `x-4,03=5,94`

`x=5,94+4,03`

`x=9,97`

b, `13,7+x=59,8`

`x=59,8-13,7`

`x=46,1`

`(x-4/3)xx7/4=5-7/6`

`(x-4/3)xx7/4=23/6`

`x-4/3=23/6:7/4`

`x-4/3=46/21`

`x=46/21+4/3`

`x=74/21`

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

Thay \(x=6-2\sqrt{5}\) vào B ta có:

\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\sqrt{6-2\sqrt{5}}-1}{\sqrt{6-2\sqrt{5}}+1}\\ =\dfrac{\sqrt{5-2\sqrt{5}+1}-1}{\sqrt{5-2\sqrt{5}+1}+1}\\ =\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}\\ =\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}\\ =\dfrac{\sqrt{5}-2}{\sqrt{5}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{5}\\ =\dfrac{5-2\sqrt{5}}{5}\)

a, `2/13xx9/14+11/13xx9/14`

`=9/14xx(2/13+11/13)`

`=9/14xx13/13`

`=9/14xx1`

`=9/14`

b, `20/17+7/4-3/17-3/4`

`=(20/17-3/17)+(7/4-3/4)`

`=17/17+4/4`

`=1+1`

`=2`

Đổi 2h45'=2,75h

Độ dài quãng đường người đó đi được là:
`42xx2,75=115,5(km)`

Đáp số: 115,5km

\(a,\left(2x-3\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=-2\end{matrix}\right.\\ b,2x-\left(3-5x\right)=4\left(x+3\right)\\ \Leftrightarrow2x-3+5x=4x+12\\ \Leftrightarrow7x-3-4x-12=0\\ \Leftrightarrow3x-15=0\\ \Leftrightarrow x=5\)

\(c,ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}-\dfrac{2}{x+1}=\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}-\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}-\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x+1-x+2-11+3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow3x-8=0\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)