Câu 5:

$\frac{20}{\sqrt{5}}=\frac{20\sqrt{5}}{5}=4\sqrt{5}$

Câu 6:

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

Câu 7:

1. ĐKXĐ: $x\neq 1; x\geq 0$

\(A=\left[\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}+1}+1\right]:\left[\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}-1\right]=(\sqrt{x}+1):(\sqrt{x}-1)\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

2.

\(A< 1\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-1}-1<0\Leftrightarrow \frac{2}{\sqrt{x}-1}<0\)

\(\Leftrightarrow \sqrt{x}-1<0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ suy ra $0\leq x< 1$

H1: x = 360^o - 130^o - 60^o - 82^o = 88^o

H2: x = 360^o - 90^o - 90^o - 72^o = 108^o

H3: x = 360^o - 90^o - 115^o - 70^o = 85^o

H4: 2x = 360^o - 71^o - 105^o = 184^o

=> x = 184^o : 2 = 62^o

\(=\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a+2\cdot\dfrac{sin^2a}{cos^2a}\cdot cos^4a\)

\(=1-2\cdot sin^2a\cdot cos^2a+2\cdot sin^2a\cdot cos^2a\)

=1

Lời giải:
a. Diện tích quét vôi bể nước là:

$5\times 7+2\times 5\times 4+2\times 7\times 4=131$ (m²)

b.

Thể tích bể là:

$7\times 5\times 4=140$ (m³)

Thể tích bể không có nước là:

$140\times (100-80):100=28$ (m³)

\(\dfrac{19}{20}-x=\dfrac{8}{5}+\dfrac{3}{4}\)
\(\dfrac{19}{20}-x=\dfrac{32}{20}+\dfrac{15}{20}\)
\(\dfrac{19}{20}-x=\dfrac{47}{20}\)
\(x=\dfrac{19}{20}-\dfrac{47}{20}\)
\(x=\dfrac{-28}{20}=\dfrac{-7}{5}\)

\(\dfrac{19}{20}-x=\dfrac{8}{5}+\dfrac{3}{4}\\ \dfrac{19}{20}-x=\dfrac{47}{20}\\ x=\dfrac{19}{20}-\dfrac{47}{20}\\ x=-\dfrac{28}{20}\\ x=-\dfrac{7}{5}.\)