Đổi 5 tấn = 5000kg; 1 hec-ta= 10000m²

Lượng thóc thu được 1 sào bắc bộ:

(360:10000) x 5000= 180(kg)

Lượng gạo thu được trên 1 sào bắc bộ:

180 x 4/5 = 144(kg)

Đ.số: 144kg gạo

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)

5a.

$\frac{1}{2}-\sqrt{x}=0$

$\Rightarrow \sqrt{x}=\frac{1}{2}$
$\Rightarrow x=(\frac{1}{2})^2=\frac{1}{4}$

5b.

$\frac{5}{11}\sqrt{x}-\frac{1}{6}=\frac{1}{3}$

$\Rightarrow \frac{5}{11}\sqrt{x}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$

$\Rightarrow \sqrt{x}=\frac{1}{2}: \frac{5}{11}=\frac{11}{10}$

$\Rightarrow x=(\frac{11}{10})^2=\frac{121}{100}$

5c.

$-\frac{4}{3}\sqrt{x}+\frac{8}{5}=\frac{1}{3}+\frac{2}{3}=1$

$\Rightarrow -\frac{4}{3}\sqrt{x}=1-\frac{8}{5}=\frac{-3}{5}$

$\Rightarrow \frac{4}{3}\sqrt{x}=\frac{3}{5}$

$\Rightarrow \sqrt{x}=\frac{3}{5}: \frac{4}{3}=\frac{9}{20}$
$\Rightarrow x=(\frac{9}{20})^2=\frac{81}{400}$

5d.

$x-6\sqrt{x}=0$

$\Rightarrow \sqrt{x}(\sqrt{x}-6)=0$

$\Rightarrow \sqrt{x}=0$ hoặc $\sqrt{x}-6=0$

$\Rightarrow \sqrt{x}=0$ hoặc $\sqrt{x}=6$

$\Rightarrow x=0$ hoặc $x=36$

5e.

$1-3x^2=7$

$3x^2=1-7=-6$

$x^2=-2<0$ (vô lý)

Do đđ không tồn tại $x$ thỏa mãn đề.

5f.

$7x^2-4=1$

$7x^2=1+4=5$

$x^2=\frac{5}{7}=(\sqrt{\frac{5}{7}})^2=(-\sqrt{\frac{5}{7}})^2$

$\Rightarrow x=\pm \sqrt{\frac{5}{7}}$

C = 1/(9.10) - 1/(8.9) - 1/(7.8) - ... - 1/(2.3) - 1/(1.2)

= 1/9 - 1/10 - 1/8 + 1/9 - 1/7 + 1/8 - ... - 1/2 + 1/3 - 1 + 1/2

= 1/9 - 1/10 + 1/9 - 1

= 2/9 - 11/10

= -79/90

\(B=\left(8-2,25+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+1\dfrac{1}{4}\right)-\left(3+0,5-1\dfrac{2}{7}\right)\\ B=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{1}{2}+\dfrac{9}{7}\\ B=\left(8+6-3\right)+\left(-\dfrac{9}{4}-\dfrac{5}{4}\right)+\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\dfrac{1}{2}\)

\(B=11-\dfrac{7}{2}+2-\dfrac{1}{2}\\ B=\left(11+2\right)+\left(-\dfrac{7}{2}-\dfrac{1}{2}\right)\\ B=13-4\\ B=9.\)

\(\dfrac{-5}{4}+x:\dfrac{12}{3}=\dfrac{-4}{3}\)

\(x:\dfrac{12}{3}=\dfrac{-4}{3}+\dfrac{5}{4}\)

\(x:4=\dfrac{-1}{12}\)

\(x=\dfrac{-1}{3}\)

\(\left(3-\dfrac{1}{2}\right)^3-\dfrac{5^4}{25}+\left[\left(\dfrac{1}{2}\right)^2\right]^3\)

\(=\left(\dfrac{5}{2}\right)^3-\dfrac{5^4}{5^2}+\left[\left(\dfrac{1}{2}\right)^2\right]^3\)

\(=\left(\dfrac{5}{2}\right)^3+\left[\left(\dfrac{1}{2}\right)^2\right]^3-5\)

\(=\left(\dfrac{5}{2}+\dfrac{1}{2}\right)\left[\left(\dfrac{5}{2}\right)^2-\dfrac{5}{2}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-5\)

\(=3\left[\dfrac{25}{4}+\dfrac{1}{4}-\dfrac{5}{4}\right]-5\)

\(=3.\dfrac{21}{4}-5\)

\(=\dfrac{63}{4}-5=\dfrac{43}{4}\)

Đính chính \(\dfrac{5^4}{5^2}=25\)

\(...=\dfrac{63}{4}-25=-\dfrac{37}{4}\)

 Theo đề bài, ta có \(AB=8\) nên ta lập bảng:

 Thử lại, ta chỉ thấy số 81 là thỏa mãn yêu cầu bài toán. Vậy hiệu \(A-B=7\)