Bộ 5 đề thi giữa kì 1 Toán 7 Cánh Diều (2023-2024) có đáp án - Đề 5

a $\frac{5}{6} + \frac{{ - 11}}{{30}} + \frac{7}{5}$

$= \frac{{25}}{{30}} + \frac{{ - 11}}{{30}} + \frac{{42}}{{30}}$

$= \frac{{56}}{{30}} = \frac{{28}}{{15}}$

b ${\left( {\frac{{ - 4}}{5}} \right)^2} \cdot \frac{{57}}{2} + \frac{{ - 7}}{2} \cdot {\left( {\frac{{ - 4}}{5}} \right)^2}$

$$= {\left( {\frac{{ - 4}}{5}} \right)^2} \cdot \left( {\frac{{57}}{2} + \frac{{ - 7}}{2}} \right)$$

$$= \frac{{16}}{{25}} \cdot \frac{{50}}{2} = 16.$$

c $\left( {\frac{7}{{25}} - 1\frac{2}{9}} \right) - \left( {\frac{{23}}{{54}} - \frac{{18}}{{25}}} \right) + \frac{{ - 31}}{{54}}$

$$= \frac{7}{{25}} - 1\frac{2}{9} - \frac{{23}}{{54}} + \frac{{18}}{{25}} + \frac{{ - 31}}{{54}}$$

$$= \left( {\frac{7}{{25}} + \frac{{18}}{{25}}} \right) + \left( {\frac{{ - 31}}{{54}} + \frac{{ - 23}}{{54}}} \right) - 1\frac{2}{9}$$

$= 1 + \left( { - 1} \right) - \frac{{11}}{9}$$= \frac{{ - 11}}{9}$.

a $\frac{5}{2} - x = \frac{{27}}{8}$

$x = \frac{5}{2} - \frac{{27}}{8}$

$x = \frac{{20}}{8} - \frac{{27}}{8}$

$x = \frac{{ - 7}}{8}$

Vậy $x = \frac{{ - 7}}{8}.$

b $\frac{{ - 1}}{3} + \frac{4}{{10}}x = 0{,}2$

$\frac{4}{{10}}x = 0{,}2 - \left( {\frac{{ - 1}}{3}} \right)$

$\frac{4}{{10}}x = \frac{8}{{15}}$

$x = \frac{8}{{15}}:\frac{4}{{10}}$

$x = \frac{4}{3}$

Vậy $x = \frac{4}{3}$

c ${\left( {3x + 2} \right)^2} = \frac{{25}}{{49}}$

$${\left( {3x + 2} \right)^2} = {\left( {\frac{5}{7}} \right)^2} = {\left( {\frac{{ - 5}}{7}} \right)^2}$$

Trường hợp 1: $3x + 2 = \frac{5}{7}$

$$3x = \frac{5}{7} - 2$$

$$3x = \frac{{ - 9}}{7}$$

$$x = \frac{{ - 9}}{7}:3$$

$$x = \frac{{ - 3}}{7}$$

Trường hợp 2: $3x + 2 = \frac{{ - 5}}{7}$

$$3x = \frac{{ - 5}}{7} - 2$$

$$3x = \frac{{ - 19}}{7}$$

$$x = \frac{{ - 19}}{7}:3$$

$$x = \frac{{ - 19}}{{21}}$$

Vậy $x \in \left\{ {\frac{{ - 3}}{7};\frac{{ - 19}}{{21}}} \right\}.$

Có: $\widehat {AOH}$ và $\widehat {AOK}$ là 2 góc kề bù

nên $\widehat {AOH} + \widehat {AOK} = 180^\circ$

$\widehat {AOH} + 120^\circ = 180^\circ$

$\widehat {AOH} = 180^\circ - 120^\circ$

$\widehat {AOH} = 60^\circ$

a Thể tích của bể bơi là: $8 \cdot 4 \cdot 0{,}5 = 16\;\left( {{{\rm{m}}^{\rm{3}}}} \right)$.

b Đổi: $16\;\left( {{{\rm{m}}^{\rm{3}}}} \right) = 16\,\,000\left( {{\rm{d}}{{\rm{m}}^{\rm{3}}}} \right) = 16\,\,000\,\,\left( l \right)$

Lượng nước máy bơm bơm vào bể sau $5$ giờ $5$ phút (bằng $305$ phút) là: $50 \cdot 305 = 15\,\,250\;\left( l \right)$

Vì: $15\,\,250 < 16\;\,000$

Vậy nên sau $5$ giờ $5$ phút bể vẫn chưa đầy nước.

$A = \frac{{{7^{2011}} + 2}}{{{7^{2012}} + 2}}$suy ra $7 \cdot A = \frac{{{7^{2012}} + 14}}{{{7^{2012}} + 2}} = 1 + \frac{{12}}{{{7^{2012}} + 2}}$

$B = \frac{{{7^{2023}} + 2}}{{{7^{2024}} + 2}}$suy ra $7 \cdot B = \frac{{{7^{2024}} + 14}}{{{7^{2024}} + 2}} = 1 + \frac{{12}}{{{7^{2024}} + 2}}$

Có ${7^{2012}} + 2 < {7^{2024}} + 2$ suy ra $\frac{{12}}{{{7^{2012}} + 2}} > \frac{{12}}{{{7^{2024}} + 2}}$

Nên $A > B.$