Bạn tham khảo!
Bài 1.
$a)$ $\dfrac{7}{4}$ $+$ $\dfrac{-3}{5}$
$=$ $\dfrac{35}{20}$ $+$ $\dfrac{-12}{20}$
$=$ $\dfrac{23}{20}$
$b)$ $2021$ $-$ $($ $\dfrac{1}{3}$ $)^{2}$ $.$ $3^{2}$
$=$ $2021$ $-$ $($ $\dfrac{1}{3}$ $.$ $3$ $)^{2}$
$=$ $2021$ $-$ $1^{2}$
$=$ $2020$
$c)$ $7,5$ $.$ $($ $\dfrac{-3}{5}$ $)$
$=$ $\dfrac{-9}{2}$
$d)$ $($ $\dfrac{-1}{4}$ $)^{2}$ $.$ $\dfrac{4}{11}$ $+$ $\dfrac{7}{11}$ $.$ $($ $\dfrac{-1}{4}$ $)^{2}$
$=$ $\dfrac{1}{16}$ $.$ $\dfrac{4}{11}$ $+$ $\dfrac{7}{11}$ $.$ $\dfrac{1}{16}$
$=$ $\dfrac{1}{16}$ $.$ $($ $\dfrac{4}{11}$ $+$ $\dfrac{7}{11}$ $)$
$=$ $\dfrac{1}{16}$ $.$ $1$
$=$ $\dfrac{1}{16}$
Bài 2.
$a)$ $x$ $+$ $5,5$ $=$ $7,5$
$→$ $x$ $=$ $7,5$ $-$ $5,5$
$→$ $x$ $=$ $2$
$b)$ $\dfrac{2}{3}$ $.$ $x$ $-$ $\dfrac{1}{2}$ $=$ $\dfrac{4}{9}$
$→$ $\dfrac{2}{3}$ $.$ $x$ $=$ $\dfrac{17}{18}$
$→$ $x$ $=$ $\dfrac{17}{12}$
Bài 3.
Gọi số học sinh giỏi, khá, trung bình của khối 7 là $a$ $;$ $b$ $;$ $c$ (hs); (a;b;c ∈ N*)
Theo bài ra ta có :
$\dfrac{a}{4}$ $=$ $\dfrac{b}{5}$ $=$ $\dfrac{c}{7}$ và $a$ $+$ $b$ $+$ $c$ $=$ $336$
Áp dụng t/c dãy tỉ số $=$ nhau ta có :
$\dfrac{a}{4}$ $=$ $\dfrac{b}{5}$ $=$ $\dfrac{c}{7}$ $=$ $\dfrac{a+b+c}{4+5+7}$ $=$ $\dfrac{336}{16}$ $=$ $21$
$→$ $\dfrac{a}{4}$ $=$ $21$ $→$ $a$ $=$ $84$ (hs)
và $\dfrac{b}{5}$ $=$ $21$ $→$ $b$ $=$ $105$ (hs)
và $\dfrac{c}{7}$ $=$ $21$ $→$ $c$ $=$ $147$ (hs)
Vậy .....
Bài 4.
$a)$ Vì p//q
AB⊥p
$→$ AB⊥q (Quan hệ từ ⊥ đến //)
$b)$
Ta có : ∠D1 = ∠D2 = $70^{o}$ (2 góc đối đỉnh)
$c)$
Vì AB⊥q
$→$ ∠B1 = $90^{o}$
Vì p//q
$→$ ∠D1 + ∠C2 = $180^{o}$ (2 góc tcp)
$→$ ∠C2 = $180^{o}$ - $70^{o}$
$→$ ∠C2 = $110^{o}$
$FbBinhne2k88$
