PHIẾU BÀI TẬP TOÁN 7 TUẦN 01
Đại số 7 : § 1: Tập hợp Q các số hữu tỉ
Hình học 7: § 1: Hai góc đối đỉnh
Bài 1: Điền các kí hiệu N, Z, Q vào dấu … (viết đầy đủ các trường hợp):
a) 2000 $\in $ … b) $\frac{4}{5}\,\,\in \,\,...$ c) $\frac{-7}{100}\,\,\in \,\,...$
d) -671 $\in $… e) $\frac{-671}{1}\,\,\in \,\,...$
Bài 2: Cho số hữu tỉ $\frac{a}{b}$ khác 0. Chứng minh:
a) Nếu a, b cùng dấu thì $\frac{a}{b}$ là số dương.
b) Nếu a, b trái dấu thì $\frac{a}{b}$ là số âm.
Bài 3: So sánh các số hữu tỉ sau:
a) $\frac{-13}{40}\,\,\,v\text{ }\!\!\grave{\mathrm{a}}\!\!\text{ }\,\,\frac{12}{-40}$ b) $\frac{-5}{6}\,\,\,v\text{ }\!\!\grave{\mathrm{a}}\!\!\text{ }\,\,\frac{-91}{104}$ c) $\frac{-15}{21}\,\,\,v\text{ }\!\!\grave{\mathrm{a}}\!\!\text{ }\,\,\frac{-36}{44}$
d) $\frac{-16}{30}\,\,\,v\text{ }\!\!\grave{\mathrm{a}}\!\!\text{ }\,\,\frac{-35}{84}$ e) $\frac{-5}{91}\,\,\,v\text{ }\!\!\grave{\mathrm{a}}\!\!\text{ }\,\,\frac{-501}{9191}$ f) $\frac{-11}{{{3}^{7}}\,.\,{{7}^{3}}}\,\,\,v\text{ }\!\!\grave{\mathrm{a}}\!\!\text{ }\,\,\frac{-78}{{{3}^{7}}{{.7}^{4}}}$
Bài 4: Tìm tất cả các số nguyên x để các phân số sau có giá trị là số nguyên:
a) $A\,=\,\,\frac{x\,+\,1}{x\,-\,2}\,\left( x\,\,\ne \,2 \right)$ b) $B\,\,=\,\,\frac{2x\,-\,1}{x\,+\,5}\,\,\left( x\,\,\ne \,\,-5 \right)$ c) $C\,\,=\,\,\frac{10\text{x}\,-\,9}{2\text{x}\,-\,3}$
Bài 5:
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Trong hình vẽ bên, $O\in \text{xx }\!\!'\!\!\text{ }$
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PHẦN HƯỚNG DẪN GIẢI
Bài 1:
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Bài 2:
Xét số hữu tỉ $\frac{a}{b}$, có thể coi b > 0.
a) Nếu a, b cùng dấu thì a > 0 và b > 0. Suy ra $\frac{a}{b}\,\,\,>\,\frac{0}{b}\,=\,\,0\,\,$, tức là $\frac{a}{b}$ dương.
b) Nếu a, b trái dấu thì a < 0 và b > 0. Suy ra $\frac{a}{b}\,\,\,<\,\,\frac{0}{b}\,=\,\,0\,\,$, tức là $\frac{a}{b}$ âm.
Bài 3:
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a) $\,\,\frac{12}{-40}\,\,=\,\,\frac{-12}{40}$ Vì -13 < -12 nên $\frac{-13}{40}\,\,\,<\,\,\,\frac{-12}{40}\,\,\Rightarrow \,\,\,\frac{-13}{40}\,\,\,<\,\,\,\frac{12}{-40}$
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b) $\frac{-5}{6}\,\,=\,\,\frac{-20}{24}$; $\frac{-91}{104}\,\,=\,\,\frac{-7}{8}\,\,=\,\,\frac{-21}{24}$ Vì $-20\,\,>\,\,-21\,\,\,\Rightarrow \,\,\,\frac{-20}{24}\,\,>\,\,\frac{-21}{24}\,\,\,\Rightarrow \,\,\frac{-5}{6}\,\,\,>\,\,\frac{-91}{104}$
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c) $\frac{-15}{21}\,\,=\,\,\,\frac{-5}{7}\,\,=\,\,\frac{-55}{77}$; $\frac{-36}{44}\,\,\,=\,\,\frac{-9}{11}\,\,\,=\,\,\frac{-63}{77}$ Vì$-55\,\,>\,\,-63\,\,\,\Rightarrow \,\,\frac{-55}{77}\,\,\,>\,\,\,\frac{-63}{77}\,\,\,\Rightarrow \,\,\,\frac{-15}{21}\,\,\,>\,\,\frac{-36}{44}$
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d) $\frac{-16}{30}\,\,\,=\,\,\frac{-8}{15}\,\,\,=\,\,\frac{-32}{60}\,\,\,;\,\,\frac{-35}{84}\,\,\,=\,\,\frac{-5}{12}\,\,\,=\,\,\frac{-25}{60}$ Vì $-32\,\,\,<\,\,\,-25\,\,\,\Rightarrow \,\,\,\frac{-32}{60}\,\,\,<\,\,\frac{-25}{60}$. Hay $\frac{-16}{30}\,\,\,<\,\,\frac{-35}{84}$
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e) $\frac{-5}{91}\,\,\,=\,\,\frac{-505}{9191}\,\,$. Vì $-505\,\,\,<\,\,-501\,\,\Rightarrow \,\,\,\frac{-505}{9191}\,\,\,<\,\,\,\frac{-501}{9191}\,\,\,\,\Rightarrow \,\,\,\frac{-5}{91}\,\,\,<\,\,\,\frac{-501}{9191}$ Vậy $\frac{-5}{91}\,\,\,<\,\,\,\,\frac{-501}{9191}$
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f) $\frac{-11}{{{3}^{7}}{{.7}^{3}}}\,\,\,=\,\,\frac{-11.7}{{{3}^{7}}{{.7}^{3}}.7}\,\,=\,\,\,\frac{-77}{{{3}^{7}}.\,{{7}^{4}}}\,\,\,\,$ Vì $-77\,\,>\,\,-78\,\,\,\Rightarrow \,\,\frac{-77}{{{3}^{7}}{{.7}^{4}}}\,\,\,>\,\,\frac{-78}{{{3}^{7}}{{.7}^{4}}}\,\,\,\Rightarrow \,\,\,\frac{-11}{{{3}^{7}}{{.7}^{3}}}\,\,>\,\,\frac{-78}{{{3}^{7}}{{.7}^{4}}}$ |
Bài 4:
a) $A\,=\,\,\frac{x\,+\,1}{x\,-\,2}\,\left( x\,\,\ne \,2 \right)$$=\,\,1\,\,+\frac{3}{x\,-\,2}$
$A\,\,\in \,\,Z\,\,\Leftrightarrow \,\,\frac{3}{x\,-\,2}\,\,\in \,\,Z\,\,\Leftrightarrow \,\,\,x\,-\,2\,\,\in \,\,$Ư(3)$\Leftrightarrow \,\,x\,\,-\,\,2\,\,\in \,\,\left\{ -3\,;\,\,-1\,\,;\,\,1\,\,;\,\,3 \right\}\,\,\,\Leftrightarrow \,\,\,x\,\,\in \,\,\left\{ -1\,;\,\,1\,\,;\,\,3;\,\,5 \right\}$
b) $B\,\,=\,\,\frac{2x\,-\,1}{x\,+\,5}\,\,\left( x\,\,\ne \,\,-5 \right)$$=\,\,2\,-\,\,\frac{11}{x\,+\,5}$
$B\,\,\in \,\,Z\,\,\Leftrightarrow \,\,\frac{11}{x\,+\,\,5}\,\,\in \,\,Z\,\,\Leftrightarrow \,\,\,x\,+\,5\,\,\in \,\,$Ư(11)
$\Leftrightarrow \,\,x\,\,+\,\,5\,\,\in \,\,\left\{ -11\,;\,\,-1\,\,;\,\,1\,\,;\,\,11 \right\}\,\,\,\Leftrightarrow \,\,\,x\,\,\in \,\,\left\{ -16\,;\,\,-6\,\,;\,\,-4;\,\,6 \right\}$
c) $C\,\,=\,\,\frac{10\text{x}\,-\,9}{2\text{x}\,-\,3}$$=\,\,\,5\,+\,\,\frac{6}{2\text{x}\,\,-\,\,3}$
$C\,\,\in \,\,Z\,\,\Leftrightarrow \,\,\frac{6}{2x\,-\,\,3}\,\,\in \,\,Z\,\,\Leftrightarrow \,\,\,2x\,-\,\,3\,\,\in \,\,$Ư(6)
$\Leftrightarrow \,\,2x\,\,-\,\,\,3\in \,\,\left\{ -6\,;\,\,-3\,\,;\,\,-2\,\,;\,\,-1\,;\,\,1\,;\,\,2\,;\,\,3\,\,;\,\,6 \right\}\,\,\,$$\Leftrightarrow \,\,\,x\,\,\in \,\,\left\{ 0\,;\,\,1\,\,;\,\,2;\,\,3 \right\}$, $\left( x\,\,\in \,\,Z \right)$
b) Hai góc $\widehat{mOn}$ và $\widehat{tOy}$ là hai góc đối đỉnh
Vì + $\widehat{xOt};\,\,\widehat{\text{nOx}'}$ là hai góc đối đỉnh $\Rightarrow $$Ot$ và $On$ là hai tia đối nhau (1)
+ Lại có: $\widehat{tOy}=\widehat{mOn}\left( ={{90}^{0}} \right)$ mà $\widehat{xOt}=\,\widehat{\text{nOx}'}$(hai góc đối đỉnh) $\Rightarrow \widehat{xOm}=\widehat{x'Oy}$ (do $\widehat{xOx'}={{180}^{0}}$). Ta có $\widehat{xOt}+\widehat{tOy}+yOx'=\widehat{xOt}+\widehat{tOy}+\widehat{xOm}={{180}^{0}}$
$\Rightarrow Om$và $Oy$ là hai tia đối nhau (2)
$\left( 1 \right)\left( 2 \right)\Rightarrow $ Hai góc $\widehat{mOn}$ và $\widehat{tOy}$ là hai góc đối đỉnh. - Hết -
