1. 下列各组中的$a,b,c,d$不能成比例的是(
A.$a = 3,b = 6,c = 2,d = 4$
B.$a = 1,b = \sqrt{2},c = \sqrt{6},d = \sqrt{3}$
C.$a = 4,b = 6,c = 5,d = 10$
D.$a = 2,b = \sqrt{5},c = \sqrt{15},d = 2\sqrt{3}$
C
)A.$a = 3,b = 6,c = 2,d = 4$
B.$a = 1,b = \sqrt{2},c = \sqrt{6},d = \sqrt{3}$
C.$a = 4,b = 6,c = 5,d = 10$
D.$a = 2,b = \sqrt{5},c = \sqrt{15},d = 2\sqrt{3}$
答案:
C
2. 已知比例外项为$m,n$,比例内项为$p,q$,则下列比例式中,正确的是(
A.$m:n = p:q$
B.$m:p = n:q$
C.$m:q = n:p$
D.$m:p = q:n$
D
)A.$m:n = p:q$
B.$m:p = n:q$
C.$m:q = n:p$
D.$m:p = q:n$
答案:
D
3. 已知$2x = 3y(y \neq 0)$,则下列说法中,正确的是(
A.$\frac{x}{y} = \frac{3}{2}$
B.$\frac{x}{3} = \frac{2}{y}$
C.$\frac{x}{y} = \frac{2}{3}$
D.$\frac{x}{2} = \frac{y}{3}$
A
)A.$\frac{x}{y} = \frac{3}{2}$
B.$\frac{x}{3} = \frac{2}{y}$
C.$\frac{x}{y} = \frac{2}{3}$
D.$\frac{x}{2} = \frac{y}{3}$
答案:
A
4. [教材 P117 例 2 改编]已知$\frac{a}{b} = \frac{c}{d}$,则下列各式中,一定成立的是(
A.$\frac{a}{c} = \frac{d}{b}$
B.$\frac{ac}{bd} = \frac{c}{b}$
C.$\frac{a + 1}{b} = \frac{c + 1}{d}$
D.$\frac{a + 2b}{b} = \frac{c + 2d}{d}$
D
)A.$\frac{a}{c} = \frac{d}{b}$
B.$\frac{ac}{bd} = \frac{c}{b}$
C.$\frac{a + 1}{b} = \frac{c + 1}{d}$
D.$\frac{a + 2b}{b} = \frac{c + 2d}{d}$
答案:
D [解析]$\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+2=\frac{c}{d}+2\Rightarrow\frac{a + 2b}{b}=\frac{c + 2d}{d}$.
5. [教材 P118 作业题 T4 改编]已知$\frac{b}{a + b} = \frac{5}{6}$,则$\frac{b}{a} = $
5
,$\frac{a - b}{b} = $−$\frac{4}{5}$
.
答案:
5 −$\frac{4}{5}$ [解析]
∵$\frac{b}{a + b}=\frac{5}{6}$,
∴5a + 5b = 6b,
∴5a = b,
∴$\frac{b}{a}=5$,$\frac{a - b}{b}=\frac{a - 5a}{5a}=-\frac{4}{5}$.
∵$\frac{b}{a + b}=\frac{5}{6}$,
∴5a + 5b = 6b,
∴5a = b,
∴$\frac{b}{a}=5$,$\frac{a - b}{b}=\frac{a - 5a}{5a}=-\frac{4}{5}$.
6. [教材 P118 作业题 T2 改编]求下列各式中的$x$.
(1)$3:x = 2:8$.
(2)$x:(x + 1) = (1 - x):4$.
(1)$3:x = 2:8$.
(2)$x:(x + 1) = (1 - x):4$.
答案:
解:
(1)由比例的基本性质,得2x = 3×8,
∴x = 12.
(2)由比例的基本性质,得(x + 1)(1 - x)= 4x,
∴x² + 4x - 1 = 0, 解得x1 = $\sqrt{5}$ - 2,x2 = -$\sqrt{5}$ - 2.
(1)由比例的基本性质,得2x = 3×8,
∴x = 12.
(2)由比例的基本性质,得(x + 1)(1 - x)= 4x,
∴x² + 4x - 1 = 0, 解得x1 = $\sqrt{5}$ - 2,x2 = -$\sqrt{5}$ - 2.
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