1. 下列各式中,属于一元一次方程的是( )
A.$1 + 2t$
B.$1 - 2x = 0$
C.$m^{2}+m = 1$
D.$\frac{4}{x}+1 = 3$
A.$1 + 2t$
B.$1 - 2x = 0$
C.$m^{2}+m = 1$
D.$\frac{4}{x}+1 = 3$
答案:
B
2. 方程$\frac{x}{2}-1 = 2$的解是( )
A.$x = 2$
B.$x = 3$
C.$x = 5$
D.$x = 6$
A.$x = 2$
B.$x = 3$
C.$x = 5$
D.$x = 6$
答案:
D
3. 设$a$,$b$,$c$为互不相等的实数,且$b=\frac{4}{5}a+\frac{1}{5}c$,则下列结论中,正确的是( )
A.$a > b > c$
B.$c > b > a$
C.$a - b = 4(b - c)$
D.$a - c = 5(a - b)$
A.$a > b > c$
B.$c > b > a$
C.$a - b = 4(b - c)$
D.$a - c = 5(a - b)$
答案:
D
4. 若关于$x$的方程$\frac{4 - x}{2}+a = 4$的解是$x = 2$,则$a$的值为______.
答案:
3【解析】把x=2代入方程$\frac{4-x}{2}+a=4$,得$\frac{4-2}{2}+a=4$,解得a=3.
5. 方程$-2(2x + 1)=x$去括号后,正确的是( )
A.$-4x + 1 = -x$
B.$-4x + 2 = -x$
C.$-4x - 1 = x$
D.$-4x - 2 = x$
A.$-4x + 1 = -x$
B.$-4x + 2 = -x$
C.$-4x - 1 = x$
D.$-4x - 2 = x$
答案:
D
6. 下列变形中,正确的是( )
A.由$7x = 4x - 3$移项,得$7x - 4x = 3$
B.由$\frac{2x - 1}{3}=1+\frac{x - 3}{2}$去分母,得$2(2x - 1)=1 + 3(x - 3)$
C.由$2(2x - 1)-3(x - 3)=1$去括号,得$4x - 2 - 3x - 9 = 1$
D.由$2(x + 1)=x + 7$去括号、移项、合并同类项,得$x = 5$
A.由$7x = 4x - 3$移项,得$7x - 4x = 3$
B.由$\frac{2x - 1}{3}=1+\frac{x - 3}{2}$去分母,得$2(2x - 1)=1 + 3(x - 3)$
C.由$2(2x - 1)-3(x - 3)=1$去括号,得$4x - 2 - 3x - 9 = 1$
D.由$2(x + 1)=x + 7$去括号、移项、合并同类项,得$x = 5$
答案:
D
7. 解方程:
(1) $4x - 3(5 - x)=6$.
(2) $\frac{x - 3}{2}+\frac{x - 1}{3}=4$.
(1) $4x - 3(5 - x)=6$.
(2) $\frac{x - 3}{2}+\frac{x - 1}{3}=4$.
答案:
解:
(1)去括号,得4x-15+3x=6.
移项、合并同类项,得7x=21.
两边同除以7,得x=3.
(2)去分母,得3(x-3)+2(x-1)=24.
去括号,得3x-9+2x-2=24.
移项,得3x+2x=24+9+2.
合并同类项,得5x=35.
两边同除以5,得x=7.
(1)去括号,得4x-15+3x=6.
移项、合并同类项,得7x=21.
两边同除以7,得x=3.
(2)去分母,得3(x-3)+2(x-1)=24.
去括号,得3x-9+2x-2=24.
移项,得3x+2x=24+9+2.
合并同类项,得5x=35.
两边同除以5,得x=7.
8. 解方程:
(1) $3(x + 1)+\frac{1}{3}(x + 1)=\frac{1}{2}(x + 1)-\frac{17}{6}$.
(2) $\frac{1}{8}\left\{\frac{1}{6}\left[\frac{1}{4}\left(\frac{x - 1}{2}+3\right)+5\right]+7\right\}=1$.
(1) $3(x + 1)+\frac{1}{3}(x + 1)=\frac{1}{2}(x + 1)-\frac{17}{6}$.
(2) $\frac{1}{8}\left\{\frac{1}{6}\left[\frac{1}{4}\left(\frac{x - 1}{2}+3\right)+5\right]+7\right\}=1$.
答案:
解:
(1)$\frac{17}{6}(x+1)=-\frac{17}{6}$,
x+1=-1,
x=-2.
(2)$\frac{1}{6}\left[\frac{1}{4}\left(\frac{x-1}{2}+3\right)+5\right]+7=8$,
$\frac{1}{4}\left(\frac{x-1}{2}+3\right)+5=6$,
$\frac{x-1}{2}+3=4$,
$\frac{x-1}{2}=1$,
x=3.
(1)$\frac{17}{6}(x+1)=-\frac{17}{6}$,
x+1=-1,
x=-2.
(2)$\frac{1}{6}\left[\frac{1}{4}\left(\frac{x-1}{2}+3\right)+5\right]+7=8$,
$\frac{1}{4}\left(\frac{x-1}{2}+3\right)+5=6$,
$\frac{x-1}{2}+3=4$,
$\frac{x-1}{2}=1$,
x=3.
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