Tìm x;
x.(x+7)=0
(x+12).(x-3)=0
(-x+5).(3-x)=0
x.(2+x).(7-x)=0
(x-1).(x+2).(-x-3)=0
tìm x
(2x-5)+17=6
10-2(4-3x)=-4
-12+3(-x+7)=-18
24 ÷(3x-2)=-3
-45 ÷(-3-2x)=3
Tìm x;
x.(x+7)=0
(x+12).(x-3)=0
(-x+5).(3-x)=0
x.(2+x).(7-x)=0
(x-1).(x+2).(-x-3)=0
tìm x
(2x-5)+17=6
10-2(4-3x)=-4
-12+3(-x+7)=-18
24 ÷(3x-2)=-3
-45 ÷(-3-2x)=3
Đáp án:
Giải thích các bước giải:
Bài 1:
`a)x(x+7)=0`
`→` \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)
Vậy `x∈{0;-7}`
`b)(x+12)(x-3)=0`
`->` \(\left[ \begin{array}{l}x+12=0\\x-3=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\)
Vậy `x∈{-12;3}`
`c)(-x+5)(3-x)=0`
`->` \(\left[ \begin{array}{l}-x+5=0\\3-x=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\)
Vậy `x∈{5;3}`
`d)x(x+2)(7-x)=0`
`->` \(\left[ \begin{array}{l}x=0\\x+2=0\\7-x=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\)
Vậy `x∈{0;-2;7}`
`e)(x-1)(x+2)(-x-3)=0`
`->` \(\left[ \begin{array}{l}x-1=0\\x+2=0\\-x-3=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\x=-2\\x=-3\end{array} \right.\)
Vậy `x∈{1;-2;-3}`
Bài 2:
`a)(2x-5)+17=6`
`->2x-5=-11`
`->2x=-6`
`->x=-3`
Vậy `x=-3`
`b)10-2(4-2x)=-4`
`->10-8+4x=-4`
`->2+4x=-4`
`->4x=-6`
`->x=-3/2`
Vậy `x=-3/2`
`c)-12+3(-x+7)=-18`
`->-12-3x+21=-18`
`->9-3x=-18`
`->3x=9-(-18)`
`->3x=27`
`->x=9`
Vậy `x=9`
`d)24:(3x-2)=-3`
`->3x-2=-8`
`->3x=-6`
`->x=-2`
Vậy `x=-2`
`e)-45:(-3-2x)=3`
`->-3-2x=-15`
`->2x=(-3)-(-15)`
`->2x=12`
`->x=6`
Vậy `x=6`
Đáp án:
Giải thích các bước giải:
bài 1
+)x(x+7)=0+)x(x+7)=0
⇔⇔[x=0x+7=0[x=0x+7=0
⇔⇔[x=0x=−7[x=0x=−7
Vậy : x∈{−7;0}x∈{-7;0}
+)(x+12)(x−3)=0+)(x+12)(x-3)=0
⇔⇔[x+12=0x−3=0[x+12=0x−3=0
⇔⇔[x=−12x=3[x=−12x=3
Vậy : x∈{−12;3}x∈{-12;3}
+)(−x+5)(3−x)=0+)(-x+5)(3-x)=0
⇔⇔[−x+5=03−x=0[−x+5=03−x=0
⇔⇔[x=5x=3[x=5x=3
Vậy : x∈{3;5}x∈{3;5}
+)x(x+2)(7−x)=0+)x(x+2)(7-x)=0
⇔⇔⎡⎢⎣x=0x+2=07−x=0[x=0x+2=07−x=0
⇔⇔⎡⎢⎣x=0x=−2x=7[x=0x=−2x=7
Vậy : x∈{−2;0;7}x∈{-2;0;7}
+)(x−1)(x+2)(−x−3)=0+)(x-1)(x+2)(-x-3)=0
⇔⇔⎡⎢⎣x−1=0x+2=0−x−3=0[x−1=0x+2=0−x−3=0
⇔⇔⎡⎢⎣x=1x=−2x=−3[x=1x=−2x=−3
Vậy : x∈{−3;−2;1}
bài2
a)(2x-5)+17=6
<=>2x-5+17=6
<=>2x=-6
<=>x=-3
b)10-2(4-3x)=-4
<=>10-8+6x=-4
<=>6x=-6
<=>x=-1
c)-12+3(-x+7)=-3
<=>-12-3x+21=-3
<=>-3x=-12
<=>x=4
d)24:(3x-2)=-3
<=>24=-3(3x-2)
<=>24=-9x+6
<=>18=-9x
<=>x=-2
e)-45:5(-3-2x)=3
<=>-9(-3-2x)=3
<=>27+18x=3
<=>18x=24
<=>x=4/3