CN
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
18 tháng 7 2017
a) -3x+4+5x=-10-x
-3x+4+5x+10+x=0
(-3x+5x+x)+10=0
3x+10=0
3x=-10
x=\(\dfrac{-10}{3}\)
Vậy x=\(\dfrac{-10}{3}\)
b)-x+1=-3x-8
-x+1+3x+8=0
(-x+3x)+(1+8)=0
2x+9=0
2x=-9
x=\(\dfrac{-9}{2}\)
Vậy x=\(\dfrac{-9}{2}\)
c)8-(x-1)=10+(x+5)
8-x+1=10+x+5
9-x=15+x
9-x-15-x=0
(9-15)-(x+x)=0
-6-2x=0
2x=-6
x=-3
Vậy x=-3
d)100+(x+7)-(-2x+3)=8+(x+100)
100+x+7+2x-3=8+x+100
(x+2x)+(100+7-3)=(8+100)+x
3x+104=108+x
3x+104-108-x=0
(3x-x)+(104-108)=0
2x-4=0
2x=4
x=2
Vậy x=2
DH
18 tháng 7 2017
e, \(\left|2x+5\right|=\left|x-1\right|\)
\(\Rightarrow\left\{{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x=-6\end{matrix}\right.\)
g, \(\left|-x+4\right|=\left|-3x-8\right|\)
\(\Rightarrow\left\{{}\begin{matrix}-x+4=3x+8\\-x+4=-3x-8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-4x=4\\2x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
h, \(\left|x+4\right|=\left|-3-8\right|\)
\(\Rightarrow\left|x+4\right|=\left|-11\right|=11\)
\(\Rightarrow\left\{{}\begin{matrix}x+4=-11\\x+4=11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-15\\x=7\end{matrix}\right.\)
Chúc bạn học tốt!!!
18 tháng 2 2020
tớ làm phần II
1, (-7) . 8 = -56
2, 11 . 5 = 55
3, (-250).(-4) = 1000
4, 15 . (-2) - (-5) = -25
5, \(\left(-2\right)^3.\left(-3\right).4=96\)
6, (26-6) . (-4) + 31.(-7-13) = 20.(-4) + 31.(-20)= -700
7, (-98) .(1-246)-246 . 98 = -98
8,[(-5) + 8 ] . (-3)\(^2\) = 27
PT
18 tháng 2 2020
I. Tìm x:
1) 8 + x = -10
\(x=-10-8=-18\)
2) (2x - 5) +17 = 6
\(2x-5=6-17=-11\)
\(2x=-11+5=-6\)
\(\Rightarrow x=\frac{-6}{2}=-3\)
3) -12 + 3. (-x + 7) = -18
\(3.\left(-x+7\right)=\left(-18\right)-\left(-12\right)=-6\)
\(-3x+21=-6\)
\(-3x=-6-21=-27\)
\(x=\frac{-27}{-3}=9\)
4) 24 : (3x -2) = -3
\(3x-2=\frac{24}{-3}=-8\)
\(3x=-8+2=-6\)
\(x=-\frac{6}{3}=-2\)
5) -45 : 5. (-3-2x) =3
\(\left(-45\right):\left(-15\right)-10x=3\)
\(\left(-15\right)-10x=\frac{-45}{3}=-15\)
\(-10x=-15+15=0\)
\(\Rightarrow x=0\)
6) 4x -7 = -42
\(4x=-42+7=-35\)
\(x=-\frac{35}{4}\)
19 tháng 12 2015
a) |2x +1| = 7
Th1: 2x + 1 = 7
<=> x = 3
Th2: 2x + 1 = -7
<=> x = -4
KL
1 tháng 8 2017
1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
12 tháng 12 2017
c)
\(4\left(3x-4\right)-2=18\)
<=> \(12x-16-2=18\)
<=> \(12x=36\)
<=> \(x=3\)
Vậy x=3
d)
\(\left(3x-10\right):10=50\)
<=> \(3x-10=500\)
<=> \(3x=510\)
<=> x= \(170\)
Vậy x= 170
f)
\(x-\left[42+\left(-25\right)\right]=-8\)
<=> \(x-17=-8\)
<=> x= \(9\)
Vậy x=9
h)
\(x+5=20-\left(12-7\right)\)
<=> \(x+5=15\)
<=> \(x=10\)
Vậy x= 10
k)
\(\left|x-5\right|=7-\left(-3\right)\)
<=> \(\left|x-5\right|=10\)
* Với \(x>=5\) ; ta được:
\(x-5=10\)
<=> x= 15 (thoả mãn điều kiện )
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=10\)
<=> \(-x+5=10\)
<=> \(-x=5\)
<=> \(x=-5\) (thoả mãn điều kiện)
Vậy x=15 ; x= -5
i)
\(\left|x-5\right|=\left|7\right|\)
<=> \(\left|x-5\right|=7\)
*Với \(x>=5\) ; ta được:
\(x-5=7\)
<=> \(x=12\) (thoả mãn)
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=7\)
<=> \(-x=2\)
<=> \(x=-2\) (thoả mãn)
Vậy x= 12; x= -2
m)
\(2^{x+1}.2^{2009}=2^{2010}\)
<=> \(2^{x+1+2009}=2^{2010}\)
<=> \(2^{x+2010}=2^{2010}\)
=> \(x+2010=2010\)
=> \(x=0\)
Vậy x=0
n)
\(10-2x=25-3x\)
<=>\(x=15\)
Vậy x=15
21 tháng 5 2022
a: -3x+4+5x=-10-x
=>2x+4=-x-10
=>3x=-14
hay x=-14/3
b: \(-x+1=-3x-8\)
=>-x+3x=-8-1
=>2x=-9
hay x=-9/2
c: \(8-\left(x-1\right)=10+\left(x+5\right)\)
=>x+15=8-x+1
=>x+15=9-x
=>2x=-6
hay x=-3
d: \(100+\left(x+7\right)-\left(-x+3\right)=8+\left(x+100\right)\)
=>x+7+x-3=8+x
=>2x+4-x-8=0
=>x=4
4 tháng 8 2017
a ) \(\left(x+1\right)^2-3\left(x+1\right)^2=-8\)
\(\Leftrightarrow\left(x+1\right)^2.\left(1-3\right)=-8\)
\(\Leftrightarrow-2\left(x+1\right)^2=-8\)
\(\Leftrightarrow\left(x+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy .......
b ) \(x^2-7x=4-7\left(x-3\right)\)
\(\Leftrightarrow x^2-7x-4+7x-21=0\)
\(\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy ........
c ) \(\left(2x+1\right)^2-3x+3=4-3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)^2-3\left(x-1\right)+3\left(x-1\right)=4\)
\(\Leftrightarrow\left(2x+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=2\\2x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy......
BH
4 tháng 8 2017
b. x2 - 7x = 4 - 7(x-3)
=> x2 - 7x = 4 - 7x +21
=> x2 - 7x + 7x = 25
=> x2 = 25
=> \(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
c.
TG
12 tháng 4 2020
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)