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a,|x-2|=3
=>x-2= 3 hoặc -3
-với x -2 =3 =>x=1
-với x-2 = -3 =>x= -1
Vậy x ∈ {1;-1}
b, |x-3|-7=13
=>|x-3|=13-7=6
=>x-3=6 hoặc -6
-với x-3 =6 => x=9
-với x-3= -6 => x=-3
Vậy x ∈ {9;-3}
c,72-3.|x+1|=9
=>3.|x+1|=72-9=63
=>|x+1|=63:3=21
=>x+1 = 21 hoặc -21
-với x+1 = 21 => x=20
-với x+1 = -21 => x= -22
Vậy x ∈ {20;-21}
d,17-(43-|x|)=45
=>43-|x|=17-45= -28
=>|x|=43-(-28)=71
=> x =71 hoăc -71
Vậy x ∈ {71;-71}
f, 3|x-1|-5=7
=>3|x-1|=7+5=12
=>|x-1|=12:3=4
=>x-1=4 hoặc -4
-với x-1=4 =>x=5
-với x-1=-4=>x=-3
Vậy x ∈ {5;-3}
CHÚC BẠN LÀM BÀI TÔT NHA
a) \(\left|x-2\right|=3\\ \Rightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3+2\\x=-3+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)Vậy \(x\in\left\{5;-1\right\}\)
b) \(\left|x-3\right|-7=13\\ \left|x-3\right|=13+7\\ \left|x-3\right|=20\\ \Rightarrow\left[{}\begin{matrix}x-3=20\\x-3=-20\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=20+3\\x=-20+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=23\\x=-17\end{matrix}\right.\)
Vậy \(x\in\left\{23;-17\right\}\)
c) \(72-3\cdot\left|x+1\right|=9\\ 3\left|x+1\right|=72-9\\ 3\left|x+1\right|=63\\ \Rightarrow\left[{}\begin{matrix}x+1=63\\x+1=-63\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=63-1\\x=-63-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=62\\x=-64\end{matrix}\right.\)
Vậy \(x\in\left\{62;-64\right\}\)
d) \(17-\left(43-\left|x\right|\right)=45\\ 43-\left|x\right|=17-45\\ 43-\left|x\right|=-28\\ \left|x\right|=43-\left(-28\right)\\ \left|x\right|=43+28\\ \left|x\right|=71\\ \Rightarrow x\in\left\{71;-71\right\}\)Vậy \(x\in\left\{71;-71\right\}\)
e) \(\left|x\right|< 3\\ \left|x\right|\in\left\{0;1;2\right\}\\ \Rightarrow x\in\left\{0;\pm1;\pm2\right\}\)Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
f) \(3\left|x-1\right|-5=7\\ 3\left|x-1\right|=7+5\\ 3\left|x-1\right|=12\\ \left|x-1\right|=12:3\\ \left|x-1\right|=4\\ \Rightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{5;-3\right\}\)
a. \(\left|x-3\right|-7=13\Leftrightarrow\left|x-3\right|=20\)
TH1: \(x-3=20\Leftrightarrow x=23\)
TH2: \(x-3=-20\Leftrightarrow x=-17\)
b. \(72-3.\left|x+1\right|=9\Leftrightarrow\left|x+1\right|=21\)
TH1: \(x+1=21\Leftrightarrow x=20\)
TH2: \(x+1=-21\Leftrightarrow x=-22\)
c. \(17-\left(43-\left|x\right|\right)=45\Leftrightarrow\left|x\right|=71\)
TH1: \(x=71\)
TH2: \(x=-71\)
d. \(3\left|x-1\right|-5=7\Leftrightarrow\left|x-1\right|=4\)
TH1: \(x-1=4\Leftrightarrow x=5\)
TH2: \(x-1=-4\Leftrightarrow x=-3\)
e. \(-12\left(x-5\right)+7\left(3-x\right)=5\Leftrightarrow-12x+60+21-7x=5\Leftrightarrow-19x=-76\Leftrightarrow x=4\)
a.
\(\left|x-3\right|-7=13\)
\(\left|x-3\right|=13+7\)
\(\left|x-3\right|=20\)
\(x-3=\pm20\)
- \(x-3=20\)
\(x=20+3\)
\(x=23\)
- \(x-3=-20\)
\(x=-20+3\)
\(x=-17\)
Vậy x = 23 hoặc x = - 17.
b.
\(72-3\times\left|x+1\right|=9\)
\(3\times\left|x+1\right|=72-9\)
\(3\times\left|x+1\right|=63\)
\(\left|x+1\right|=63\div3\)
\(\left|x+1\right|=21\)
\(x+1=\pm21\)
- \(x+1=21\)
\(x=21-1\)
\(x=20\)
- \(x+1=-21\)
\(x=-21-1\)
\(x=-22\)
Vậy x = 20 hoặc x = - 22.
c.
\(17-\left(43-\left|x\right|\right)=45\)
\(43-\left|x\right|=17-45\)
\(43-\left|x\right|=-28\)
\(\left|x\right|=43-\left(-28\right)\)
\(\left|x\right|=71\)
\(x=\pm71\)
Vậy x = 71 hoặc x = - 71.
d.
\(3\times\left|x-1\right|-5=7\)
\(3\times\left|x-1\right|=7+5\)
\(3\times\left|x-1\right|=12\)
\(\left|x-1\right|=12\div3\)
\(\left|x-1\right|=4\)
\(x-1=\pm4\)
- \(x-1=4\)
\(x=4+1\)
\(x=5\)
- \(x-1=-4\)
\(x=-4+1\)
\(x=-3\)
Vậy x = 5 hoặc x = - 3.
e.
\(-12\times\left(x-5\right)+7\times\left(3-x\right)=5\)
\(-12\times x+\left(-12\right)\times\left(-5\right)+7\times3-7\times x=5\)
\(-12\times x+60+21-7\times x=5\)
\(-19\times x+81=5\)
\(-19\times x=5-81\)
\(-19\times x=-76\)
\(x=\frac{-76}{-19}\)
\(x=4\)
Chúc bạn học tốt
a, 2x - ( -17) = 15
=> 2x + 1 7 = 15
=> 2x = 15 -17
=> 2x = -2
=> x = -1
a, 2x - ( -17) = 15
=> 2x + 1 7 = 15
=> 2x = 15 -17
=> 2x = -2
=> x = -1
mình chỉ biết làm phần a thôi mong các bn thông cảm
1) -5x - (-3) = 13
\(\Rightarrow-5x+3=13\)
\(\Rightarrow-5x=10\)
\(\Rightarrow x=-2\)
Vậy: \(x=-2\)
2) 15 - ( x - 7) = -21
\(\Rightarrow15-x+7=-21\)
\(\Rightarrow15+7-x=-21\)
\(\Rightarrow22-x=-21\)
\(\Rightarrow x=-22-21=-43\)
Vậy: \(x=-43\)
3) 45 - ( x - 9) = -35
\(\Rightarrow45-x+9=-35\)
\(\Rightarrow45+9-x=-35\)
\(\Rightarrow54-x=-35\)
\(\Rightarrow x=-54-35=-89\)
Vậy: \(x=-89\)
4)/x-2/=3\(\Rightarrow\)ta có 2 t/h:
t/h1: x-2=3\(\Rightarrow\)x=2+3=5
t/h2: x-2=-3\(\Rightarrow\)x=2+(-3)=-1
Vậy x=5 hoặc -1
a)=20
b)=-1
c)=-1 hoặc 5
d)=-17 hoặc 23
e)=20 hoặc=-22
i)=71 hoặc -71
=5 hoặc -3
1) x - 2 = -6
x = -6 + 2
x = -4
2) -5 . x - ( -3 ) =13
-5 . x = 13 + ( -3 )
-5 . x = 10
x = 10 : ( -5 )
x = -2
a)-12(x-5)+7(3-x)=5
-12x-(-60)+21-7x=5
-12x+60+21-7x=5
-12x-7x=5-60-21
-19x=-76
x=-76/-19
x=4. Vậy x=4
ặc, bạn ghi ít ít thôi, ghi thế này nhiều quá
a.\(\Rightarrow\)\(\left\{{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
b.\(\Rightarrow\)\(\left\{{}\begin{matrix}x-3=20\\x-3=-20\end{matrix}\right.\)
\(\Rightarrow\)\(\left\{{}\begin{matrix}x=23\\x=-17\end{matrix}\right.\)
c.
a) \(\left|x-2\right|=3\)
\(\circledast\) TH1: x-2=3
x=3+2
x=5
\(\circledast\) TH2: x-2=-3
x=-3+2
x=-1
Vậy x\(\in\){-1;5}.
b) \(\left|x-3\right|-7=13\)
\(\left|x-3\right|=13+7\)
\(\left|x-3\right|=20\)
\(\circledast\) TH1: x-3=20
x=20+3
x=23
\(\circledast\) TH2: x-3=-20
x=-20+3
x=-17
Vậy x\(\in\){-17;23}.
c) \(72-3\cdot\left|x+1\right|=9\)
\(3\cdot\left|x+1\right|=72-9\)
\(3\cdot\left|x+1\right|=63\)
\(\left|x+1\right|=\dfrac{63}{3}\)
\(\left|x+1\right|=21\)
\(\circledast\) TH1: x+1=21
x=21-1
x=20
\(\circledast\) TH2: x+1=-21
x=-21-1
x=-22
Vậy x\(\in\){-22;20}.
d) \(17-\left(43-\left|x\right|\right)=45\)
\(43-\left|x\right|=17-45\)
\(43-\left|x\right|=-28\)
\(\left|x\right|=43-\left(-28\right)\)
\(\left|x\right|=71\)
x=71 hoặc x=-71
Vậy x\(\in\){-71;71}.
e) \(3\left|x-1\right|-5=7\)
\(3\left|x-1\right|=7+5\)
\(3\left|x-1\right|=12\)
\(\left|x-1\right|=\dfrac{12}{3}\)
\(\left|x-1\right|=4\)
\(\circledast\) TH1: x-1=4
x=4+1
x=5
\(\circledast\) TH2: x-1=-4
x=-4+1
x=-3
Vậy x\(\in\){-3;5}.