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a, \(\left|x+3\right|=15\)
TH1 : \(x+3=15\Leftrightarrow x=12\)
TH2 : \(x+3=-15\Leftrightarrow x=-18\)
b, \(\left|x-7\right|+13=15\Leftrightarrow\left|x-7\right|=2\)
TH1 : \(x-7=2\Leftrightarrow x=9\)
TH2 : \(x-7=-2\Leftrightarrow x=5\)
c, \(\left|x-3\right|-16=-4\Leftrightarrow\left|x-3\right|=12\)
TH1 : \(x-3=12\Leftrightarrow x=15\)
TH2 : \(x-3=-12\Leftrightarrow x=-9\)
d, \(26-\left|x+9\right|=-13\Leftrightarrow\left|x+9\right|=39\)
TH1 : \(x+9=39\Leftrightarrow x=30\)
TH2 : \(x+9=-39\Leftrightarrow x=-48\)
`1, -2/9 xx 15/17 + (-2/9) xx 2/17`
`= -2/9 xx (15/17 + 2/17)`
`= -2/9 xx 17/17`
`=-2/9xx1`
`=-2/9`
__
`-5/3 xx 6/5 + (-7/9) xx 3/10`
`= -30/15 + (-21/90)`
`= -2 + (-7/30)`
`=-60/30 +(-7/30)`
`=-67/30`
__
`15/20 xx 7/5 + (-9/7) xx (-6/4)`
`=3/4 xx7/5 + (-9/7) xx(-6/4)`
`= 21/20 + 54/28`
`= 21/20 + 27/14`
`=417/140`
__
`-25/13 xx 5/19 + (-25/13) xx 14/19`
`=-25/13 xx (5/19 +14/19)`
`=-25/13 xx 19/19`
`= -25/13 xx 1`
`=-25/13`
__
`-7/13 xx 13/5 + (-9/7) xx 5/3`
`=-7/5 +(-15/7)`
`=-124/35`
a,\(x+3=15\)
\(=>x=15-3=12\)
b,\(x-7+3=25\)
\(=>x=25+7-3=29\)
c,\(x-3-6=-4\)
\(=>x=-4+6+3=5\)
d,\(26-x+9=-13\)
\(=>26-x=-13-9=-22\)
\(=>26+22=x\)
\(=>x=48\)
x+3=15
x=15-3
x=12
Vậy x=12
x-7+3=25
x-7=25-3
x-7=22
x=22+7
x=29
Vậy x=29
x-3-6=-4
x-3=-4+6
x-3=2
x=2+3
x=5
Vậy x=5
26-x+9=-13
26-x=-13-9
26-x=22
x=26-22
x=4
Vậy x=4
1: =-2/9(15/17+2/17)=-2/9
2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)
=-2-7/30=-67/30
3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
1: =-2/9(15/17+2/17)=-2/9
2: =−63+−2190=−63+−2190
=-2-7/30=-67/30
3: =34⋅75+97⋅32=34⋅75+97⋅32
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
a) \(\left|x+3\right|=15\)
\(\Rightarrow\orbr{\begin{cases}x=15-3\\x=-15-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=-18\end{cases}}\)
b) Đề sai
c) Đề sai
d) \(26-\left|x+9\right|=-13\)
\(\Rightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Rightarrow\left|x+9\right|=39\)
\(\Rightarrow\orbr{\begin{cases}x=39-9\\x=-39-9\end{cases}\Rightarrow\orbr{\begin{cases}x=30\\x=-48\end{cases}}}\)
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
a)|x-7|+13=25
|x-7| = 25-13
|x-7| = 12
|x| = 12+7
=> x = 19
b)|x-3|-16=-4
|x-3| = -4+16
|x-3| = 12
|x| = 12+3
=> x = 15
c)26-|x+9|=-13
|x+9| = -13-26
|x+9| = -39
|x| = -39 -9
=> x = 48
a) -x + 8 = -17
⇔ -x = -17 - 8
⇔ -x = -25
⇔ x = 25
b) 35 - x = 37
⇔ x = 35 - 37
⇔ x = -2
c) -19 - x = -20
⇔ x = -19 + 20
⇔ x = 1
d) x - 45 = -17
⇔ x = -17 + 45
⇔ x = 28
e) |x + 3| = 15
⇔ \(\left[{}\begin{matrix}x+3=15\\x+3=-15\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=15-3\\x=-15-3\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=12\\x=-18\end{matrix}\right.\)
f) |x - 3| - 16 = -4
⇔ |x - 3| = -4 + 16
⇔ |x - 3| = 12
⇔ \(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=12+3\\x=-12+3\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
k) 26 - |x + 9| = -13
⇔ |x + 9| = 26 + 13
⇔ |x + 9| = 39
⇔ \(\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=39-9\\x=-39-9\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=30\\x=-48\end{matrix}\right.\)
a,-x + 8 = -17
=> -x=-17-8
=> -x=-25
=> x=25
vậy x=25
b, 35 - x = 37
=> x=35-37
=> x=-2
vậy x=-2
c, -19 -x = -20
=> x=-19+20
=> x=1
vậy x=1
d, x - 45 = -17
=> x=-17+45
=> x=28
vậy x=28
e, | x+ 3| = 15
th1 x+3=15
=> x=15-3
=> x=12
th2 x+3=-15
=> x=-15-3
=> x=-18
vậy x=12 hoặc x=-18
g, |x-7| +13 = 25
=> \(\left|x-7\right|=25-13\)
=> \(\left|x-7\right|=12\)
th1 x-7=12
=> x=12+7
=> x=19
th2 x-7=-12
=>x=-12+7
=> x=-5
vậy x=19 hoặc x=-5
các câu sau tt
a: \(\left|x-7\right|+13=25\)
=>|x-7|=12
=>x-7=12 hoặc x-7=-12
=>x=19 hoặc x=-5
b: 26-|x+9|=-13
=>|x+9|=39
=>x+9=39 hoặc x+9=-39
=>x=30 hoặc x=-48