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a) \(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\)
\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)
b) \(5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\)
\(\Leftrightarrow3x=48\Leftrightarrow x=16\)
c) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(\Leftrightarrow3x-12-8+x=12\)
\(\Leftrightarrow4x-20=12\)
\(\Leftrightarrow4x=32\Leftrightarrow x=8\)
d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)
1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
a,|x|+3=5
\(\Leftrightarrow\left|x\right|=5-3=2\)
\(\Rightarrow x=\left\{{}\begin{matrix}2\\-2\end{matrix}\right.\)
b,|x+3|=5
\(\Rightarrow\left\{{}\begin{matrix}x+3=5\\x+3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\\-8\end{matrix}\right.\)
c,|x-7|+13=25
<=>|x-7|=25-13=12
\(\Rightarrow\left\{{}\begin{matrix}x-7=12\\x-7=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=19\\x=-5\end{matrix}\right.\)
d,\(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)=39\)
\(\Rightarrow\left\{{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=30\\x=-48\end{matrix}\right.\)
e,8-|x|=15
<=>|x|=8-15=-7
\(\Rightarrow\left\{{}\begin{matrix}x=-7\\x=7\end{matrix}\right.\)
f,6-|-3+x|=-15
\(\Leftrightarrow\left|-3+x\right|=6-\left(-15\right)=21\)
\(\Rightarrow\left\{{}\begin{matrix}-3+x=21\\-3+x=-21\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=24\\x=-18\end{matrix}\right.\)
a, |x| + 3 = 5
|x| = 5 - 3
|x| = 2
|x| = 2 hoặc |x| = -2
Vậy |x| thuộc {2; -2}
b,|x + 3| = 5
|x + 3| = 5 hoặc |x + 3| = -5
x = 5 - 3 x = (-5) - 3
x = 2 x = -8
Vậy x thuộc {2; -8}
c,|x - 7| + 13 = 25
|x - 7| = 25 - 13
|x - 7| = 12
|x - 7| = 12 hoặc |x - 7| = -12
x = 12 + 7 x = (-12) + 7
x = 19 x = -5
Vậy x thuộc {19 ; -5}
Lời giải:
a)
\(-3\frac{5}{8}+\left(-\frac{3}{8}+\frac{9}{4}\right)\)
\(=-\frac{29}{8}+\left(-\frac{3}{8}+\frac{18}{8}\right)\)
\(=-\frac{29}{8}+\frac{15}{8}=-\frac{14}{8}=-\frac{7}{4}\)
b) \(\frac{\left(-9\right)\cdot11+32\cdot\left(-9\right)}{\left(-43\right)\cdot15+12\cdot\left(-43\right)}=\frac{\left(-9\right)\left(11+32\right)}{\left(-43\right)\left(15+12\right)}=\frac{\left(-9\right)\cdot43}{\left(-43\right)\cdot27}=\frac{\left(-1\right)\cdot1}{\left(-1\right)\cdot3}=\frac{1}{3}\)
c) Thay \(x=\frac{2011}{2012}\)vào biểu thức \(x\cdot\frac{1}{3}+2x\cdot\frac{3}{6}-3x\cdot\frac{4}{9}\)ta có :
\(\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{3}{6}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)
\(=\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{1}{2}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)
\(=\frac{2011}{6036}+\frac{2011}{2012}-\frac{2011}{1509}\)
\(=\frac{2011}{6036}+\frac{6033}{6036}-\frac{8044}{6036}=\frac{2011+6033-8044}{6036}=0\)
a) \(\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left[{}\begin{matrix}x-9=1\\x-9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=10\\x=9\end{matrix}\right.\)
b) \(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(\Rightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{3}\\x=\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left(x-8\right)^3=\left(x-8\right)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x-8=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=9\end{matrix}\right.\)
a, `(x-9)^4=(x-9)^7`
`(x-9)^4-(x-9)^7=0`
`(x-9)^4 . [(1-(x-9)^3]=0`
TH1: `(x-9)^4=0`
`x-9=0`
`x=9`
TH2: `1-(x-9)^3=0`
`(x-9)^3=1^3`
`x-9=1`
`x=10`
b, `(3x-15)^10=(3x-15)^15`
`(3x-15)^10 . [1-(3x-15)^5]=0`
TH1: `(3x-15)^10=0`
`3x-15=0`
`x=5`
TH2: `1-(3x-15)^5=0`
`(3x-15)^5=1^5`
`3x-15=1`
`x=16/3` (Loại)
c, `(x-8)^3=(x-8)^6`
`(x-8)^3 .[1-(x-8)^3]=0`
TH1: `(x-8)^3=0`
`x=8`
TH2: `1-(x-8)^3=0`
`x-8=1`
`x=9`
\(a,\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left(x-9\right)=\left(x-9\right)^2\)
\(\Rightarrow\left(x-9\right)^3\)
\(\Rightarrow x=9\)