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a)
\(5+\left(-7\right)+9+\left(-11\right)+13+\left(-15\right)\)
\(=\left[5+\left(-7\right)\right]+\left[9+\left(-11\right)\right]+\left[13+\left(-15\right)\right]\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)=-6\)
b)
\(\left(-6\right)+8+\left(-10\right)+12+\left(-14\right)+16\)
\(=\left[\left(-6\right)+8\right]+\left[\left(-10\right)+12\right]+\left[\left(-14\right)+16\right]\)
\(=2+2+2=6\)
a, (19x+2.52) : 14 = (13-8)2 - 42
(19x + 2.25) : 14 = 52 - 42
(19x + 50) : 14 = 25 - 16
(19x + 50) : 14 = 9
19 x + 50 = 9.14
19x + 50 = 126
19x = 126 - 50
19x = 76
x = 76 : 19
x = 4
vậy____
b) x + (x + 1) + (x + 2) + (x + 3)+.....+(x+30) = 1240
(x+x+x+...+x) + (1+2+3+...+30) = 1240
31x + 465 = 1240
31x = 1240 - 465
31x = 775
x = 775 : 31
x = 25
vậy____
c) |x + 7| = 20 + 5.(-3)
|x + 7| = 20 + (-15)
|x + 7| = 5
\(\Rightarrow\orbr{\begin{cases}x+7=5\\x+7=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5-7\\x=-5-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=-12\end{cases}}\)
vậy_____
a) 24.5 - [ 131. ( 13 - 4 )2 ]
=120 - [ 131 . 92 ]
=120 - [ 131 . 81 ]
=120 - 10611
= - 10491
b) 100 : {230:[450−(4−53−52.25)]}
= 100 : \(\left\{230:\left[450-\left(4-125-25.25\right)\right]\right\}\)
= \(100:\left\{230:\left[450-\left(4-125-625\right)\right]\right\}\)
= \(100:\left\{230:\left[450-\left(-746\right)\right]\right\}\)
=\(100:\left\{230:1196\right\}\)
= 100 : \(\dfrac{5}{26}\)= 520
\(a = \left( { - 2} \right).\left( { - 3} \right) = 2.3 = 6\)
\(b = \left( { - 15} \right).\left( { - 6} \right) = 15.6 = 90\)
\(c = \left( { + 3} \right).\left( { + 2} \right) = 3.2 = 6\)
\(d = \left( { - 10} \right).\left( { - 20} \right) = 10.20 = 200\).
giúp mik mik đang cần gấp
nhưng phả có lời giải đừng cho mỗi đáp án
a:Ta có: \(\left(x-9\right)^7=\left(x-9\right)^4\)
\(\Leftrightarrow\left(x-9\right)^4\cdot\left[\left(x-9\right)^3-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
b: ta có: \(\left(3x-15\right)^{15}=\left(3x-15\right)^{10}\)
\(\Leftrightarrow\left(3x-15\right)^{10}\cdot\left[\left(3x-15\right)^5-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{16}{3}\end{matrix}\right.\)
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) (−30)+(−5)= - (30+5)= -35
b) (−7)+(−13)= -(7+13)= -20
c) (−15)+(−235)= -(15+235)= -250
a) (-30) + (-5)
= -35
b) (-7) + (-13)
= -20
c) (-15) + (-235)
= -250