15 +35 =
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\(\left(-35\right):63=-15:x\\ \dfrac{-35}{63}=\dfrac{-15}{x}\\ x=\dfrac{-15.63}{-35}=\dfrac{3.\left(-5\right).7.9}{-35}=\dfrac{3.\left(-35\right).9}{-35}=3.9=27\)
(-35) : 63 = -15 : \(x\)
\(x\) = (-15) : [ (-35) : 63]
\(x\) = -15 x \(\dfrac{63}{-35}\)
\(x\) = 27
`@` `\text {Ans}`
`\downarrow`
`18*15 + 35 + 15*32 + 35*49`
`= 15 (18+32) + 35(1+49)`
`= 15*50 + 35*50`
`= 50*(15+35)`
`= 50*50`
`= 2500`
\(18.15+35+15.32+35.4+9\)
\(=\left(18.15+15.32\right)+\left(35+35.4\right)+9\)
\(=\left[15.\left(18+32\right)\right]+\left[35.\left(1+4\right)\right]+9\)
\(=\left[15.50\right]+\left[35.5\right]+9\)
\(=750+175+9\)
\(=934\)
Bài 6. Tìm x ϵ N biết
a) (x –15) .15 = 0
b) 32 (x –10 ) = 32
c) ( x – 5)(x – 7) = 0
d) (x – 35).35 = 35
A.\(\left(x-15\right).15=0\)
\(x-15=0:15\)
\(x-15=0\)
\(x=15+0\)
\(x=15\)
B.\(32\left(x-10\right)=32\)
\(x-10=32:32\)
\(x-10=1\)
\(x=10+1\)
\(x=11\)
`a) `
`(x-15)xx15=0`
`<=> x-15 = 0 : 15`
`<=> x-15 = 0`
`<=> x = 0 + 15`
`<=> x =15`
`b)`
`32.(x-10)=32`
`<=> x - 10 = 32:32`
`<=>x-10=1`
`<=> x = 1+10`
`<=> x =11`
`c)`
`(x-5).(x-7)=0`
`<=>` \(\left[ \begin{array}{l}x-5 = 0\\x-7=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
`d)`
`(x-35)xx35=35`
`<=> x - 35 = 35:35`
`<=> x - 35 = 1`
`<=> x = 1+35`
`<=> x = 36`
(75+35+15)-(15-75+35)
= 75 + 35 + 15 - 15 + 75 - 35
= (35 - 35) + (15 - 15) + (75 + 75)
= 0 + 0 + 150
= 150
= 75+35+15-15+75-35
=(75+75) + ( 35-35)+ (15-15)
= 150+0+0=150
50, k mk nha
50
duyệt nha