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Bài 1.
\(a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}\)
\(=\left(16\cdot3\cdot25\right):\left\{450:\left[450- \left(4\cdot125-8\cdot25\right)\right]\right\}\)
\(=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}\)
\(=1200:\left[450:\left(450-300\right)\right]\)
\(=1200:\left(450:150\right)\)
\(=1200:3\)
\(=400\)
\(---\)
\(b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}\)
\(=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}\)
\(=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}\)
\(=675-20\left[90-\left(164-2\cdot50\right)\right]\)
\(=675-20\left[90-\left(164-100\right)\right]\)
\(=675-20\left(90-64\right)\)
\(=675-20\cdot26\)
\(=675-520\)
\(=155\)
\(---\)
\(c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0\)
\(=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1\)
\(=\left(1\cdot2022-1986\right)\cdot5-169\)
\(=\left(2022-1986\right)\cdot5-169\)
\(=36\cdot5-169\)
\(=180-169\)
\(=11\)
Bài 2.
\(a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\\Rightarrow (2^x+1)^2+3\cdot5=40\\\Rightarrow (2^x+1)^2+15=40\\\Rightarrow (2^x+1)^2=40-15\\\Rightarrow (2^x+1)^2=25\\\Rightarrow (2^x+1)^2= (\pm 5)^2\\\Rightarrow \left[\begin{array}{} 2^x+1=5\\ 2^x+1=-5 \end{array} \right.\\ \Rightarrow \left[\begin{array}{} 2^x=4\\ 2^x=-6 (vô.lí) \end{array} \right. \\ \Rightarrow 2^x=2^2\\\Rightarrow x=2\)
Vậy \(x=2\).
\(---\)
\(b)3\cdot(x-7)+2\cdot(x+5)=41\\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\\Rightarrow 3x-21+2x+10=41\\\Rightarrow (3x+2x)+(-21+10)=41\\\Rightarrow 5x-11=41\\\Rightarrow 5x=41+11\\\Rightarrow 5x=52\\\Rightarrow x=\dfrac{52}{5}\)
Vậy \(x=\dfrac{52}{5}\).
\(Toru\)
a, 450 - 5x = 75
5x = 450 - 75
5x = 375
x = 75
Vậy...
b, ( 450 - 5x ) . 15 = 75
450 - 5x = 5
5x = 445
x = 89
Vậy ...
c, ( 450 - x ) : 15 = 150
450 - x = 2250
x = 450 - 2250
x = -1800
Vậy...
a) 450 -5 x X =75 b) ( 450 - 5 x X ) x15 =75 c) ( 450 - X ) : 15 =150
5 x X =450-75 450 - 5 x X =75 : 15 450 - X =150x15
5 x X =375 450 - 5 x X =5 450 - X =2250
X=375:5 5 x X =450:5 X =450-2250
X=75 5 x X =90 X =-1800
X =90:5
X =18
b)
`15-3(3x+18)=36`
`=> 3(3x+18)=15-36`
`=> 3(3x+18)=-21`
`=> 3x+18=-21:3`
`=> 3x+18=-7`
`=> 3x=-7-18`
`=> 3x=-25`
`=> x=-25:3`
`=> x=-25/3`
c)
`-12+(3-x)=25`
`=>-12+3-x=25`
`=>-x=25+12-3`
`=>-x=34`
`=>x=-34`
d)
`100-90xx(x:2-1)=-10`
`=>90xx(x:2-1)=100-(-10)`
`=>90xx(x:2-1)=110`
`=>x:2-1=110:90`
`=>x:2-1=11/9`
`=>x:2=11/9 +1`
`=> x:2=20/9`
`=> x=20/9 xx2`
`=> x=40/9`
Gọi số học sinh của mỗi tổ là a
24 : a suy ra a thuộc Ư (24)
28 : 4 suy ra a thuộc Ư (28)
Suy ra a thuộc ƯC ( 24 ; 28 )
24 = 23 . 3
28 = 22 . 7
Suy ra ƯCLN ( 24 ; 28 ) = 22 = 4
Vậy có thể chia được 4 tổ
số h/s nam trong 1 tổ là : 28 : 4 = 7
số h/s nữ trong 1 tổ là : 24 : 4 = 6
x ⋮ 18 và x ⋮ 15
⇒ x ∈ BC(18; 15)
Ta có:
\(18=3^2\cdot2\)
\(15=3\cdot5\)
\(\Rightarrow BCNN\left(18;15\right)=3^2\cdot2\cdot5=90\)
\(\Rightarrow BC\left(18;15\right)=\left\{0;90;180;270;360;450;540;630;...\right\}\)
Mà: 90 < x ≤ 450
\(\Rightarrow x\in\left\{180;270;360;450\right\}\)
\(18=3^2\cdot2;15=3\cdot5\)
=>\(BCNN\left(18;15\right)=3^2\cdot2\cdot5=90\)
\(x⋮18;x⋮15\)
=>\(x\in BC\left(18;15\right)\)
=>\(x\in B\left(90\right)\)
=>\(x\in\left\{90;180;270;360;450;540;...\right\}\)
mà 90<x<=450
nên \(x\in\left\{180;270;360;450\right\}\)