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Bài 3:
a: Ta có: 60-3(x-2)=51
\(\Leftrightarrow x-2=3\)
hay x=5
b: Ta có: \(4x-20=25:2^2\)
\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)
hay \(x=\dfrac{105}{16}\)
c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
a)36:32+23.22-33.3
= 34+25-34
=25
=32
b)38:34-95:93
= 34-92
= 34-34
= 0
c)23.15+23.35
= 23(15+35)
= 8.50
= 400
a: \(3^6:3^2+2^3\cdot2^2-3^3\cdot3\)
\(=3^4+32-3^4\)
=32
b: \(3^8:3^4-9^5:9^3\)
\(=3^4-9^2\)
=0
c: Ta có: \(2^3\cdot15+2^3\cdot35\)
\(=8\cdot50\)
=400
a: \(\left(-2\right)^3-45:\left(-3\right)^2+\left(-2019\right)^0\cdot1^{2019}\)
\(=-8-45:9+1\)
\(=-8-5+1\)
=-13+1
=-12
b: \(11^{25}:11^{13}-3^5:\left(1^{10}+2^3\right)-60\)
\(=11^{25-13}-3^5:3^2-60\)
\(=11^{12}-27-60\)
\(=11^{12}-87\)
`a, = 53 xx (12+172 -84)`
`= 53 xx 100`
`=5300`
`b,= 2018 xx (91-45-46)`
`= 2018 xx 0`
`=0`
b: =-45-27+8=-72+8=-64
c: =12:6=2
d: =-24:8=-3
e: =-36:9=-4
f: =14:7=2
\(25\cdot69+31\cdot25-150\)
\(=25\cdot\left(69+31\right)-150\)
\(=25\cdot100-150\)
\(=2500-150\)
\(=2350\)
a)31.65+31.35-500
=31.(65+35)-500
=31.100-500
=3100-500
=2600