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a, \(\left|2x+1\right|=3\)
=> 2x + 1 = 3 hoặc 2x + 1 = -3
=> 2x = 3 - 1 hoặc 2x = -3 - 1
=> 2x = 2 hoặc 2x = -4
=> x = 1 hoặc x = -2
b, \(2\frac{1}{2}x+1\frac{1}{2}=2\frac{2}{3}\)
=> \(\frac{5}{2}x+\frac{3}{2}=\frac{8}{3}\)
=> \(\frac{5}{2}x=\frac{8}{3}-\frac{3}{2}\)
=> \(\frac{5}{2}x=\frac{16-9}{6}\)
=> \(\frac{5}{2}x=\frac{7}{6}\)
=> \(x=\frac{7}{6}:\frac{5}{2}=\frac{7}{6}\cdot\frac{2}{5}=\frac{7}{3}\cdot\frac{1}{5}=\frac{7}{15}\)
c, \(3\cdot5^{x-3}+1=16\)
=> 3 . 5x-3 = 16 - 1
=> 3 . 5x-3 = 15
=> 5x-3 = 15 : 3
=> 5x-3 = 5
=> x - 3 = 5 : 5
=> x - 3 = 1
=> x = 1 + 3 = 4
d, \((x-1)^2=25\)
=> \((x-1)^2=5^2\)
=> x - 1 = 5 hoặc x - 1 = -5
=> x = 6 hoặc x = -4
e, \((-2)^2+\left|3x+1\right|=(-28)\cdot7\)
=> 4 + |3x + 1| = -196
=> |3x + 1| = -196 - 4 = -200
=> |3x + 1| = -200
Không thỏa mãn điều kiện
a: \(=\dfrac{-3}{5}\cdot\dfrac{5}{7}+\dfrac{-3}{5}\cdot\dfrac{3}{7}+\dfrac{-3}{5}\cdot\dfrac{6}{7}\)
\(=\dfrac{-3}{5}\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)=\dfrac{-3}{5}\cdot2=-\dfrac{6}{5}\)
b: \(=\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{5}{11}-\dfrac{2}{13}=\dfrac{3}{13}-\dfrac{2}{13}=\dfrac{1}{13}\)
c: =>1/2x+1+3/8=7/16
=>1/2x=-15/16
=>x=-15/8
d: =>5/2x-1/3=1/6*(-9)/2=-9/12=-3/4
=>5/2x=-3/4+1/3=-9/12+4/12=-5/12
=>x=-1/6
1; 5.22 + (\(x\) + 3) = 52
5.4 + (\(x\) + 3) = 25
20 + (\(x\) + 3) = 25
\(x\) + 3 = 25 - 20
\(x+3\) = 5
\(x\) = 5 - 3
\(x\) = 2
Vậy \(x=2\)
2; 23 + (\(x\) - 32) = 53 - 43
8 + (\(x\) - 9) = 125 - 64
8 + (\(x\) - 9) = 61
\(x\) - 9 = 61 - 8
\(x\) - 9 = 53
\(x\) = 53 + 9
\(x\) = 62
Vậy \(x\) = 62
\(6^2-\left(x+3\right)=45\)
\(36-\left(x+3\right)=45\)
\(x+3=35-45\)
\(x+3=-10\)
\(x=-13\)
<=>{x2-[62(64-63)3-35]3-15}3=1
<=>[x2-(62.13-35)3-15]3=1
<=>[x2-(36-35)3-15]3=1
<=>(x2-1-15)3=1
<=>(x2-16)=\(\sqrt{1}=1\)
=>x2=17=>x=\(\sqrt{17}\)
{ x2 - [ 62 - ( 82 -9.7 )3 -7.5 ]3 -5.3 }3 =1 ... {x2 - [62 - 13 -7.5]3 - 5.3}3=1. {x2 - [36 - 1 -7.5]3 - 5.3}3=1. {x2 - [36 - 1 - 35]3 - 5.3}3=1. {x2 - [35 - 35]3 -5.3}3=1. {x2 - 03 - 5.3}3=1. {x2 - 0 -5.3}3=1. {x2 - 0 - 15}3=1. {x.x - 0 - 15}3=1. {x.x + 0+15}3=1. {x.x +15}3=1.
1)Tính:
4.{624:[5+7.(72:23-6)]}-5.3
=4.{624:[5+7.(72:8-6]}-15
=4.{624:[5+7.(9-6)]}-15
=4.{624:[5+7.3]}-15
=4.{624:26}-15
=4.24-15
=81
2) Tìm x biết
a)108-x=32.7
108-x=224
x=108-224
x=-116
b)15-(5x-4):3=23
(5x-4):3=15-23
(5x-4):3=-8
5x-4=-8.3
5x-4=-24
5x=-24+4
5x=-20
x=-20:5
x=-4
Bài 1 :
4 . {624 :[5 + 7 . (72 : 23 - 6)]} - 5 . 3
= 4 . {624 :[ 5 + 7 . (72 : 8 - 6)]} - 15
= 4 . {624 : [5 + 7 . (9 - 6)]} - 15
= 4 . [624 : (5 + 7 . 3)] - 15
= 4 . [624 : (5 + 21)] - 15
= 4 . (624 : 26) - 15
= 4 . 24 - 15
= 96 - 15
= 81