a)91-5.(5+x)=61 ;b)[(x+34)-50].2=56;c)1045-[2015-(3x-24)]=5;d)[195-(3x-27)].39=4212;e)30-3(x-2)=18
91-5.(5+x)=61
\(91-5\left(5+x\right)=61\)
\(\Rightarrow5\left(5+x\right)=91-61\)
\(\Rightarrow5\left(5+x\right)=30\)
\(\Rightarrow5+x=\dfrac{30}{5}\)
\(\Rightarrow5+x=6\)
\(\Rightarrow x=6-5\)
\(\Rightarrow x=1\)
91-5.(5+x)=61 [195-(3x-27)].39=4212
\(a,91-5.\left(5+x\right)=61\\ \Rightarrow=5.\left(5+x\right)=30\\ \Rightarrow5+x=6\\ \Rightarrow x=1.\\ b,\left[195-\left(3x-27\right)\right].39=4212\\ \Rightarrow195-\left(3x-27\right)=108\\ \Rightarrow3x-27=87\\ \Rightarrow3x=114\\ \Rightarrow x=38.\)
\(91-5.\left(5+x\right)=61\)
\(\Rightarrow5.\left(5+x\right)=91-61\)
\(\Rightarrow5.\left(5+x\right)=30\)
\(\Rightarrow5+x=\dfrac{30}{5}=6\)
\(\Rightarrow x=6-5=1\)
\(\left[195-\left(3x-27\right)\right].39=4212\)
\(\Rightarrow195-3x+27=\dfrac{4212}{39}=108\)
\(\Rightarrow222-3x=108\)
\(\Rightarrow3x=222-108=114\)
\(\Rightarrow x=\dfrac{114}{3}=38\)
\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
=> x = - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
Bài 1. Tìm số tự nhiên x biết:
a, 91 - 5 . ( 5 + x ) = 61
b, ( x+ 34 ) - 50 = 56 : 2
a, 91 - 5 . ( 5 + x ) = 61
5 . (5+x)=91-61
5 . (5+x)=30
5+x=30:5
5+x=6
x=6-5
x=1
Bài 1. Tìm số tự nhiên x biết:
a, 91 - 5 . ( 5 + x ) = 61
5(5 + x) = 91 - 61
5(5 + x) = 30
5 + x = 30 : 5
5 + x = 6
x = 6-5
x= 1
b, ( x+ 34 ) - 50 = 56 : 2
( x + 34 ) - 50 = 28
x + 34 = 28 + 50
x + 34 = 68
x= 78 - 34
x = 44
Tk mk nha
\(91-5\left(5+x\right)=61\Leftrightarrow5\left(5+x\right)=91-61\)
\(5\left(5+x\right)=30\Leftrightarrow5+x=30:5\Leftrightarrow5+x=15\)
\(x=15-5\Leftrightarrow x=10\)
\(\left(x+34\right)-50=56:2\Leftrightarrow\left(x+34\right)-50=28\)
\(x+34=28+50\Leftrightarrow x+34=78\Leftrightarrow x=78-34\)
\(x=44\)
Giải các phương trình sau ;
a) x-4/96 + x-7/93 + x-8/92 + x-10/90 + x-15/85 = 5
b) x+3/97 + x+5/95 + x+9/91 = x+91/98 + x+92/93 + x+61/99
\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
Ta có\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
<=> \(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+9}{91}+1\right)=\left(\frac{x+91}{9}+1\right)+\left(\frac{x+92}{8}+1\right)+\left(\frac{x+61}{39}+1\right)\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}=\frac{x+100}{9}+\frac{x+100}{8}+\frac{x+100}{39}\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}-\frac{x+100}{9}-\frac{x+100}{8}-\frac{x+100}{39}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\right)=0\)
Do \(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\ne0\)
Nên x+100=0 => x=-100
3 x 3
6 x 6
3 x 3
9 x 9
5 x 5
18 x 18
31 x 31
54 x 54
61 x 61
68 x 68
72 x 72
79 x 79
86 x 86
91 x 91
100 x 100
Tích của nhưng phép nhân này gọi là số ................. ( a) tự nhiên | b) chính phương | c) thập phân )
Tích của những phép nhân này gọi là số chính phương
mik chọn câu B nha
TL:
B) chính phương
HT.
1.Tìm x biết:
a) 91-5.(5+x)=61
b) [ (x+34)-50].2=56
2.Cho tập hợp A= {1;2;3;4}
Viết các tập hợp con của A sao cho mỗi tập hợp có 2 phần tử
Giúp mk nhé!
a) 91 - 5(5+x) = 61
5(5+x) = 30
5+x=6
x=1
b) [(x+34) - 50]2 = 56
(x+34) - 50 = 28
x+34 = 78
x= 44
Tập hợp con của A có 2 phần tử:
{1;2} {1;3} {1;4}
(2;3} {2;4}
{3;4}
a) 91 - 5 . ( 5 + x ) = 61
5 . ( 5 + x ) = 61
5 . ( 5 + x ) = 30
5 + x = 30 : 5
5 + x = 6
x = 6 - 5
x = 1.
b) [ ( x + 34 ) - 50 ] . 2 = 56
[ ( x + 34 ) - 50 ] = 56 : 2
[ ( x + 34 ) - 50 ] = 28
x + 34 = 28 + 50
x + 34 = 78
x = 78 - 34
x = 44.
2.Cho tập hợp A = { 1 ; 2 ; 3 ; 4 }
B = { 1 ; 2 }
C = { 1 ; 3 )
D = { 1 ; 4 }
E = { 2 ; 3 }
F = { 2 ; 4 }
1.Thực hiện các phép tính sau
a)42.57+43.42-600
b)22.52-64:23
c)120-[100-(5-2)3]
d)(102+112+122) : (132+142)
2. Tìm x biết
a)91-5.(5+x)=61
b)[(x+34)-50].2=56
a)
42.57 + 43.42 - 600
=42.(57+43)-600
=42.100-600
=4200-600
=3700