\(a,91-5.\left(5+x\right)=61\\ \Rightarrow=5.\left(5+x\right)=30\\ \Rightarrow5+x=6\\ \Rightarrow x=1.\\ b,\left[195-\left(3x-27\right)\right].39=4212\\ \Rightarrow195-\left(3x-27\right)=108\\ \Rightarrow3x-27=87\\ \Rightarrow3x=114\\ \Rightarrow x=38.\)

\(91-5.\left(5+x\right)=61\)

\(\Rightarrow5.\left(5+x\right)=91-61\)

\(\Rightarrow5.\left(5+x\right)=30\)

\(\Rightarrow5+x=\dfrac{30}{5}=6\)

\(\Rightarrow x=6-5=1\)

\(\left[195-\left(3x-27\right)\right].39=4212\)

\(\Rightarrow195-3x+27=\dfrac{4212}{39}=108\)

\(\Rightarrow222-3x=108\)

\(\Rightarrow3x=222-108=114\)

\(\Rightarrow x=\dfrac{114}{3}=38\)

$91-5.\left(5+x\right)=61$

$\Rightarrow 5.\left(5+x\right)=91-61$

$\Rightarrow 5.\left(5+x\right)=30$

⇒5+�=305=6⇒5+x=530​=6

$\Rightarrow x=6-5=1$

$\left[195-\left(3x-27\right)\right].39=4212$

⇒195−3�+27=421239=108⇒195−3x+27=
​4212/39=108

$\Rightarrow 222-3x=108$

$\Rightarrow 3x=222-108=114$

⇒�=1143=38⇒x=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}

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 

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

\(91-5\left(5+x\right)=61\Leftrightarrow5\left(5+x\right)=91-61\)

\(5\left(5+x\right)=30\Leftrightarrow5+x=30:5\Leftrightarrow5+x=15\)

\(x=15-5\Leftrightarrow x=10\)

\(\left(x+34\right)-50=56:2\Leftrightarrow\left(x+34\right)-50=28\)

\(x+34=28+50\Leftrightarrow x+34=78\Leftrightarrow x=78-34\)

\(x=44\)

a,91-5(5+x)=61

=>25+5x=91-61=30

=>5x=30-25=5

=>x=1

b,\([\left(x+34\right)-50]\)x2=56

=>(x+34)-50=56:2=28

=>x+34=28+50=78

=>x=78-34=44.

c,1045-\([2015-\left(3x-24\right)]\)=5

=>2015-(3x-24)=1045-5=1040

=>3x-24=2015-1040=975

=>3x=975+24=999

=>x=999:3=333

d,\([195-\left(3x-27\right)]\)x39=4212

=>195-(3x-27)=4212:39=108

=>3x-27=195-108=87

=>3x=87+27=117

=>x=39

e,30-3(x-2)=18

=>30-3x+6=18

=>30-3x=18-6=12

=>3x=30-12=18

=>x=18:3=6

a) \(...\Rightarrow5\left(5+x\right)=91-61=30\)

\(\Rightarrow\left(5+x\right)=30:5=6\Rightarrow x=6-5=1\)

b) \(...\Rightarrow\left(x+34\right)-50=56:2=28\)

\(\Rightarrow\left(x+34\right)=28+50=78\Rightarrow x=78-34=44\)

c) \(...\Rightarrow2015-\left(3x-24\right)=1045-5=1040\)

\(\Rightarrow\left(3x-24\right)=2015-1040=975\)

\(\Rightarrow3x=975+24=999\Rightarrow x=999:3=333\)

d) \(...\Rightarrow195-\left(3x-27\right)=4212:39=108\)

\(\Rightarrow\left(3x-27\right)=195-108=87\)

\(\Rightarrow3x=87+27=114\Rightarrow x=114:3=38\)

e) \(...\Rightarrow3\left(x-2\right)=30-18=12\Rightarrow x-2=12:3=4\)

\(\Rightarrow x=4+2=6\)

Tích của những phép nhân này gọi là số chính phương

mik chọn câu B nha

TL:
B) chính phương

HT.

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 }

con truc la con nao ?