`@` `\text {Ans}`

`\downarrow`

`a)`

`2 \times x + 34 = 168`

`=> 2x = 168 - 34`

`=> 2x = 134`

`=> x = 134 \div 2`

`=> x = 67`

Vậy, `x=67`

`b)`

`x \times 11 + 51 = 273`

`=> 11x = 273 - 51`

`=> 11x = 222`

`=> x = 222 \div 11`

`=> x = 222/11`

`c)`

`x \div 120 - 45 = 155`

`=> x \div 120 = 155 + 45`

`=> x \div 120 = 200`

`=> x = 200 \times 120`

`=> x = 24000`

Vậy, `x = 24000`

`d)`

`320 \div x - 127 = 33`

`=> 320 \div x = 33+127`

`=> 320 \div x = 160`

`=> x = 320 \div 160`

`=> x = 2`

Vậy, `x=2.`

`@` `\text {Kaizuu lv uuu}`

a)2 x x +34 =168

2 x X = 168 - 34

2 x X = 134

X= 134:2

X= 67

b) x x 11 + 51 = 273

X x 11 = 273 - 51

X x 11= 222

X= 222 :11

X= 222/11

c) x : 120 – 45 = 155

x:120= 155+45

x:120=200

x=200:120

x=5/3

d) 320 : x – 127 = 33

320: x= 33+127

320:x=160

x=320:160

x=2

c,65

like nha b

A

A. 41; 43; 45; 47; 49.

minh ko lam duoc'

Đề có 1 chút lỗi nhỏ nhé, mình sửa lại luôn.

\(321⋮x\) và 15 < x < 320

\(321⋮x\Rightarrow x\inƯ\left(321\right)=\left\{\pm1;\pm3;\pm107;\pm321\right\}\)

Vì 15 < x < 320 \(\Rightarrow\) x = 107

a,  70 - 5 ( x - 3 ) = 45

⇔ 5 . x - 3 = 70 - 45

⇔ 5 x - 3 = 35

⇔ x - 3 = 35 : 5 ⇔ x - 3 = 7

⇔ x = 10

b,  10 + 2 x = 4 5 : 4 3

⇔ 10 + 2 x = 4 2 ⇔ 10 + 2 x = 16

⇔ 2 x = 4 ⇔ x = 2

c,  60 - 3 x - 2 = 51

⇔ 3 x - 2 = 60 - 51

⇔ 3 x - 2 = 9

⇔ x - 2 = 3 ⇔ x = 5

d,  4 x - 20 = 2 5 : 2 3

⇔ 4 x - 20 = 2 2 ⇔ 4 x - 20 = 4

⇔ 4 x = 24 ⇔ x = 6

\(3B=5+\left(\dfrac{5}{3}\right)+\left(\dfrac{5}{3}\right)^2+...+\left(\dfrac{5}{3}\right)^{20}\)

=>\(2B=5-\left(\dfrac{5}{3}\right)^{21}=\dfrac{5\cdot3^{21}-5}{3^{21}}\)

=>\(B=\dfrac{5\cdot3^{21}-5}{3^{21}\cdot2}\)

=> 
B
=
5
⋅
3
21
−
5
3
21
⋅
2

\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)

\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)

=>100-x=0

hay x=100

ta có: (59-x)/41  +(57-x)/43  +(55-x)/45  +(53-x)/47  +(51-x)/49 =-5

       <=>[(59-x)/41  +1 ]  +[(57-x)/43  +1]  +[(55-x)/45  +1]  +[(53-x)/47  +1]   +[(51-x)/49  +1] =0

      <=>(59-x-41)/41      + (57-x-43)/43     +(55-x-45)/45     +(53-x-47)/47      +(51-x-49)/49    =0

      <=>(100-x)/41   +   (100-x)/43       +    (100-x)/45     +(100-x)/47      +   (100-x)/49   =0

     <=>(100-x).( 1/41   +  1/43  +  1/45   +   1/47   +  1/49  )  =0

mà   (1/41  +  1/43  +  1/45  +  1/47  +  1/49) khác 0  nên 100-x =0 <=>x=100

vậy nghiệm của pt là x=100