=> \(\left[\begin{array}{} x=\sqrt{15}\\ x=-\sqrt{15} \end{array} \right.\)

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a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

1. Ta có :

27⁵. 49⁸ = 3¹⁵ . 7¹⁶ 

21¹⁵ = 7¹⁵.3¹⁵

Do  3¹⁵ . 7¹⁶ > 7¹⁵.3¹⁵

Nên 27⁵. 49⁸ > 21¹⁵ 

Câu 2 đang làm

21^15 và 27^5 . 49^8

27^5 . 49^8 = (3^3)^5 . (7^2)^8 = 3^15 . 7^16 = 3^15 . 7^15 . 7 = 21^15 . 7

Vì 21^15 < 21^15 . 7 nên 21^15 < 27^5 . 49^8

Ta có :

x¹⁰ = 1¹⁰

x¹⁰ = 1

=> x = 1

Nếu có gì sai thì mình sorry.

\(a,\left(x-15\right)\times7-270=169\times45\\ \left(x-15\right)\times7-270=7605\\ \left(x-15\right)\times7=7605+270\\ \left(x-15\right)\times7=7875\\ x-15=7875:7\\ x-15=1125\\ x=1125+15\\ x=1140\\ b,64-\left(34\times x-8\right)=4\\ 34\times x-8=64-4\\ 34\times x-8=60\\ 34\times x=60+8\\ 34\times x=68\\ x=68:34\\ x=2\)

60-[15*X+4]=15/2:1/2

60-[15*X+4]=15

15*X+4=60-15

15*X+4=45

15*X=45-4

15*X=41

X=41:15

X=41/15

ko ghi đề

\(60-\left(15.x+4\right)=\frac{15}{2}.\frac{2}{1}\)

\(60-\left(15.x+4\right)=15\)

\(15.x+4=60-15\)

\(15.x+4=45\)

\(15.x=45-4\)

\(15.x=41\)

\(x=41:15\)

\(x=\text{2.7333}\)

à nhầm bạn ơi 

\(41:15=\frac{41}{15}=2\frac{11}{15}\)

Sửa đề: \(\dfrac{x+2}{2014}+\dfrac{x+1}{2015}=\dfrac{x+2001}{15}+\dfrac{x+2014}{2}\)
Ta có: \(\dfrac{x+2}{2014}+\dfrac{x+1}{2015}=\dfrac{x+2001}{15}+\dfrac{x+2014}{2}\)

\(\Leftrightarrow\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}+1=\dfrac{x+2001}{15}+1+\dfrac{x+2014}{2}+1\)

\(\Leftrightarrow\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}=\dfrac{x+2016}{15}+\dfrac{x+2016}{2}\)

\(\Leftrightarrow\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}-\dfrac{x+2016}{15}-\dfrac{x+2016}{2}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{15}-\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{15}-\dfrac{1}{2}\ne0\)

nên x+2016=0

hay x=-2016

Vậy: S={-2016}

40= 2³. 5

24= 2³ . 3

ƯCLN(40, 24) = 2³= 6

ƯC( 40,24) = ước của 6 = 1, 2, 3, 6

Sao lại = ... hả bạn? Nếu $35\times x+45$ không có giá trị nào thì cũng không tìm được $x$ nhé.