(x-5)(x-7)=0

=> x-5=0     hoặc x-7=0

=> x=5        hoặc x=7

541+(218-x)=735

=> 218-x=735-541

=> 218-x=194

=> x=218-194

=> x=24

96-3(x+1)=42

=>3(x+1)=96-42

=>3(x+1)=54

=>x+1=54:3

=>x+1=18

=>x=18-1

=>x=17

(x-47)-115=0

=>x-47=0+115

=>x-47=115

=>x=115+47

=>x=162

(x-36):18=12

=>x-36=12.18

=>x-36=216

=>x=216+36

=>x=252

(x-5).(x-7)=0

x-5=0+7

x=7+5

x=12

541+(218-x)=735

218-x=735-541

x=218-196

x=12

96-3(x+1)=42

3(x+1)=96-42

x+1=54:3

x=17-1

x=16

(x-47)-115=0

x-47=0+115

x=115+47

x=162

(x-36):12=18

x-36=12*18

x-36=216

x=216+36

x=252

a)= \(3^{x-1}=3^4\)

=>\(x-1=4\)

\(x=1+4\)

\(x=5\)

b)

\(3^{2.x}=36\)

\(x=2\)

x - 36 : 18 = 12

x - 2 = 12

x = 12 + 2

x = 14

(x : 23 + 45) . 67 = 8911

x : 23 + 45 = 8911 : 67

x : 23 + 45 = 133

x : 23 = 133 - 45

x : 23 = 88

x = 88 . 23

x = 2024

(x - 5)(x - 7) = 0

=> \(\orbr{\begin{cases}x-5=0\\x-7=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=5\\x=7\end{cases}}\)

\(\left(x-5\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=7\end{cases}}}\)

tíc mình nha

dễ mà bạn

Bài 1:

b: \(\frac{\frac37-\frac{3}{13}+\frac{3}{17}}{\frac57-\frac{5}{13}+\frac{5}{17}}\)

\(=\frac{3\left(\frac17-\frac{1}{13}+\frac{1}{17}\right)}{5\left(\frac17-\frac{1}{13}+\frac{1}{17}\right)}=\frac35\)

c: \(\frac{5^4\cdot4^4\cdot5^4}{5^{10}\cdot4^5}=\frac{5^8\cdot4^4}{5^{10}\cdot4^5}=\frac{1}{5^2\cdot4}=\frac{1}{100}\)

d: \(\left|x-2007\right|+\left|2y-2020\right|^{2021}=0\)

=>x-2007=0 và 2y-2020=0

=>x=2007 và 2y=2020

=>x=2007 và y=1010

Bài 2:

b: \(A=\frac{1}{2^2}+\frac{1}{2^4}+\cdots+\frac{1}{2^{100}}\)

=>\(4A=1+\frac{1}{2^2}+\cdots+\frac{1}{2^{98}}\)

=>4A-A=\(1+\frac{1}{2^2}+\cdots+\frac{1}{2^{98}}-\frac{1}{2^2}-\frac{1}{2^4}-\cdots-\frac{1}{2^{100}}\)

=>3A=\(1-\frac{1}{2^{100}}\)

=>3A<1

=>A<1/3

c: \(H=\left|x-3\right|+\left|4+x\right|\)

\(=\left|x+4\right|+\left|3-x\right|\ge\left|x+4+3-x\right|=7\forall x\)

Dấu '=' xảy ra khi (x-3)(x+4)<=0

=>-4<=x<=3

1a) (x-3)(2y+1)= 7

Vì \(x;y\in Z\Rightarrow\left(x-3\right);\left(2y+1\right)\inƯ\left(7\right)\)

Ta có bảng sau:

Vậy x= -4 ; y= -1

      x=2 ; y= -4

       x=4; y=3

      x= 10 ; y=0

1b) (2x+1)(3y-2) = -55

 Vì \(x;y\in Z\Rightarrow2x+1;3y-2\inƯ\left(-55\right)\)

Ta có bảng sau 

Vậy x=-28 ; y=1

      x=2 ; y=-3

       x= 5 ; y=-1

 bạn nhớ thử lại nha( ra giấy nháp)

a. \(7\left(x+3\right)=5\left(x+7\right)\)

\(7x+21=5x+35\)

\(2x=14\)

\(x=7\)

b.\(3x+1=3x+3\)

\(0x=2\)

\(x=\varnothing\)