a. x:(1/2+2/3)=6/5

=>x:7/6=6/5

=>x=6/5*7/6=>x=7/5

b.(x-1/2)-5(x-2/3)=3/2x

=>x-1/2-5x+10/3=3/2x

=>-4x+17/6=3/2x

=>17/6x=3/2x--4x

=>17/6=x(3/2+4)=>17/6=11/2x=>x=17/33

c.-5(x+1/5)-1/2(x-2/3)=x

=>-5x-1-1/2x+1/3=x=>-11/2x-2/3=x

=>-2/3=x+11/2x=>-2/3=x(1+11/2)=>-2/3=13/2x

=>x=-4/39

a) x : (1/2 + 2/3) = 6/5

=> x : 7/6 = 6/5

=> x = 6/5 x 7/6

      x = 7/5

b) , c) ko bít hihi

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)

1: (x+1)(y+2)=5

mà y+2>=2(do y là số tự nhiên)

nên (x+1;y+2)∈(1;5)

=>(x;y)∈(0;3)

2: (x+1)(y+2)=6

mà x+1>=1 và y+2>=2(do x,y là các số tự nhiên)

nên (x+1;y+2)∈{(3;2);(2;3);(1;6)}

=>(x;y)∈{(2;0);(1;1);(0;4)}

3: (x+2)(y+3)=6

mà x+2>=2 và y+3>=3(do x,y là các số tự nhiên)

nên (x+2;y+3)∈{(2;3)}

=>(x;y)∈(0;0)

4: (x-1)(y+3)=6

mà y+3>=3(do y là số tự nhiên)

nên (x-1;y+3)∈{(2;3);(1;6)}

=>(x;y)∈{(3;0);(2;3)}

5: (x-1)(y-3)=5

=>(x-1;y-3)∈{(1;5);(5;1)}

=>(x;y)∈{(4;8);(6;4)}

6: (x-2)(y-1)=3

=>(x-2;y-1)∈{(1;3);(3;1)}

=>(x;y)∈{(3;4);(5;2)}

7: (x-2)(y-1)=5

=>(x-2;y-1)∈{(1;5);(5;1)}

=>(x;y)∈{(3;6);(7;2)}

8: (x-3)(y+1)=7

mà y+1>=1(do y là số tự nhiên)

nên (x-3;y+1)∈{(1;7);(7;1)}

=>(x;y)∈{(4;6);(10;0)}

1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2

2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2

7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7

3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0

8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9

4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2

9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9

5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8

10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4

1) 2(x + 5) + 3(x + 7) = 41

2x + 10 + 3x + 21 = 41

5x + 31 = 41

5x = 41 - 31

5x = 10

x = 10 : 5

x = 2

2) 5(x + 6) + 2(x - 3) = 38

5x + 30 + 2x - 6 = 38

7x + 24 = 38

7x = 38 - 24

7x = 14

x = 14 : 7

x = 2

3) 7(5 + x) + 2(x - 10) = 15

35 + 7x + 2x - 20 = 15

9x + 15 = 15

9x = 15 - 15

9x = 0

x = 0

4) 3(x + 4) + (8 - 2x) = 22

3x + 12 + 8 - 2x = 22

x + 20 = 22

x = 22 - 20

x = 2

5) 4(x + 5) + 3(7 - x) = 49

4x + 20 + 21 - 3x = 49

x + 41 = 49

x = 49 - 41

x = 8

6) 7(x - 1) + 5(3 - x) = 11x - 10

7x - 7 + 15 - 5x = 11x - 10

2x - 11x + 8 = -10

-9x = -10 - 8

-9x = -18

x = -18 : (-9)

x = 2

7) 4(2 + x) + 3(x - 2) = 12

8 + 4x + 3x - 6 = 12

7x + 2 = 12

7x = 12 - 2

7x = 10

x = 10/7

8) 5(2 + x) + 4(3 - x) = 10x - 15

10 + 5x + 12 - 4x = 10x - 15

10x - 15 = x + 22

10x - x = 22 + 15

9x = 37

x = 37/9

9) 7(x - 2) + 5(3 - x) = 11x - 6

7x - 14 + 15 - 5x = 11x - 6

11x - 6 = 2x + 1

11x - 2x = 1 + 6

9x = 7

x = 7/9

10) 5(3 - x) + 5(x + 4) = 6 + 4x

15 - 5x + 5x + 20 = 6 + 4x

6 + 4x = 35

4x = 35 - 6

4x = 29

x = 29/4

a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

=>\(x< -\dfrac{3}{2}\)

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)