a) 2x.4=128

   2x     =128:4

   2x      =   32

     x       =  32:2

     x       =   16

\(2^x.4=128\)

\(\Rightarrow2^x=32\left(\text{cùng chia cho 4}\right)\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

\(x^{15}=x\)

\(\Rightarrow x.x.x.x.x.x....x=x\)

\(\Rightarrow x=1\)

a) (2^x).4=128

2^x = 128:4

2^x = 32

mà 32=2^5=>x=5

b) ta có: x^15=x

theo quy ước: 0^15=0;1^15=1

=> x=1

4 câu còn lại mai mình sẽ giải nhé

a) x¹⁵= x.

=> x¹⁵- x= 0.

=> x( x¹⁴- 1)= 0.

=> \(\orbr{\begin{cases}x=0.\\x^{14}-1=0.\end{cases}}\)

=> \(\orbr{\begin{cases}x=0.\\x^{14}=1.\end{cases}}\)

=> \(\orbr{\begin{cases}x=0.\\x=1.\end{cases}}\)

Vậy x\(\in\) { 0; 1}

b) 16^x< 128.

Nếu x= 0 thì 16^x= 16⁰= 0( chọn)

Nếu x= 1 thì 16^x= 16¹= 16( chọn)

Nếu x= 2 thì 16^x= 16²= 256( loại)

Vậy x\(\in\) { 0; 1}

c) 5^x. 5^{x+ 1}. 5^{x+ 2}\(\le\) 1000...00: 2¹⁸( 18 chữ số 0)

=> 5^{x+ x+ 1+ x+ 2}\(\le\) 10¹⁸: 2¹⁸.

=> 5^{3x+ 3}\(\le\) 5¹⁸.

=> 3x+ 3\(\le\) 18.

=> 3x\(\le\) 15.

=> x\(\le\) 5.

=> x\(\in\){ 0; 1; 2; 3; 4; 5}

Vậy x\(\in\){ 0; 1; 2; 3; 4; 5}

d) 2^x.( 2²)²=( 2³)².

=> 2^x. 2⁴= 2⁶.

=> 2^x= 2⁶: 2⁴.

=> 2^x= 2².

=> x= 2.

Vậy x= 2.

e)( x⁵)¹⁰= x.

=> x⁵⁰- x= 0.

=> x( x⁴⁹- 1)= 0.

=> \(\orbr{\begin{cases}x=0.\\x^{49}-1=0.\end{cases}}\)

=> \(\orbr{\begin{cases}x=0.\\x^{49}=1.\end{cases}}\)

=> \(\orbr{\begin{cases}x=0.\\x=1.\end{cases}}\)

Vậy x\(\in\) { 0; 1}

\(x^{15}=x\)

\(\Rightarrow x^{15}-x=0\)

\(\Rightarrow x\left(x^{14}-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x^{14}-1=0\Rightarrow x=\pm1\end{cases}}\)

A) x=0 hoặc x= 1

B)x=1

\(a,2^x\cdot4=128\)

\(2^x=128:4=32\)

\(2^x=2^5\)

\(x=5\)

\(b,x^{15}=x\)

\(x^{15}-x=0\)

\(x\left(x^{14}-1\right)=0\)

\(\left\{{}\begin{matrix}x=0\\x^{14}-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(c,16^x< 128\)

\(2^{4x}< 2^7\)

\(4x< 7\)

\(x=1\)

d,\(5^x\cdot5^{x+1}\cdot5^{x+2}< 1000000000000000000:2^{18}\)

\(5^{x+x+1+x+2}< 10^{18}:2^{18}\)

\(5^{3x+3}< 5^{18}\)

\(3x+3< 18\)

\(3\left(x+1\right)< 18\)

\(x+1< 6\)

\(x< 5\)

\(e,2^x\cdot\left(2^2\right)^2=\left(2^3\right)^2\)

\(2^x\cdot2^4=2^6\)

\(2^{x+4}=2^6\)

\(x+4=6\)

\(x=2\)

\(f,\left(x^5\right)^{10}=x\)

\(x^{50}=x\)

\(x^{50}-x=0\)

\(x\left(x^{49}-1\right)=0\)

\(\left\{{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

a) (x - 15) × 7 - 270 : 45 = 169

(x - 15) × 7 - 6 = 169

(x - 15) × 7 = 169 + 6

(x - 15) × 7 = 175

x - 15 = 175 : 7

x - 15 = 25

x = 25 + 15

x = 40

b) [(4x + 28) × 3 + 55] : 5 = 35

(4x + 28) × 3 + 55 = 35 × 5

(4x + 28) × 3 + 55 = 175

(4x + 28) × 3 = 175 - 55

(4x + 28) × 3 = 120

4x + 28 = 120 : 3

4x + 28 = 40

4x = 40 - 28

4x = 12

x = 12 : 4

x = 3

c) (455 × x : 2 × 6) : 5 = 31

455 × x : 2 × 6 = 31 × 5

455 × x : 2 × 6 = 155

x × 455 : 2 × 6 = 155

x × 1365 = 155

x = 155 : 1365

x = 31/273

d) 128 × x - 12 × x - 16 × x = 520800

(128 - 12 - 16) × x = 520800

100 × x = 520800

x = 520800 : 100

x = 5208

e) (x × 0,25 + 2022) × 2023 = (50 + 2022) × 2023

(x × 0,25 + 2022) × 2023 = 2072 × 2023

(x × 0,25 + 2022) × 2023 = 4191656

x × 0,25 + 2022 = 4191656 : 2023

x × 0,25 + 2022 = 2072

x × 0,25 = 2072 - 2022

x × 0,25 = 50

x = 50 : 0,25

x = 200

f) 4 × x + 100 = x + 280

4 × x - x = 280 - 100

(4 - 1) × x = 180

3 × x = 180

x = 180 : 3

x = 60

g) (x + 1) + (x + 2) + (x + 3) + ... + (x + 100) = 7450

x + 1 + x + 2 + x + 3 + ... + x + 100 = 7450

100 × x + 100 × 101 : 2 = 7450

100 × x + 5050 = 7450

100 × x = 7450 - 5050

100 × x = 2400

x = 2400 : 100

x = 24

Bạn chỉ gửi 1 bài thôi chứ nhiều quá làm mỏi tay lắm

Làm bài 1 trước

\(4\cdot(-5)^2+2\cdot(-5)-20\)

\(=4\cdot25+2\cdot(-5)-20\)

\(=100+(-10)-20=100-30=70\)

\(35\cdot(14-10)-14\cdot(35-10)\)

\(=35\cdot14-35\cdot10-14\cdot35-14\cdot10\)

\(=35\cdot14-14\cdot35-35\cdot10-14\cdot10\)

\(=35\cdot10-14\cdot10=(35-14)\cdot10=210\)

\(3\cdot(-5)^2+2\cdot(-5)-20\)

Tương tự như ở câu trên

\(34\cdot(15-10)-15\cdot(34-10)\)

Tương tự như câu thứ 2

Câu cuối tự làm

\(a.2^x.4=128\)

\(\Rightarrow2^x=32\)

\(\Rightarrow x=5\)

\(b.x^{15}=x\)

\(\Rightarrow x\in \left\{1;0\right\}\)

Tương tự