a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

2^x+2^x+3=144

2^x+2^x.2^3=144

2^x(1+2^3)=144

2^x.9=144

2^x=16

2^x=2^4=>x=4

a, 3^x . 3 = 243

3^x+1 = 3⁵ 

=> x + 1 = 5 => x = 4

b, 7^x . 49  = 7²⁷

7^x . 7² = 7²⁷

7^x+2 = 7²⁷

=> x + 2 = 27 => x = 25

c, 5^x+1 = 125

5^x+1 = 5³

=> x + 1 = 3 => x = 2

`#3107.101107`

a)

\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)

Vậy, `x = 4`

b)

\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)

Vậy, `x = 3`

c)

\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)

Vậy, `x = 4.`

a) \(27< 3^x< 243\)

\(\Rightarrow3^3< 3^x< 3^5\)

\(\Rightarrow3< x< 5\)

c) \(3^x+3^{x+2}=810\)

\(\Rightarrow3^x\left(1+3^2\right)=810\)

\(\Rightarrow3^x.10=810\)

\(\Rightarrow3^x=810:10\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

a) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

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

\(\Rightarrow x=5\)

b) \(27.3^x=243\)

\(3^x=243:27\)

\(3^x=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

c) \(3^x+25=26.2^2+2.3^0\)

\(3^x+25=104+2\)

\(3^x+25=106\)

\(3^x=106-25\)

\(3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

d) \(\left(7x-11\right)^3=2^5.5^2+200\)

\(\left(7x-11\right)^3=800+200\)

\(\left(7x-11\right)^3=1000\)

\(\Rightarrow\left(7x-11\right)^3=10^3\)

\(\Rightarrow7x-11=10\)

\(7x=10+11\)

\(7x=21\)

\(x=21:7\)

\(x=3\)

e) \(3^x.3=27\)

\(3^x=27:3\)

\(3^x=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

a) \(2^x-15=17\)

\(\Leftrightarrow2^x=17+15\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy x = 5

b) \(27.3^x=243\)

\(\Leftrightarrow3^x=243:27\)

\(\Leftrightarrow3^x=9\)

\(\Leftrightarrow3^x=3^2\)

\(\Leftrightarrow x=2\)

Vậy x = 2

c) \(\left(7x-11\right)^3=2^5.5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=800+200\)

\(\Leftrightarrow\left(7x-11\right)^3=1000\)

\(\Leftrightarrow\left(7x-11\right)^3=10^3\)

\(\Leftrightarrow7x-11=10\)

\(\Leftrightarrow7x=10+11\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=21:7\)

\(\Leftrightarrow x=3\)

Vậy x = 3

d) \(3^x.3=27\)

\(\Leftrightarrow3^x=27:3\)

\(\Leftrightarrow3^x=9\)

\(\Leftrightarrow3^x=3^2\)

\(\Leftrightarrow x=2\)

Vậy x = 2

_Chúc bạn học tốt_

Bài 4:

\(a,2^{30}=\left(2^3\right)^{10}=8^{10};3^{20}=\left(3^2\right)^{10}=9^{10}\\ Vì:8^{10}< 9^{10}\left(Vì:8< 9\right)\Rightarrow2^{30}< 3^{20}\\ b,9^{10}.27^5=\left(3^2\right)^{10}.\left(3^3\right)^5=3^{20}.3^{15}=3^{35}\\ 243^7=\left(3^5\right)^7=3^{35}\\ Vì:3^{35}=3^{35}\Rightarrow243^7=9^{10}.27^5\)

Bài 5:

100< 5^2x-3 < 59

Đề vầy hả em?

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Lời giải:

a) 

$3^{2x+1}.7^y=9.21^x=3^2.(3.7)^x=3^{2+x}.7^x$

Vì $x,y$ là số tự nhiên nên suy ra $2x+1=2+x$ và $y=x$

$\Rightarrow x=y=1$

b) \(\frac{27^x}{3^{2x-y}}=\frac{3^{3x}}{3^{2x-y}}=3^{x+y}=243=3^5\Rightarrow x+y=5(1)\)

\(\frac{25^x}{5^{x+y}}=\frac{5^{2x}}{5^{x+y}}=5^{x-y}=125=5^3\Rightarrow x-y=3\) $(2)$

Từ $(1);(2)\Rightarrow x=4; y=1$