a) \(3^x+3^{x+2}=810\)

\(=>3^x+3^x\cdot9=810\)

\(=>3^x\left(1+9\right)=810\)

\(=>3^x\cdot10=810\)

\(=>3^x=810:10=81\)

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

\(=>x=4\)

a) \(3^x+3^{x+2}=810\)

\(\Leftrightarrow3^x\left(1+3^2\right)=810\)

\(\Leftrightarrow3^x\cdot10=810\)

\(\Leftrightarrow3^x=81\)

\(\Leftrightarrow x=4\)

b)\(2^x+2^{x+3}=144\)

\(\Leftrightarrow2^x\left(1+2^3\right)=144\)

\(\Leftrightarrow2^x\cdot9=144\)

\(\Leftrightarrow2^x=16\)

\(\Leftrightarrow x=4\)

a) \(3^x+3^{x+2}=810\)

\(\Rightarrow3^x+3^x.3^2=810\)

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

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

\(\Rightarrow3^x=81\)

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

\(\Rightarrow x=4\)

Vậy x = 4

b) \(2^x+2^{x+3}=144\)

\(2^x+2^x.2^3=144\)

\(2^x.\left(1+2^3\right)=144\)

\(2^x.9=144\)

\(2^x=144:9\)

\(2^x=16\)

\(2^x=2^4\)

\(\Rightarrow x=4\)

Vậy x = 4

\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)

\(C=\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)\)

\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]\)

\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[x^2+6x+4x+24\right]=\left[\left(x+2\right)\left(x+8\right)\right]\left(x^2+10x+24\right)\)

\(\frac{A}{B}=\frac{\left[\left(x+2\right)\left(x+8\right)\right]\left(x^2+10x+24\right)+2016}{\left(x^2+10x+24\right)}=\left(x+2\right)\left(x+8\right)+\frac{2016}{\left(x^2+10x+24\right)}\)

C=(x+2)(x+4)(x+6)(x+8)  chia het cho B

=> A chia B dư 2016

bạn có thể trình bày bài toán ra ko?

a) 4x - 20 = 0 <=> 4x = 20  <=>  x = 5

b) 2x + x + 12 = 0 <=> 3x + 12 = 0 <=>  3x = -12 <=> x = -4

c) x - 5 = 3 - x <=> 2x = 8 <=> x = 4

d) 7 - 3x = 9 - x <=> 2x = -2 <=> x = -1

a) 4x-20=0

4x=20

x=5

b)2x+x+12=0

3x=-12

x=-4

c) x-5=3-x

2x=8

x=4

d) 7-3x= chin -x

-2x=16

x=-8

Phương pháp đặt nhân tử ưu tiên nha bạn :

\(4x-20=0\Leftrightarrow4\left(x-5\right)=0\Leftrightarrow x=5\)

\(x-5=3-x\Leftrightarrow2x-8=0\Leftrightarrow2\left(x-4\right)=0\Leftrightarrow x=4\)

mình ko bít gì hết

a, TC: (x+1)+8 chia hết x+1

Vì (x+1) chia hết cho x+1=> 8 chia hết cho x+1

=>(x+1) thuộc Ư(8)

=>x thuộc{ -9;-5;-3;-1;0;1;3;7}

b, tương tự

a, (x+9) chia hết (x+1)
Vì x+9 = x+1+8
Mà(x+1) chia hết (x+1) với mọi x
=> 8 chia hết cho (x+1)
Ta có :Ư(8)=(1;-1;2;-2;4;-4;8;-8)
=> x+1=(1;-1;2;-2;4;-4;8;-8)
x=(0;-2;1;-3;3;-5;7;-9)
b, (4x+3) chia hết (x-2)
Vì 4x+3=4x-2+5
Mà(4x-2)chia hết (x-2) với mọi x
=>5 chia hết cho (x-2)
Ta có :Ư(5)=(1;-1;5;-5)
=>x-2=(1;-1;5;-5)
x=(3;1;7;-3)