a) 2^{x + 2} - 2^x = 96

=> 2^x . 4 - 2^x = 96

=> 2^x . (4 - 1) = 96

=> 2^x . 3 = 96

=> 2^x = 96 : 3

=> 2^x = 32

=> 2^x = 2⁵

=> x = 5

a) 2^{x + 2} - 2^x = 96

=> 2^x. 2² - 2^x = 96

=> 2^x( 4 - 1 ) = 96

=> 2^x = 96 : 3

=> 2^x = 2⁵

=> x = 5

b) 7^{x + 2} + 2 . 7^{x - 1} = 345

=> 7^{x - 1} . ( 7³ + 2 ) = 345

=> 7^{x - 1} . 345 = 345

=> 7^{x - 1} = 345 : 345

=> 7^{x - 1} = 1 = 7⁰

=> x - 1 = 0

=> x = 1

k mik nhé

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

c.\(4-x=2\left(x-4\right)^2\)

\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)

\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x=4 hoặc x=7/2

d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)

\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\) 

a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\)

\(\Leftrightarrow x^3+9x+2=x^3+8\)

\(\Leftrightarrow x^3+9x=x^3+8-2\)

\(\Leftrightarrow x^3+9x=x^3+6\)

\(\Leftrightarrow x^3+9x=x^3+6x-x^3\)

\(\Leftrightarrow\frac{2}{3}\)

b) \(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-4=8x-16\)

\(\Leftrightarrow x^4-4=8x-16+16\)

\(\Leftrightarrow x^2+12=8x\)

\(\Leftrightarrow x^2+12=8x-8x\)

\(\Leftrightarrow x^2-8x+12=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

\(\frac{1}{2}\cdot2^x+2^x\cdot2^2=2^8+2^5\)

\(2^x\left(\frac{1}{2}+4\right)=2^8+2^5\)

\(2^x\cdot\frac{9}{2}=288\)

\(2^x=64\)

\(2^x=2^6\)

\(x=6\)

\(9^x:3^x=3^7\)

\(3^{2x}:3^x=3^7\)

\(3^x=3^7\)

\(x=7\)

\(7^{x+2}+2\cdot7^{x-1}=345\)

\(7^x\cdot7^2+2\cdot7^x:7=345\)

\(7^x\left(7^2+\frac{2}{7}\right)=345\)

\(7^x\cdot\frac{345}{7}=345\)

\(7^x=7\)

\(x=1\)

a) 1/2.2^x + 2^x+2 = 256 + 32

1/2.2^x + 2^2.2^x=288

2^x(1/2+4)= 288

2^x.4,5=288

2^x= 288:4,5

2^x=64=2^6

x=6

(2x+1)^2=5^2

2x+1=5

x=2

(x-1)^3=-125

x-1=-5

x=-4

a)

(2x+1)²=25

^=> \(\left[\begin{array}{nghiempt}2x+1=5\\2x+1=-5\end{array}\right.\)

=>\(\left[\begin{array}{nghiempt}2x=4\\2x=-6\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)

d)

(x-1)³=-125

=> x-1=-5

=> x=-4

còn câu b và c bạn viết đề rõ hơn nha

a.

\(7^{x+2}+2\times7^{x-1}=345\)

\(7^x\times7^2+2\times7^x\div7=345\)

\(7^x\times\left(7^2+\frac{2}{7}\right)=345\)

\(7^x\times\frac{345}{7}=345\)

\(7^x=345\div\frac{345}{7}\)

\(7^x=345\times\frac{7}{345}\)

\(7^x=7\)

\(x=1\)

b.

\(2^{x+2}-2^x=96\)

\(2^x\times\left(2^2-1\right)=96\)

\(2^x\times3=96\)

\(2^x=\frac{96}{3}\)

\(2^x=32\)

\(2^x=2^5\)

\(x=5\)

\(a,7^{x+2}+2.7^{x-1}=345\Rightarrow7^x.49+\frac{2}{7}.7^x=345\Rightarrow7^x\left(49+\frac{2}{7}\right)=345\Rightarrow7^x.\frac{345}{7}=345\Rightarrow7^x=345:\frac{345}{7}=7^1\Rightarrow x=1\)

\(b,2^{x+2}-2^x=96\Rightarrow2^x.4-2^x=96\Rightarrow2^x\left(4-1\right)=96\Rightarrow2^x.3=96\Rightarrow2^x=96:3=32\Rightarrow2^x=2^5\Rightarrow x=5\)

\(a,7^{x+2}+2.7^{x-1}=345=>7^{x-1+3}+2.7^{x-1}=345=>7^{x-1}.7^3+2.7^{x-1}=345\)

\(=>\left(7^3+2\right).7^{x-1}=345=>345.7^{x-1}=345=>7^{x-1}=1=7^0=>x-1=0=>x=1\)

\(b,2^{x+2}-2^x=96=>2^x.2^2-2^x=96=>2^x.\left(4-1\right)=96=>2^x.3=96=>2^x=32=2^5=>x=5\)

Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

help me

Bài 2:

a) \(\dfrac{3}{10}+\dfrac{2}{5}:\dfrac{4}{15}=\dfrac{3}{10}+\dfrac{3}{2}=\dfrac{9}{5}\)

b) \(\dfrac{4}{7}.3+\dfrac{2}{3}.2=\dfrac{12}{7}+\dfrac{4}{3}=\dfrac{64}{21}\)