a) \(2^x\cdot4=16\)

\(\Rightarrow2^x=16:4\)

\(\Rightarrow2^x=4\)

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

\(\Rightarrow x=2\)

b) \(3^x\cdot3=243\)

\(\Rightarrow3^x=243:3\)

\(\Rightarrow3^x=81\)

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

\(\Rightarrow x=4\)

a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)

\(\Rightarrow2x^2+x-6-4x^2+22x-10=-16\)

\(\Rightarrow2x^2-23x=0\Rightarrow x\left(2x-23\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)

b) \(7x^2-7=x^2-2x+1\)

\(\Rightarrow7\left(x^2-1\right)-\left(x^2-2x+1\right)=0\)

\(\Rightarrow7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)\left(7x+7-x+1\right)=0\Rightarrow2\left(x-1\right)\left(3x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{3}\end{matrix}\right.\)

a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)

 \(2x^2+x-6-4x^2+22x-10=-16\)

 \(-2x^2+23x-16=-16\)

\(23x-2x^2=0\)

\(x\left(23-2x\right)=0\)

⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)

b) \(7x^2-7=x^2-2x+1\)

\(7\left(x^2-1\right)=\left(x-1\right)^2\)

\(7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)

\(\left(7x+7\right)\left(x-1\right)-\left(x-1\right)^2=0\)

\(\left(x-1\right)\left(7x+7-x+1\right)=0\)

\(\left(x-1\right)\left(6x+8\right)=0\)

⇔ \(\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(a,2.\left|x+1\right|-3=5\)

\(\Rightarrow2.\left|x+1\right|=5+3\)

\(\Rightarrow2.\left|x+1\right|=8\)

\(\Rightarrow\left|x+1\right|=8:2\)

\(\Rightarrow\left|x+1\right|=4\)

\(\Rightarrow\orbr{\begin{cases}x+1=4\\x+1=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Vậy : x = 3 hoặc x = -5

b) Để A có giá trị nguyên thì n + 1 \(⋮\)n - 2

Ta có : n + 1 = ( n - 2 ) + 3

=> n + 1 \(⋮\)n - 2

khi ( n - 2 ) + 3 \(⋮\) n - 2

=> 3 \(⋮\)n - 2

=> n - 2 \(\in\)Ư ( 3 ) = { 1 ; -1 ; 3 ; -3 }

Với n - 2 = 1 => n = 3

Với n - 2 = -1 => n = 1

Với n - 2 = 3 => n = 5

Với n - 2 = -3 => n = -1 

Vậy : n \(\in\){ 3 ; 1 ; 5 ; -1 }

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\\ b,x:2=y:\left(-5\right)\Rightarrow\dfrac{x}{2}=\dfrac{y}{-5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)

ĐK : \(x\ne2\); \(x\ne-2\)

a) \(A=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)

\(=\frac{x^3-x.\left(x+2\right)-2.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{x^3-x^2-2x-2x+4}{\left(x+2\right).\left(x-2\right)}=\frac{x^3-x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2.\left(x-1\right)-4.\left(x-1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x-1\right).\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x-1\right)\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=x-1\)

b)  -  Để A > 0 thì   x - 1 > 0  =>  x > 1

     -  Để A < 0 thì   x - 1 < 0  =>  x < 1

c) Để  | A | = 5 thì   | x-1 | = 5

+ Nếu \(x-1\ge0\) thì \(x\ge1\) , ta có phương trình

x - 1 = 5 => x = 6 ( thỏa mãn ) 

+ Nếu x - 1 < 0 thì x < 1 , ta có phương trình : 

-x + 1 = 5  < = >  -x = 4  <=>  x = -4  ( thỏa mãn )

Vậy tập nghiệm của phương trình là S = { -4 ; 6 }

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4