a) PT \(\Leftrightarrow x^2-x-x^2+2x=5\) \(\Rightarrow x=5\)

  Vậy ...

b) PT \(\Leftrightarrow8x=16\) \(\Rightarrow x=2\)

  Vậy ...

a: Ta có: \(x\left(x-1\right)-x^2+2x=5\)

\(\Leftrightarrow x^2-x-x^2+2x=5\)

hay x=5

b: Ta có: \(2x\left(3x+4\right)-6x^2=16\)

\(\Leftrightarrow6x^2+8x-6x^2=16\)

\(\Leftrightarrow8x=16\)

hay x=2

`@` `\text {Ans}`

`\downarrow`

`2^x = 16`

`=> 2^x = 2^4`

`=> x = 4`

Vậy, `x = 4.`

____

`2^x*16 = 1024`

`=> 2^x =`\(2^{10}\div2^4\)

`=> 2^x = 2^6`

`=> x = 6`

Vậy, `x = 6`

______

`2^x - 26 = 6`

`=> 2^x = 6 + 26`

`=> 2^x = 32`

`=> 2^x = 2^5`

`=> x = 5`

Vậy, `x = 5`

`3^x*3 = 243`

`=> 3^x * 3 = 3^5`

`=> 3^x = 3^5 \div 3`

`=> 3^x = 3^4`

`=> x = 4`

Vậy, `x = 4.`

Tử \(x^4+2x^3+8x+16\)

\(=x^4-2x^3+4x^2+4x^3-8x^2+16x+4x^2-8x+16\)

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

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

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

Mẫu \(x^4-2x^3+8x^2-8x+16\)

\(=x^4-2x^3+4x^2+4x^2-8x+16\)

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

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

Thay tử và mẫu vào ta có:\(\frac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(x^2+4\right)\left(x^2-2x+4\right)}=\frac{\left(x+2\right)^2}{x^2+4}\ge0\)

Dấu "=" khi \(\left(x+2\right)^2=0\Leftrightarrow x=-2\)

Vậy Min=0 khi x=-2

-16 + 23 + x = -16

          7 + x = -16 

               x = -16 - 7

               x = -23

2x + 35 = -15

        2x = -15 - 35

         2x = -50

           x = -25

-13 x |x | = -26

         x = -26 : (-13)

          x = 2

Vậy x = 2 hoặc -2

|2x - 5| = 13

      2x = 13 + 5

       2x = 18

         x = 9

(x - 3) x (x + 2) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

2)2x+35=-15

2x      =-15-35

2x      = -50

x        =-25

1)-16+23+x=-16

7+x          =-16

x             =-16-7

x             =-23

3)-13 * lxl =-26

lxl         =  -26:-13

lxl           =2

<=>x=-2 hoac x=2

5) (x-3).(x+2) =0

<=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

Đề:........

<=> x². (2x + 7) - 16. (2x + 7) = 0

<=> (2x + 7). (x² - 16) = 0

<=> (2x+ 7). (x - 4). (x + 4) = 0

=> \(\hept{\begin{cases}2x+7=0\\x-4=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}2x=-7\\x=4\\x=-4\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{-7}{2}\\x=4\\x=-4\end{cases}}\)

Vậy...........

\(x^2\left(2x+7\right)=16\left(2x+7\right)\)

\(x^2\left(2x+7\right)-16\left(2x+7\right)=0\)

\(\left(2x+7\right)\left(x^2-16\right)=0\)

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

\(\hept{\begin{cases}x=-\frac{7}{2}\\x=-4\\x=4\end{cases}}\)

(2x + 4) - (x - 16) = -12

2x + 4 - x + 16 = -12

x + 20 = -12

x = -12 - 20

x = -32