Phương trình \(\Leftrightarrow\left(2^x\right)^2+10.2^x+16=0.\)

Đăt \(y=2^x>0\)

\(\Rightarrow y^2+10y+16=0\)

Giải phương trình bậc 2 tìm y từ đó suy ra x

Ta có 4^x-10.2^x+16=0

<=> (2^x)²-10.2^x+25-9=0

<=> (2^x-5)²-9=0

<=> (2^x-5+3)(2^x-5-3)=0

<=> (2^x-2)(2^x-8)=0

=> 2^x-2=0 hoặc 2^x-8=0

• 2^x-2=0 => 2^x=2 => x=1

• 2^x-8=0 => 2^x=8 => x=3

Vậy ...

a, pt <=> (x^3+x^2)-(4x^2-4) = 0

<=> (x+1).(x^2-4x+4) = 0

<=> (x+1).(x-2)^2 = 0

<=> x+1=0 hoặc x-2=0

<=> x=-1 hoặc x=2

b, pt <=> (x^4-x^3)+(2x^3-2x^2)-(2x^2-2x)+(3x-3) = 0

<=> (x-1).(x^3+2x^2-2x+3) = 0

<=> (x-1).[(x^3+3x^2)-(x^2+3x)+(3x+3)] = 0

<=> (x-1).(x+3).(x^2-3x+3) = 0

<=> x-1=0 hoặc x+3=0 ( vì x^2-3x+3 > 0 )

<=> x=1 hoặc x=-3

c, pt <=> (4^x-10.2^x+25)-9 =0

<=> (2^x-5)^2-9 = 0

<=> (2^x-5-3).(2^x-5+3) = 0

<=> (2^x-8).(2^x-2) = 0

<=> 2^x-8=0 hoặc 2^x-2=0

<=> x=3 hoặc x=1

Tk mk nha

a)   \(x^3-3x^2+4=0\)

\(\Leftrightarrow\)\(x^3+x^2-4x^2+4=0\)

\(\Leftrightarrow\)\(x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-4x+4\right)=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

Vậy....

a, pt <=> (x^3+x^2)-(4x^2-4) = 0

<=> (x+1).(x^2-4x+4) = 0

<=> (x+1).(x-2)^2 = 0

<=> x+1=0 hoặc x-2=0

<=> x=-1 hoặc x=2

b, pt <=> (x^4-x^3)+(2x^3-2x^2)-(2x^2-2x)+(3x-3) = 0

<=> (x-1).(x^3+2x^2-2x+3) = 0

<=> (x-1).[(x^3+3x^2)-(x^2+3x)+(3x+3)] = 0

<=> (x-1).(x+3).(x^2-3x+3) = 0

<=> x-1=0 hoặc x+3=0 ( vì x^2-3x+3 > 0 )

<=> x=1 hoặc x=-3

c, pt <=> (4^x-10.2^x+25)-9 =0

<=> (2^x-5)^2-9 = 0

<=> (2^x-5-3).(2^x-5+3) = 0

<=> (2^x-8).(2^x-2) = 0

<=> 2^x-8=0 hoặc 2^x-2=0

<=> x=3 hoặc x=1

Tk mk nha

a, Đặt \(2^x=t,t>0\)

Pt trở thành: \(t^2-10t+16=0\Leftrightarrow\left(t-2\right)\left(t-8\right)=0\Leftrightarrow\orbr{\begin{cases}t=2\\t=8\end{cases}\left(tm\right)}\)

Nếu t=2 => x=1

nếu t=8=> x=3

Vậy x=...

b, Đặt: \(2x^2-3x-1=t\)

pt trở thành: \(t^2-3\left(t-4\right)-16=0\Leftrightarrow t^2-3t-4=0\Leftrightarrow\left(t+1\right)\left(t-4\right)=0\Leftrightarrow\orbr{\begin{cases}t=-1\\t=4\end{cases}}\)

* Nếu t=-1 <=> \(2x^2-3x-1=-1\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

* Nếu t=4 <=> \(2x^2-3x-1=4\Leftrightarrow2x^2-3x-5=0\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{2}\end{cases}}\)

Vậy x=...

Đặt \(2^x=a>0\Rightarrow4^x=\left(2^2\right)^x=\left(2^x\right)^2=a^2\) pt trở thành:

\(a^2-10a+16=0\Rightarrow\left[{}\begin{matrix}a=8\\a=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2^x=8\\2^x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2^x=2^3\\2^x=2^1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

4^x-10.2^x+16=0

\(\Leftrightarrow\)2(2^x-5.2^x)=-16

\(\Leftrightarrow\)2^x(1-5)=-8

\(\Leftrightarrow\)-4.2^x=-8

\(\Leftrightarrow\)2^x=2

\(\Leftrightarrow\)x=1

=> S={1}

a: \(\Leftrightarrow5\sqrt{x+3}-4\sqrt{x+3}=3\sqrt{x-2}-3\sqrt{x-2}+2\)

\(\Leftrightarrow\sqrt{x+3}=2\)

=>x+3=4

hay x=1

c: \(\Leftrightarrow\left(x^2+4x\right)\left(x^2+4x-5\right)=84\)

\(\Leftrightarrow\left(x^2+4x\right)^2-5\left(x^2+4x\right)-84=0\)

\(\Leftrightarrow\left(x^2+4x\right)^2-12\left(x^2+4x\right)+7\left(x^2+4x\right)-84=0\)

\(\Leftrightarrow x^2+4x-12=0\)

=>(x+6)(x-2)=0

=>x=-6 hoặc x=2

x=3

b,Dat an 2x^2-3x-1=a la dc

a, \(4^x-10.2^x+16=0\Leftrightarrow\left(2^x\right)^2-10.2^x+16=0\)

Đặt \(2^x=t\Rightarrow t^2-10t+16=0\Leftrightarrow\orbr{\begin{cases}t=8\\t=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

b. Đặt \(2x^2-3x-1=t\Rightarrow t^2-3\left(t-4\right)-16=0\)

\(\Leftrightarrow t^2-3t-28=0\Leftrightarrow\orbr{\begin{cases}t=7\\t=-4\end{cases}}\)

Thế vào rồi giải tiếp em nhé.

camon ạ

b: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

c: \(\Leftrightarrow x\left(x^2-4x+5\right)=0\)

=>x=0

d: \(\Leftrightarrow2\cdot2^x-10\cdot2^x=-16\)

\(\Leftrightarrow-8\cdot2^x=-16\)

\(\Leftrightarrow2^x=2\)

hay x=1