\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

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=...

a) \(x^3-3x^2+4=0\)

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

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

b) \(\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-5\right)-16=0\)

\(\Leftrightarrow4x^4-12x^3+7x^2+3x=0\)

\(\Leftrightarrow x\left(2x-3\right)\left(2x^2-3x-1\right)=0\)

\(\Leftrightarrow2x-3=0\)

\(\Leftrightarrow2x=0+3\)

\(\Leftrightarrow2x=3\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

a)  \(x^3-3x^2+4=0\)

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

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

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

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

Vậy....

a) \(x^3+x^2+2x-16\ge0\)

\(\Leftrightarrow x^3-2x^2+3x^2-6x+8x-16\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+8\right)\ge0\)

Mà \(x^2+3x+8>x^2+3x+2,25=\left(x+1,5\right)^2\ge0\)

Cho nên \(x-2\ge0\)

\(\Leftrightarrow x\ge2\)

a,x^3-2x^2+3x^2-6x+8x-16>=0

(x^2+3x+8)(x-2)>=0

x^2+3x+8>0

=> để lớn hơn hoac bang 0 thì x-2 phải>=0

=>x>=2

b,hình như là vô nghiệm ko chắc chắn lắm

a) \(x^3+x^2+2x-16\ge0\)

\(\Leftrightarrow x^3-2x^2+3x^2-6x+8x-16\ge0\)

\(\Leftrightarrow\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(8x-16\right)\ge0\)

\(\Leftrightarrow x^2\left(x-2\right)+3x\left(x-2\right)+8\left(x-2\right)\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+8\right)\ge0\)

Để \(\left(x-2\right)\left(x^2+3x+8\right)\ge0\)thì \(x-2\ge0\left(x^2+3x+8>0\forall x\right)\)

\(\Rightarrow x\ge2\)

Đặt \(t=2^x\left(t>0\right)\) thì phương trình trở thành 

\(4t^2-2t.4-\left(t^4+2t^3\right)=0\)

Bây giờ coi 4=u là một ẩn của phương trình, còn t là số đã biết. Phương trình trở thành phương trình bậc 2 đối với ẩn u. Tính \(\Delta'\)

ta có :

\(\Delta'=\left(-t\right)^2+\left(t^4+2t^3\right)=\left(t^2+t\right)^2\)

Do đó :

\(\begin{cases}u=t-t\left(t+1\right)\\u=t+t\left(t+1\right)\end{cases}\) \(\Leftrightarrow\begin{cases}4=-t^2\\4=t^2+2t\end{cases}\) \(\Leftrightarrow t^2+2t-4=0\)

                             \(\Leftrightarrow\begin{cases}t=-1-\sqrt{5}\\t=-1+\sqrt{5}\end{cases}\)

Suy ra \(2^x=\sqrt{5}-1\Leftrightarrow x=\log_2\left(\sqrt{5}+1\right)\)

\(\left(\sqrt{2x+5}-\left(x+1\right)\right)^2+\left(\sqrt{3\left(x+1\right)}-\sqrt{x+7}\right)^2=0.\\ \)
Đến đây chắc biết phải làm gì =))
 

Link : Hoc24

`3x+7=0`

`<=>3x=-7`

`<=>x=-7/3`

Vậy `S={-7/3}`

______________________

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

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

`<=>2x(3-2x)=0`

`@TH1:2x=0<=>x=0`

`@TH2: 3-2x=0<=>2x=3<=>x=3/2`

Vậy `S={0;3/2}`

3x+7=0

\(\Leftrightarrow3x=-7\Leftrightarrow x=-\dfrac{7}{3}\)

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

\(\Leftrightarrow2x\left(x-2+5-3x\right)=0\)

\(\Leftrightarrow2x\left(-2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-3}{-2}=\dfrac{3}{2}\end{matrix}\right.\)

3x + 7 = 0

\(\Leftrightarrow\) 3x = -7

\(\Leftrightarrow\) x = -7/3

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

\(\Leftrightarrow\) 2x(x - 2 + 5 - 3x) = 0

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\-2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

giúp mình với

\(a,\Leftrightarrow\left(4-5x\right)\left(4+5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(3x+1-2x\right)\left(3x+1+2x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{5}\end{matrix}\right.\\ d,Sửa:\left(4x+1\right)^2-\left(x-2\right)^2=0\\ \Leftrightarrow\left(4x+1-x+2\right)\left(4x+1+x-2\right)=0\\ \Leftrightarrow\left(3x+3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{5}\end{matrix}\right.\\ e,\Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)