số mủ của 1 dù lớn đến bao hiêu thì cũng bằng 1 suy ra x=1 vậy thay x=1 vào biễu thức là ra

x²-4x+7 = 0 ⇔ x² -4x + 4 + 3 = 0 

⇔ (x-2)²+3=0 ⇔ (x-2)²=-3 (vô lí)

Vậy pt vô nghiệm

*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm

Ta có: \(x^2-4x+7=0\)

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

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

mà \(\left(x-2\right)^2+3\ge3>0\forall x\)

nên \(x\in\varnothing\)(đpcm)

ta có : \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)(=)\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)(=)\(\left(4x-1\right)^{20}\left[\left(4x-1\right)^{10}-1\right]=0\)(=)\(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left[\left(4x-1\right)^{10}-1\right]=0\end{cases}}\)(=)\(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)(=)\(\orbr{\begin{cases}4x=1\\\begin{cases}4x-1=1\\4x-1=-1\end{cases}\end{cases}}\)(=)\(\orbr{\begin{cases}x=\frac{1}{3}\\\begin{cases}4x=2\\4x=0\end{cases}\end{cases}}\)\(\orbr{\begin{cases}x=\frac{1}{4}\\\begin{cases}x=\frac{1}{2}\\x=0\end{cases}\end{cases}}\)

\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(4x^{30}-1^{30}=4x^{20}-1^{20}\)

\(4x^{30}-4x^{20}=-1+1\)

\(4x^{20}\left(x^{10}-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4x^{20}=0\\x^{10}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x^{20}=0\\x^{10}=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)

hok tốt!!

Ta có \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

<=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

<=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)

<=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=1;4x-1=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\4x=2;4x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2;x=0\end{cases}}\)

Vậy \(x\in\left\{0;2;\frac{1}{4}\right\}\)

T k chắc bài t nhưng t chắc bạn ღTĭểυ Tɦưღ lm sai ròi )) lũy thừa thì lm j cs cái ct đó

Hc tốt

Răng lại -1ở đó ✰Nanamiya Yuu⁀ᶜᵘᵗᵉ

\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\Leftrightarrow\left(4x-1\right)^{20}\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=-1;1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=\frac{1}{2}\end{cases}}\)

Vậy \(x=\frac{1}{4};0;\frac{1}{2}\)

P/s : phần \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=\frac{1}{2}\end{cases}}\)   thay dấu \(\hept{\begin{cases}\\\\\end{cases}}\) thành dấu \(\orbr{\begin{cases}\\\end{cases}}\) nhé!
\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\end{cases}}\)

\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\Leftrightarrow\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\4x-1=\pm1\end{cases}}\)

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

Vậy x = 1/4 hoặc 1/2 hoặc 0 

Ta có:

\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

Xét \(4x-1=0\)

\(\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\), thỏa mãn

Xét \(4x-1\ne0\)

\(\Rightarrow\left(4x-1\right)^{30}:\left(4x-1\right)^{20}=1\)

\(\Rightarrow\left(4x-1\right)^{10}=1\Rightarrow\left[{}\begin{matrix}4x-1=1\Rightarrow x=\frac{1}{2}\\4x-1=-1\Rightarrow x=0\end{matrix}\right.\)

Vậy....

\(\Rightarrow124-\left(20-4x\right)=120\\ \Rightarrow20-4x=24\\ \Rightarrow4x=-4\\ \Rightarrow x=-1\)

[ 124 - ( 20 - 4x ) ] : 30 = 4 

=> [ 124 - ( 20 - 4x ) ] : 30 = 4

=> 124 - ( 20 - 4x ) = 120

=> 20 - 4x = 124 - 120

=> 20 - 4x = 4

=>4x = 20 - 4

=>4x = 16

=> x = 16 : 4

=> x = 4

Vậy x = 4 .

\(\left[124-\left(20-4x\right)\right]:30=4\)

\(\Leftrightarrow20-4x=4\)

hay x=4