vãi fgdgfd

vãi fgdgfd

đặt 

\(A=15.\left(4^2+1\right)\left(4^4+1\right)..\left(4^{32}+1\right)\)

ta có 

\(A=\left(4^2-1\right)\left(4^2+1\right)..\left(4^{32}+1\right)\)

\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)..\left(4^{32}+1\right)=..=4^{64}-1\)

Đặt \(A=15\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)

\(=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)

\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)

\(=\left(4^{32}-1\right)\left(4^{32}+1\right)=4^{64}-1\)

a, \(\left(x-4\right)\left(x+2\right)\ge0\)

th1 : \(\hept{\begin{cases}x-4\ge0\\x+2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge4\\x\ge-2\end{cases}\Rightarrow}x\ge4}\)

th2 : \(\hept{\begin{cases}x-4\le0\\x+2\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le4\\\le-2\end{cases}\Rightarrow}x\le-2}\)

vậy x ≥ 4 hoặc x ≤ -2

b, \(x^2-6x+5=\left(x-1\right)\left(x+5\right)< 0\)  

th1 : \(\hept{\begin{cases}x-1< 0\\x+5>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x>-5\end{cases}\Rightarrow}-5< x< 1}\)

th2 : \(\hept{\begin{cases}x-1>0\\x+5< 0\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x< 5\end{cases}\left(voli\right)}}\)

vậy -5<x<1

b, \(x^2-6x+5< 0\Leftrightarrow\left(x-1\right)\left(x-5\right)< 0\)

Vì \(x-5< x-1\)

\(\hept{\begin{cases}x-1>0\\x-5< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x< 5\end{cases}\Leftrightarrow1< x< 5}\)

Vậy bft có tập nghiệm S = { x | 1 < x < 5 } 

a, \(2\left(x-1\right)^2+\left(x+5\right)\left(3-2x\right)=-22\)

\(\Leftrightarrow2\left(x^2-2x+1\right)+3x-2x^2+15-10x=-22\)

\(\Leftrightarrow2x^2-4x+2-7x-2x^2+15=-22\)

\(\Leftrightarrow-11x=-39\Leftrightarrow x=\frac{39}{11}\)

b, \(\left(x-3\right)^2-4x+12=0\Leftrightarrow\left(x-3\right)^2-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\Leftrightarrow x=3;x=7\)

c, \(2021x\left(x-2022\right)-x+2022=0\)

\(\Leftrightarrow2021x\left(x-2022\right)-\left(x-2022\right)=0\Leftrightarrow\left(x-2022\right)\left(2021x-1\right)=0\Leftrightarrow x=\frac{1}{2021};x=2022\)

\(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)\left(x-2-9\right)=0\\\left(x-2\right)\left(x-11\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-11=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=11\end{cases}}\)

bác nào giúp em đi em k

Phân tích:

\(4x^2-36x+56\)

\(=\left(4x^2-28x\right)-\left(8x-56\right)\)

\(=4x\left(x-7\right)-8\left(x-7\right)\)

\(=\left(x-7\right)\left(4x-8\right)\)

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

Tìm x:

\(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-9\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-11=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=11\end{cases}}\)