Tim x biet rang:
a) 4/4^x + 3.4^2-x = 832
b) (3x - 2)^2002 = (3x - 2)^2004
c) /1 - 2x/ + x + 2 = 0
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\(x^4-2x^3-2x^2+3x+2=0\)
\(\Leftrightarrow x^4-2x^3-2x^2+4x-x+2=0\)
\(\Leftrightarrow\left(x^4-2x^3\right)-\left(2x^2-4x\right)-\left(x-2\right)=0\)
\(\Leftrightarrow x^3\left(x-2\right)-2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-2x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x-x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x^3-x\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x^2-x\right)\left(x+1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x-1\right)=0\)
Đến đây ez r
b, Ta co: \(x^3+xy^2-x^2y-y^3+3\)
\(=\left(x^3-y^3\right)+\left(xy^2-x^2y\right)+3\)
\(=\left(x-y\right)^3+3xy\left(x-y\right)-xy\left(x-y\right)+3\)
= 3 ( vì x-y = 0)
Goi y
B1 X+3 chia het cho 5 7 9
B2 a ; Nhan x-1 vs 2 Roi tru cho nhau
b ; nhan x+1 vs 3
B3 nhan 3n +4 vs 4 ; 4n +5 vs3 roi tru
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)