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Bài 2:
a, \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
Vậy...
b, \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy...
c, \(x^3-\dfrac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy...
Bài 3:
1, Đặt \(A=x^2+\dfrac{1}{2}x+\dfrac{1}{16}=x^2+\dfrac{1}{4}.x.2+\dfrac{1}{16}\)
\(=\left(x+0,25\right)^2\)
Thay x = 49,75 vào A ta có:
\(A=50^2=2500\)
2, tương tự
1. x5 + x + 1
=x5-x2+x2+x+1
=(x5-x2)+(x2+x+1)
=x2(x3-1)+(x2+x+1)
=x2(x-1)(x2+x+1)+(x2+x+1)
=(x2+x+1)[x2(x-1)+1]
=(x2+x+1)(x3-x2+1)
2. x5 + x4 +1
=x5+x4+x3-x3+1
=(x5+x4+x3)-(x3-1)
=x3(x2+x+1)-(x-1)(x2+x+1)
=(x2+x+1)[x3-(x-1)]
=(x2+x+1)(x3-x+1)
1.
= 4x\(^{^{ }2}\)-4x-9x+9
=4x(x-1)-9(x-1)
=(4x-9)(x-1)
e: Ta có: \(x^5-5x^3+4x\)
\(=x\left(x^4-5x^2+4\right)\)
\(=x\left(x^2-1\right)\left(x^2-4\right)\)
\(=x\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)