1997^2-1996×1998-1 giúp mình với ạ!
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\(3x\left(\dfrac{4}{3x}+1\right)-4x\left(x-2\right)=10\\ \Rightarrow4+3x-4x^2+8x-10=0\\\Rightarrow -4x^2+11x-6=0\\ \Rightarrow-4x^2+8x+3x-6=0\\\Rightarrow \left(-4x+3\right)\left(x-2\right)=0\\\Rightarrow \left[{}\begin{matrix}x=2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Ta có :
x2+2x-y2+2y = x2 - y2 + 2x+2y
= (x-y)(x+y) + 2(x+y)
= (x-y+2)(x+y)
a, \(\dfrac{2-5x}{3}+\dfrac{2x^2-x}{2}>\dfrac{x\left(3x-1\right)}{3}-\dfrac{5x}{4}\)
\(\Rightarrow8-20x+12x^2-6x>4x\left(3x-1\right)-15x\)
\(\Leftrightarrow8-26x+12x^2>12x^2-19x\Leftrightarrow8>7x\Leftrightarrow x< \dfrac{8}{7}\)
b, \(\Rightarrow20x+10x+5>30x-2\Leftrightarrow5>-2\)(luôn đúng)
Vậy bft luôn có nghiệm
Ta có \(\left(x-5\right)^4+\left(x-2\right)^4=1^4+2^4=2^4+1^4\)
TH1 \(\left\{{}\begin{matrix}\left(x-5\right)^4=1^4\\\left(x-2\right)^4=2^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\\\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\\\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\end{matrix}\right.\)
TH2 \(\left\{{}\begin{matrix}\left(x-5\right)^4=2^4\\\left(x-2\right)^4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\end{matrix}\right.\)
A)x2-(x-1)2=15
=>(x+x-1)(x-x+1)=15
=>(2x-1)1=15
=>2x-1=15
=>2x=16
=>x=8
B)16x-(4x-5)=15
=>(4x)-(4x-5)=15
=>(4x+4x-5)(4x-4x+5)=15
=>(8x-5)5=15
=>8x-5=3
=>8x=8
=>x=1
19972-1996.1998-1
= 1997.1997-(1997-1).(1997+1)-1
= 1997.1997-(1997.1997+1997-1997-1)-1
= 1997.1997-1997.1997-1997+1997+1-1
= 0