a) -Thay \(x=a\) vào K ta được:

\(K=\dfrac{16}{\left(a^2+2\right)+4}\)

-Thay \(x=-a\) vào K ta được:

\(K=\dfrac{16}{\left(\left(-a\right)^2+2\right)+4}=\dfrac{16}{\left(a^2+2\right)+4}\)

-Vậy tại x=a và x=-a (a∈R) thì 2 giá trị của K bằng nhau.

b) -Không có GTNN, chỉ có GTLN:

\(K=\dfrac{16}{\left(x^2+2\right)^2+4}\le\dfrac{16}{2^2+4}=2\)

\(K_{max}=2\Leftrightarrow x=0\)

K=\(\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{2x-10}{x+2\sqrt{x}-3}ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-2x+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{x-1-2x+3\sqrt{x}-2\sqrt{x}-1-6+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)

Để K>0 thì :\(\frac{1}{\sqrt{x}-1}>0\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)

Với x>1 thoả mãn yêu cầu.

@Arakawa White

@DƯƠNG PHAN KHÁNH DƯƠNG

@Nguyễn Việt Lâm

@Nguyễn Huy Tú

giúp với ạ !

@Trần Trung Nguyên

a) \(K=\dfrac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{x+y}{\sqrt{xy}}\)

\(K=\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x+y\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(K=\dfrac{x^2-x\sqrt{xy}-y^2-y\sqrt{xy}-x^2+y^2}{\sqrt{xy}\left(x-y\right)}\)

\(K=\dfrac{-\sqrt{xy}\left(x-y\right)}{\sqrt{xy}\left(x-y\right)}=-1\)

Có gì đó hơi sai sai

 f(x) = x + x³ + x^9 + x^27 + x^81 
a) f(x) = (x-1).g(x) + r 
f(1) = 1+1+1+1+1+1 = 0.g(1) + r 
=> dư là r = 5 

b) f(x) = (x²-1).h(x) + ax+b 
{ f(1) = 5 = 0 + a + b <=> { a = 5 
{ f(-1) = -5 = 0 -a + b ------ { b = 0 
vậy dư là r(x) = 5x 

a) ĐKXĐ \(\hept{\begin{cases}x-1\ne0\\x+1\ne0\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne1\\x\ne-1\\x\ne0\end{cases}}\)

b)\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right)\frac{x+2003}{x}\)

\(=\frac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-4x-1}{\left(x-1\right).\left(x+1\right)}.\frac{x+2003}{x}\)

\(\frac{\left(x+1-x+1\right)\left(x+1+x-1\right)+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)

\(\frac{4x+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)

\(=\frac{x^2-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)

\(=\frac{x+2003}{x}\)

c) Ta có \(K=\frac{x+2003}{x}\)

Để K nguyên thì x + 2003 ⋮ x

Ta có x ⋮ x => 2003 ⋮ x

=> x thuộc Ư(2003) = { 1; -1; 2003; -2003 }

Vậy khi x thuộc { 1; -1; 2003; -2003 } thì K nguyên

1) \(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)

\(=\left(x+3\right).x^2-5\left(x+3\right)+\left(x+4\right)\left(x-1x^2\right)\)

\(=x^3+3x^2-5x-15+\left(x+4\right)\left(x-x^2\right)\)

\(=x^3+3x^2-5x-15-x^3+x^2-4x^2+4x\)

\(=3x^2-5x-15-3x^2+4x\)

\(=-x-15\)

2) Đặt đa thức là \(N\left(x\right)\)ta được: \(3x^3+2x^2-x+k=N\left(x\right)\left(x-1\right)\)

Để \(3x^3+2x^2-x+K⋮x-1\Leftrightarrow x=1\)

Thay vào ta được

\(\Rightarrow3.1^3+2.1^2-1+K=0\)

\(\Rightarrow3+2-1+K=0\)

\(\Rightarrow K=-4\)

3) \(Q=x^2-x+3\)

\(=x^2-x+\frac{1}{4}+\frac{11}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\)

Với \(\forall x\) ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow N=\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\)

a: \(K=\dfrac{3x+3-4x-2-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-x^2+x}{\left(x-1\right)\left(x+1\right)}=\dfrac{-x}{x+1}\)

Ta có: x+y+z=0

\(\Leftrightarrow\left(x+y+z\right)^2=0\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)(1)

Ta có: \(K=\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)

\(=\dfrac{x^2+y^2+z^2}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2}\)

\(=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2-x^2-y^2-z^2-2xy-2yz-2xz}\)

\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x^2+y^2+z^2+2xy+2yz-2xz\right)}\)

\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)

Vậy: \(K=\dfrac{1}{3}\)

\(K=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}\)

\(K=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2}=\dfrac{1}{3}\)

\(a,F=\dfrac{x^2+x+4x^2+2-x^2+3x-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x}{x-1}\\ b,\left|x+2\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1-2=-1\left(ktm\right)\\x=-1-2=-3\end{matrix}\right.\Leftrightarrow x=-3\\ \Leftrightarrow F=\dfrac{-12}{-4}=3\\ c,K=F\left(x-1\right)-x^2-2021=4x-x^2-2021\\ K=-\left(x^2-4x+4\right)-2017=-\left(x-2\right)^2-2017\le-2017\\ K_{max}=-2017\Leftrightarrow x=2\left(tm\right)\)