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1: ĐKXĐ: -2/2x-2>=0
=>2x-2<0
=>x<1
2: ĐKXĐ: 2/3x-1>=0
=>3x-1>0
=>x>1/3
3: ĐKXĐ: 2x-2/(-2)>=0
=>2x-2<=0
=>x<=1
4: ĐKXĐ: (3x-2)/5>=0
=>3x-2>=0
=>x>=2/3
5: ĐKXĐ: (x-2)/(x+3)>=0
=>x>=2 hoặc x<-3
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
Xét A= \(\dfrac{x}{\sqrt{x+2yz}}\).\(\dfrac{1}{\sqrt{2}}\)=\(\dfrac{x}{\sqrt{2x+4yz}}\)=\(\sqrt{\dfrac{x.x}{2x+4yz}}\)
ta có x+y+z=\(\dfrac{1}{2}\)=> 2x+2y+2z= 1=> 2x+4yz= 4yz+1-2y-2z=(2y-1)(2z-1)
từ đó A= \(\sqrt{\dfrac{x}{2y-1}.\dfrac{x}{2z-1}}\)=\(\sqrt{\dfrac{x}{2y-2x-2y-2z}.\dfrac{x}{2z-2x-2y-2z}}\)
=\(\sqrt{\dfrac{x}{-2\left(x+y\right)}\dfrac{x}{-2\left(x+z\right)}}\)=\(\sqrt{\dfrac{1}{4}.\dfrac{x}{x+z}.\dfrac{x}{x+y}}\)=\(\dfrac{1}{2}\sqrt{\dfrac{x}{x+y}.\dfrac{x}{x+z}}\)
Áp dụng cô si \(\sqrt{ab}\)≤\(\dfrac{a+b}{2}\) =>\(\dfrac{1}{2}\sqrt{ab}\)≤\(\dfrac{a+b}{4}\)ta được
A≤\(\dfrac{1}{4}\).(\(\dfrac{x}{x+y}\)+\(\dfrac{x}{x+z}\))
cmmt thì \(\dfrac{P}{\sqrt{2}}\)≤ \(\dfrac{1}{4}\).\(\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{y+x}+\dfrac{y}{y+z}+\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)\)
\(\dfrac{P}{\sqrt{2}}\)≤\(\dfrac{3}{4}\)=>P≤\(\dfrac{3.\sqrt{2}}{4}\)=\(\dfrac{3}{2\sqrt{2}}\)
Dấu"=" xảy ra <=> x=y=z=\(\dfrac{1}{6}\)
Câu 1:
1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{7}{x^2+3}\)
Câu 2a đề sai, pt này ko giải được
2b.
\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)
Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)
\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)
3a.
\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)
\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)
3b
Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6
\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6
\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)
Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)
\(\Rightarrow2021^{2022}-1⋮6\)
\(\Rightarrow S-1⋮6\)
