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Câu 1:
$A+2=\frac{2}{2-x^2}+\frac{2}{x^2+1}=2(\frac{1}{2-x^2}+\frac{1}{x^2+1})$
$\geq 2.\frac{4}{2-x^2+x^2+1}=\frac{8}{3}$ (áp dụng BĐT Cauchy-Schwarz)
$\Rightarrow A\geq \frac{2}{3}$
Vậy $A_{\min}=\frac{2}{3}$ khi $x=\frac{1}{\sqrt{2}}$
Mặt khác:
\(A-1=\frac{2(x^2-1)}{2-x^2}+\frac{1-x^2}{1+x^2}=\frac{3x^2(x^2-1)}{(2-x^2)(x^2+1)}\leq 0\) với mọi $0\leq x\leq 1$
$\Rightarrow A\leq 1$
Vậy $A_{\max}=1$ khi $x=0$ hoặc $x=1$
Lời giải:
Gọi cạnh hình vuông là $a$
a) Áp dụng định lý Pitago cho các tam giác vuông sau:
Tam giác $ADM$: $AM=\sqrt{AD^2+DM^2}=\sqrt{a^2+(\frac{a}{2})^2}=\frac{\sqrt{5}}{2}a$
$AH=\sqrt{AB^2+BH^2}=\sqrt{a^2+(\frac{a}{3})^2}=\frac{\sqrt{10}}{3}a(1)$
$AB\parallel DM$ nên theo định lý Talet:
$\frac{AN}{NM}=\frac{AB}{DM}=2$
$\Rightarrow \frac{AN}{AM}=\frac{2}{3}$
$\Rightarrow AN=\frac{\sqrt{5}}{3}a(2)$
Mặt khác:
$\frac{BN}{DN}=\frac{AB}{DM}=2=\frac{BK}{KC}$ nên $NK\parallel DC$ (theo Talet đảo)
$\Rightarrow NK\perp BC$
$\frac{NK}{DC}=\frac{BK}{BC}=\frac{2}{3}\Rightarrow NK=\frac{2}{3}a$
Áp dụng định lý Pitago: $NH=\sqrt{NK^2+KH^2}=\sqrt{(\frac{2}{3}a)^2+(\frac{a}{3})^2}=\frac{\sqrt{5}}{3}a(3)$
Từ $(1);(2);(3)$ kết hợp Pitago đảo suy ra $ANH$ vuông cân tại $N$.
b)
Cho $AC$ cắt $NK$ tại $Q$
Theo định lý Talet:
$\frac{NQ}{MC}=\frac{AQ}{AC}=\frac{BK}{BC}=\frac{2}{3}$
$\Rightarrow \frac{NQ}{a}=\frac{1}{3}(4)$
$\frac{QK}{a}=\frac{QK}{AB}=\frac{KC}{BC}=\frac{1}{3}(5)$
Từ $(4);(5)\Rightarrow \frac{NQ}{a}=\frac{QK}{a}$
$\Rightarrow NQ=QK$ nên $Q$ là trung điểm $NK$
Do đó ta có đpcm.
\(P=\dfrac{1}{\left(x+1\right)^2+5}\le\dfrac{1}{5}\)
\(P_{max}=\dfrac{1}{5}\) khi \(x+1=0\Rightarrow x=-1\)
\(Q=\dfrac{x^2+x+1}{x^2+2x+1}=\dfrac{4x^2+4x+4}{4\left(x+1\right)^2}=\dfrac{3\left(x^2+2x+1\right)+x^2-2x+1}{4\left(x+1\right)^2}=\dfrac{3}{4}+\dfrac{\left(x-1\right)^2}{4\left(x+1\right)^2}\)
\(Q_{min}=\dfrac{3}{4}\) khi \(x-1=0\Rightarrow x=1\)
1: \(x^2+2x+6=x^2+2x+1+5=\left(x+1\right)^2+5>=5\forall x\)
=>\(P=\dfrac{1}{x^2+2x+6}< =\dfrac{1}{5}\forall x\)
Dấu '=' xảy ra khi x+1=0
=>x=-1
Lời giải:
Áp dụng BĐT AM-GM:
$1=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}$
$P=x^2y^2+\frac{1}{x^2y^2}+2-\frac{17}{6}$
$=x^2y^2+\frac{1}{x^2y^2}-\frac{5}{6}$
$=(x^2y^2+\frac{1}{256x^2y^2})+\frac{255}{256x^2y^2}-\frac{5}{6}$
$\geq 2\sqrt{\frac{1}{256}}+\frac{255}{256.\frac{1}{4^2}}-\frac{5}{6}=\frac{731}{48}$
Vậy $P_{\min}=\frac{731}{48}$ khi $x=y=\frac{1}{2}$
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
2)
Theo hệ quả của bất đẳng thức Cauchy ta có
\(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
Do \(x^2+y^2+z^2\le3\)
\(\Rightarrow3\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow1\ge xy+yz+xz\)
\(\Rightarrow4\ge xy+yz+xz+3\)
\(\Rightarrow\dfrac{9}{4}\le\dfrac{9}{3+xy+xz+yz}\) ( 1 )
Ta có \(C=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow C=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{9}{3+xy+yz+xz}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow C=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{9}{4}\)
Vậy \(C_{min}=\dfrac{9}{4}\)
Dấu " = " xảy ra khi \(x=y=z=\sqrt{\dfrac{1}{3}}\)
a/ \(M=\dfrac{x^2-x+1}{x^2+2x+1}=\dfrac{1}{4}+\dfrac{3x^2-6x+3}{x^2+2x+1}=\dfrac{1}{4}+\dfrac{3\left(x-1\right)^2}{x^2+2x+1}\ge\dfrac{1}{4}\)
b/ \(N=\dfrac{3x^2+4x}{x^2+1}=4-\dfrac{x^2-4x+4}{x^2+1}=4-\dfrac{\left(x-2\right)^2}{x^2+1}\le4\)
\(P=\dfrac{x^2}{2-x^2}+\dfrac{1-x^2}{1+x^2}\)
\(P+2=\dfrac{x^2}{2-x^2}+1+\dfrac{1-x^2}{1+x^2}+1\)
\(P+2=\dfrac{2}{2-x^2}+\dfrac{2}{1+x^2}\)
\(P+2=2\cdot\left(\dfrac{1}{2-x^2}+\dfrac{1}{1+x^2}\right)\)
\(P+2\ge2\cdot\dfrac{4}{2-x^2+1+x^2}=2\cdot\dfrac{4}{3}=\dfrac{8}{3}\)(AM-GM)
\(P\ge\dfrac{2}{3}\)
\(\Rightarrow MINP=\dfrac{2}{3}\Leftrightarrow x=\dfrac{\sqrt{2}}{2}\)(thỏa đk)
x^2 =t => 0<=t<=1
\(P=\dfrac{t}{2-t}+\dfrac{1-t}{1+t}=\dfrac{2-\left(2-t\right)}{2-t}+\dfrac{2-\left(t+1\right)}{1+t}\)
\(P=\dfrac{2}{2-t}-1+\dfrac{2}{1+t}-1\)
\(\dfrac{P}{2}+1=\dfrac{1}{2-t}+\dfrac{1}{1+t}=1+t+2-t=\dfrac{3}{\left(2-t\right)\left(1+t\right)}\)
\(\dfrac{P}{2}+1=\dfrac{3}{2+t-t^2}=\dfrac{3}{2+\dfrac{1}{4}-\left(\dfrac{1}{2}-t\right)^2}=\dfrac{3}{\dfrac{9}{4}-\left(\dfrac{1}{2}-t\right)^2}\ge\dfrac{3}{\dfrac{9}{4}}=\dfrac{4}{3}\)
\(\dfrac{P}{2}+1\ge\dfrac{4}{3}\Rightarrow P\ge2\left(\dfrac{4}{3}-1\right)=\dfrac{2}{3}\)
khi \(t=\dfrac{1}{2}\Rightarrow x=\pm\sqrt{\dfrac{1}{2}}=\pm\dfrac{\sqrt{2}}{2};x\in\left[0;1\right]\Rightarrow x=\dfrac{\sqrt{2}}{2}\) thủaman
GTNN P =2/3