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Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
a) \(-ĐKXĐ:x\ne\pm2;1\)
Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)
b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)
\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)
Vậy với mọi x thỏa mãn x>1 thì A > 0
c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy x = -1;-2
\(T=\frac{1}{a^2+b^2+3}+\frac{1}{2ab}\)
\(T=\frac{1}{a^2+b^2+3}+\frac{1}{5ab}+\frac{3}{10ab}\)
Ta có: \(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\ge\frac{2}{\frac{x+y}{2}}=\frac{4}{x+y}\left(x,y>0\right)\)
\(2ab\le a^2+b^2\Leftrightarrow4ab\le\left(a^2+b^2+2ab\right)\Leftrightarrow2ab\le\frac{\left(a+b\right)^2}{2}\)
Áp dụng:
\(T\ge\frac{4}{a^2+b^2+3+5ab}+\frac{3}{5.\frac{\left(a+b\right)^2}{2}}\ge\frac{4}{\left(a+b\right)^2+3+1,5.\frac{\left(a+b\right)^2}{2}}+\frac{3}{5.\frac{2^2}{2}}=\frac{4}{2^2+3+1,5.\frac{2^2}{2}}+\frac{3}{5.2}=\frac{4}{10}+\frac{3}{10}=\frac{7}{10}\)Dấu " = " xảy ra \(\Leftrightarrow a=b=1\)( lát giải thích sau )
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2+b^2=2ab\\\frac{1}{a^2+b^2+3}=\frac{1}{5ab}\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\a^2+b^2+3=5ab\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\\left(a+b\right)^2-2ab+3=5ab\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=b\\4+3=5ab+2ab\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\7=7ab\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\ab=1\end{cases}}\Leftrightarrow a=b=1\)
Bổ sung thêm:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)( x,y>0)
Dấu " = '" xảy ra <=> x=y
\(2ab\le a^2+b^2\)
Dấu " = '" xảy ra <=> a=b