Ôn tập chương 1: Căn bậc hai. Căn bậc ba

tách như nầy nè

\(\dfrac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\le\dfrac{1}{2ab+2b+2}=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}\right)\)

\(P=x-2\sqrt{xy}+3y-2\sqrt{x}+1\)

\(\Leftrightarrow3P=3x-6\sqrt{xy}+9y-6\sqrt{x}+3\)

\(=\left(x-6\sqrt{xy}+9y\right)+\left(2x-\dfrac{2.\sqrt{2}.3.\sqrt{x}}{\sqrt{2}}+\dfrac{9}{2}\right)-\dfrac{3}{2}\)

\(=\left(\sqrt{x}-3\sqrt{y}\right)^2+\left(\sqrt{2x}-\dfrac{3}{\sqrt{2}}\right)^2-\dfrac{3}{2}\ge-\dfrac{3}{2}\)

\(\Rightarrow P\ge-\dfrac{1}{2}\)

Vậy GTNN là \(P=-\dfrac{1}{2}\) đạt được khi \(\left\{{}\begin{matrix}x=\dfrac{9}{4}\\y=\dfrac{1}{4}\end{matrix}\right.\)

\(\left\{a;b\right\}=\left\{2;3\right\};c=17\)

Ta có :x+y\(\ge2\sqrt{xy}\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+2\sqrt{x}\sqrt{y}\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)(luôn đúng với mọi x,y\(\ge0\))

Dấu"+" xảy ra khi:\(\sqrt{x}=\sqrt{y}\Leftrightarrow x=y\)

Vậy với mọi x,y\(\ge0\) thì x+y\(\ge2\sqrt{xy}\)

đong 2 bạn đổi lại dấu +\(2\sqrt{xy}\) thành -\(2\sqrt{xy}\) giùm mình

Ta có :\(2x+yz=\left(x+y+z\right)x+yz=x^2+xy+xz+yz=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)

\(\Rightarrow\sqrt{2x+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{\left(x+y\right)+\left(x+z\right)}{2}\)(bất đẳng thức cô si)

Cm tương tự :\(\sqrt{2y+xz}\le\dfrac{\left(y+x\right)+\left(y+z\right)}{2}\)

\(\sqrt{2z+xy}\le\dfrac{\left(z+y\right)+\left(z+x\right)}{2}\)

Do đó :P\(\le\dfrac{4\left(x+y+z\right)}{2}=2\left(x+y+z\right)=2\times2=4\)

Dấu "=" xảy ra khi :x=y=z=\(\dfrac{2}{3}\)

Vây giá trị lớn nhất của P=\(\sqrt{2x+yz}+\sqrt{2y+xz}+\sqrt{2z+xy}\) với x+y+z=2 và x,y,z\(\ge0\) là 4 khi x=y=z=\(\dfrac{2}{3}\)

M = \(\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)

= \(\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}-\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)

= \(2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)

= \(5\sqrt{6}\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+2xy+y^2}+\dfrac{1}{2.\dfrac{\left(x+y\right)^2}{4}}=4+2=6\)

ĐKXĐ: \(a\ne1\)

a/ P = \(\dfrac{3a+3\sqrt{a}-3-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)+\sqrt{a}-1-\left(a+\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

= \(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

= \(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

= \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)

b/ \(|P|=1\Leftrightarrow P=\pm1\)

* Với P = 1 thì \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=1\Leftrightarrow\sqrt{a}+1=\sqrt{a}-1\) (loại)

* Với P = -1 thì \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=-1\Leftrightarrow\sqrt{a}+1=1-\sqrt{a}\Leftrightarrow2\sqrt{a}=0\Leftrightarrow a=0\left(tm\right)\)

c/ P = \(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=\dfrac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\dfrac{2}{\sqrt{a}-1}\)

Để P \(\in N\) thì \(2⋮\sqrt{a}-1\) \(\Leftrightarrow\sqrt{a}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Đối chiếu với đk: a \(\ne1\) ta thấy a = 0; 4 và 9

Vậy để P \(\in N\) thì a = 0; a = 4; a = 9.

a, Có:\(C=\sqrt{x}-x\\ \Rightarrow C>0\Leftrightarrow\sqrt{x}-x>0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\Rightarrow0< \sqrt{x}< 1\Rightarrow0< x< 1\)

b, \(C=\sqrt{x}-x\Rightarrow C=-\left(\sqrt{x}-\dfrac{1}{2}\right)+\dfrac{1}{4}\ge\dfrac{1}{4}\)

Tự lm nốt nhé.

\(VP=\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\)

Áp dụng BĐT \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) có:

\(\dfrac{4a}{b+c}\le\dfrac{1}{4}\left(\dfrac{4a}{b}+\dfrac{4a}{c}\right)=\dfrac{4a}{b}\cdot\dfrac{1}{4}+\dfrac{4a}{c}\cdot\dfrac{1}{4}=\dfrac{a}{b}+\dfrac{a}{c}\)

Tương tự cho 2 BĐT còn lại ta có:

\(\dfrac{4b}{a+c}\le\dfrac{b}{a}+\dfrac{b}{c};\dfrac{4c}{a+b}\le\dfrac{c}{a}+\dfrac{c}{b}\)

Cộng theo vế 3 BĐT trên ta có:

\(\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\le\left(\dfrac{a}{c}+\dfrac{b}{c}\right)+\left(\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{c}{b}+\dfrac{a}{b}\right)\)​

\(\Leftrightarrow\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\le\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)

​

Ta có: \(\dfrac{a+b}{c}\)+\(\dfrac{b+c}{a}\)+\(\dfrac{c+a}{b}\)=\(\dfrac{a+b}{c}\)+\(\dfrac{c}{a+b}\)+\(\dfrac{b+c}{a}\)+\(\dfrac{a}{b+c}\)+\(\dfrac{c+a}{b}\)+\(\dfrac{b}{c+a}\)-(\(\dfrac{c}{a+b}\)+\(\dfrac{a}{c+b}\)+\(\dfrac{b}{a+c}\))

a/d bdt cosi cho...........................ta có

A\(\ge\)2\(\sqrt{\dfrac{a+b}{c}\times\dfrac{c}{a+b}}\)+

2\(\sqrt{\dfrac{b+c}{a}\times\dfrac{a}{b+c}}\)+2\(\sqrt{\dfrac{a+c}{b}\times\dfrac{b}{a+c}}\)

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