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

Bài này dễ mà bn chỉ cần chuyển vế sau đó bình phương lên là lm đc thôi

(4x + 2y + 2z - \(\sqrt{4xy}-\sqrt{4xz}+2\sqrt{yz}\) )+(y - \(6\sqrt{y}\) + 9)+(z- \(10\sqrt{z}\) + 25) = 0

<=> (\(2\sqrt{x}-\sqrt{y}-\sqrt{z}\))² + (\(\sqrt{y}-3\))² + (\(\sqrt{z}-5\))² = 0 (1)

Vì VP \(\ge0\) => để (1) có n₀ thì

\(\left\{{}\begin{matrix}2\sqrt{x}-\sqrt{y}-\sqrt{z}=0\left(x\right)\\\sqrt{y}-3=0\left(xx\right)\\\sqrt{z}-5=0\left(xxx\right)\end{matrix}\right.\)

Từ(xx) => \(\sqrt{y}=3\) <=> y = 9

Từ (xxx) => \(\sqrt{z}=5\) <=> z = 25

Từ (x) => \(2\sqrt{x}=8\) <=> \(\sqrt{x}=4\) <=> x = 16

=> M = (16 - 15)² + (9 - 8)² + (25 - 24)² = 1 + 1 + 1 = 3

ĐKXĐ:x\(\ge\)0

Ta có:\(\sqrt{x}\ge0\forall x\in R\)

=>-5\(\sqrt{x}\le0\forall x\in R\)

=>2-5\(\sqrt{x}\le2\forall x\in R\)

\(\sqrt{x}\ge0\forall x\in R\)

=>\(\sqrt{x}+3\ge3\forall x\in R\)

=>A\(=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\le\dfrac{2}{3}\)

=>GTLN của A bằng \(\dfrac{2}{3}\) xảy ra khi và chỉ khi \(\sqrt{x}=0\)<=>x=0

Vậy...

Dự đoán dấu = xảy ra khi x=y=\(\dfrac{z}{2}\)

ta có: \(VT=3+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+\dfrac{y^2}{z^2}+\dfrac{z^2}{y^2}+\dfrac{x^2}{z^2}+\dfrac{z^2}{x^2}\)

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

Áp dụng BĐT AM-GM: \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\ge2\)

Áp dụng BĐT bunyakovsky:\(\dfrac{y^2}{z^2}+\dfrac{x^2}{z^2}\ge\dfrac{1}{2}\left(\dfrac{y}{z}+\dfrac{x}{z}\right)^2=\dfrac{1}{2}.\dfrac{\left(x+y\right)^2}{z^2}\)

\(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\ge\dfrac{1}{2}\left(\dfrac{z}{x}+\dfrac{z}{y}\right)^2\ge\dfrac{1}{2}\left(\dfrac{4z}{x+y}\right)^2=\dfrac{8z^2}{\left(x+y\right)^2}\)(AM-GM)

do đó \(VT\ge5+\dfrac{1}{2}\dfrac{\left(x+y\right)^2}{z^2}+\dfrac{8z^2}{\left(x+y\right)^2}\)

Đặt \(\dfrac{z}{x+y}=a\)(a>0)thì \(a\ge1\)do \(z\ge x+y\)

\(VT\ge8a^2+\dfrac{1}{2a^2}+5=\dfrac{a^2}{2}+\dfrac{1}{2a^2}+\dfrac{15}{2}a^2+5\ge\dfrac{a^2}{2}+\dfrac{1}{2a^2}+\dfrac{25}{2}\)

Áp dụng BĐT AM-GM: \(\dfrac{a^2}{2}+\dfrac{1}{2a^2}\ge2\sqrt{\dfrac{a^2}{4a^2}}=1\)

do đó \(VT\ge1+\dfrac{25}{2}=\dfrac{27}{2}\)(đpcm)

Dấu = xảy ra khi a=1 hay \(x=y=\dfrac{z}{2}\)

\(B=\dfrac{\left(2+\sqrt{2}-\sqrt{3}\right)+\left(2-\sqrt{6}+2\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}}\\ B=\dfrac{\left(2+\sqrt{2}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{2}-\sqrt{3}+2\right)}{2+\sqrt{2}-\sqrt{3}}\\ B=1+\sqrt{2}\)

a,\(ĐKXĐ:x\in R|x>0\)( Vì cả 2 mẫu đều >0)

Xét:\(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}+1\right)=x+\sqrt{x}\)

\(\Rightarrow\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{x\sqrt{x}+\sqrt{x}-2x^2}{x+1}=\dfrac{3x^2+x}{x+1}=P\\ \)

b, \(P< 2\Leftrightarrow\dfrac{3x^2+x}{x+1}< 2\Rightarrow3x^2-x-2< 0\Rightarrow\left(3x+2\right)\left(x-1\right)< 0\Rightarrow-\dfrac{2}{3}< x< 1\\ \)c, \(P\in Z\Leftrightarrow\dfrac{3x^2+x}{x+1}\in Z\Leftrightarrow3x^2+x⋮x+1\)

Tự lm nốt nhé.

ĐKXĐ: \(x\ge0,x\ne1\)

a/ \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

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

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

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

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

b/

Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)

Để A nhận giá trị nguyên \(\Leftrightarrow1+\dfrac{2}{\sqrt{x}-1}\in Z\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\in Z\) ( vì \(1\in Z\) )

\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(2\right)}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=2\\\sqrt{x}-1=-2\\\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-1\left(ktm\right)\\\sqrt{x}=2\\\sqrt{x}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=4\\x=0\end{matrix}\right.\) (tm)

Vậy để A nhận giá trị nguyên thì \(x=\left\{9;4;0\right\}\)

Đặt A= x+y+\(\dfrac{1}{2x}+\dfrac{2}{y}\)

Ta có: A= \(x+y+\dfrac{1}{2x}+\dfrac{2}{y}\) = \(\left(\dfrac{x}{2}+\dfrac{1}{2x}\right)+\left(\dfrac{y}{2}+\dfrac{2}{y}\right)+\dfrac{x}{2}+\dfrac{y}{2}\)

\(\geq\) \(2\sqrt{\dfrac{x}{2}.\dfrac{1}{2x}}+2\sqrt{\dfrac{y}{2}.\dfrac{2}{y}}+\dfrac{x+y}{2}\) \(\geq\) \(1+2+\dfrac{3}{2}=\dfrac{9}{2}\)

Dấu = xảy ra khi x=1;y=2

a, \(ĐKXĐ:a;b>0;a\ne2b\\ \)

Xét: \(\dfrac{2\left(a+b\right)}{\sqrt{a^3}-2\sqrt{2b^3}}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{2\left(a+b\right)}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{a+2b+\sqrt{2ab}}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}=\dfrac{1}{\sqrt{a}-\sqrt{2b}}\)\(\dfrac{\sqrt{a^3}+2\sqrt{2b^3}}{2b+\sqrt{2ab}}-\sqrt{a}=\dfrac{\left(\sqrt{a}+\sqrt{2b}\right)\left(a-\sqrt{2ab}+2b\right)}{\sqrt{2b}\left(\sqrt{a}+\sqrt{2b}\right)}-\sqrt{a}=\dfrac{\left(\sqrt{a}-\sqrt{2b}\right)^2}{\sqrt{2b}}\)\(\Rightarrow P=\dfrac{\sqrt{a}-\sqrt{2b}}{\sqrt{2b}}=\sqrt{\dfrac{a}{2b}}-1\)

b, Tự lm nhé.