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Bài 1:
Ta có a là số dương
\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\\dfrac{1}{a}>0\end{matrix}\right.\)
Áp dụng BĐT Cauchy cho hai số dương, có:
\(a+\dfrac{1}{a}\ge2\sqrt{a.\dfrac{1}{a}}\)
\(\Leftrightarrow a+\dfrac{1}{a}\ge2\sqrt{1}\)
\(\Leftrightarrow a+\dfrac{1}{a}\ge2\left(đpcm\right)\)
Vậy ...
làm giúp mình câu 1, còn câu 2 mình biết làm rồi
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
1.Ta có :\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^2-xy+y^2\) (do x+y=1)
\(=\dfrac{3}{4}\left(x-y\right)^2+\dfrac{1}{4}\left(x+y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)\(=\dfrac{1}{4}.1=\dfrac{1}{4}\)
Dấu "=" xảy ra khi :\(x=y=\dfrac{1}{2}\)
Vậy \(x^3+y^3\ge\dfrac{1}{4}\)
2.
a) Sửa đề: \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng vì \(a,b\ge0\))
Đẳng thức xảy ra \(\Leftrightarrow a=b\)
b) Lần trước mk giải rồi nhá
3.
a) Áp dụng BĐT Cauchy-Schwarz dạng Engel\(P=\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{\left(1+1+1\right)^2}{\left(x+y+z\right)+3}=\dfrac{9}{3+3}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=\dfrac{1}{y+1}=\dfrac{1}{z+1}\\x+y+z=3\end{matrix}\right.\Leftrightarrow x=y=z=1\)
b) \(Q=\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1}\le\dfrac{x}{2\sqrt{x^2.1}}+\dfrac{y}{2\sqrt{y^2.1}}+\dfrac{z}{2\sqrt{z^2.1}}\)
\(=\dfrac{x}{2x}+\dfrac{y}{2y}+\dfrac{z}{2z}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)
B=\(\dfrac{\sqrt{a.6}}{\sqrt{6.6}}+\dfrac{\sqrt{2a.3}}{\sqrt{3.3}}+\dfrac{\sqrt{3a.2}}{\sqrt{2.2}}\)
=\(\dfrac{\sqrt{6a}}{6}+\dfrac{\sqrt{6a}}{3}+\dfrac{\sqrt{6a}}{2}\)
=\(\dfrac{\sqrt{6a}+2\sqrt{6a}+3\sqrt{6a}}{6}\)
=\(\dfrac{6\sqrt{6a}}{6}=\sqrt{6a}\)
b: \(B=\dfrac{\sqrt{6}}{6}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{3}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{2}\cdot\sqrt{a}\)
\(=\sqrt{a}\cdot\sqrt{6}=\sqrt{6a}\)
e: \(=2-x-x=2-2x\)
i: \(=\left|x-\left(1-x\right)\right|-2x=\left|x-1+x\right|-2x\)
\(=\left|2x-1\right|-2x\)
=1-2x-2x=1-4x
\(VT=\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\)
\(\ge\dfrac{4}{x^2+2xy+y^2}\)
\(=\dfrac{4}{\left(x+y\right)^2}>4\)
Cách khác.
Ta có: \(A=\dfrac{1}{x\left(x+y\right)}+\dfrac{1}{y\left(x+y\right)}=\dfrac{1}{x+y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(=\dfrac{1}{x+y}.\dfrac{x+y}{xy}=\dfrac{1}{xy}\)
Áp dụng BĐT cho các số x,y >0 , ta có:
\(x+y\ge2\sqrt{xy}\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow\dfrac{\left(x+y\right)^2}{4}\ge xy\)
Và x+y \(\le\)1 \(\Rightarrow xy\le\dfrac{1}{4}\) \(\Rightarrow A\ge\dfrac{1}{\dfrac{1}{4}}=4\)
Dấu ''='' xảy ra khi x = y =0,5
Từ giả thiết suy ra:
\(\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x^2-xy+y^2\right)+\left(x^2+2xy+y^2\right)+4\left(x+y\right)+4=0\)
\(\Leftrightarrow\left(x^2-xy+y^2\right)\left(x+y+2\right)+\left(x+y+2\right)^2=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+y+2\right)\left(2x^2+2y^2-2xy+2x+2y+4\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+y+2\right)\left[\left(x-y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2+2\right]=0\)
\(\Rightarrow x+y=-2\)
Mà xy>0 nên x,y cùng nhỏ hơn 0
Áp dụng AM-GM,ta có: \(\sqrt{\left(-x\right)\left(-y\right)}\le\dfrac{-x-y}{2}=1\)
\(\Rightarrow xy\le1\Rightarrow\dfrac{-2}{xy}\le-2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{-2}{xy}\le-2\)
a/\(x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{\left(x-3\right)^2}=x+3+\left|x-3\right|=x+3+3-x=6\)
b/ \(\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{\left(x+2\right)^2}-\left|x\right|=\left|x+2\right|-\left|x\right|=-x-2-\left(-x\right)=-x-2+x=-2\)
c/ \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\cdot\left(x-1\right)=\sqrt{x^2-2x+1}=\sqrt{\left(x-1\right)^2}=\left|x-1\right|\)
d/ \(\left|x-2\right|+\dfrac{\sqrt{x^2-4x+4}}{x-2}=2-x+\dfrac{\sqrt{\left(x-2\right)^2}}{x-2}=2-x+\dfrac{\left|x-2\right|}{x-2}=2-x+\dfrac{-\left(x-2\right)}{x-2}=2-x-1=1-x\)
\(BDT\Leftrightarrow\dfrac{1}{x^2\left(4-3x\right)}-x\ge0\)
\(\Leftrightarrow\dfrac{1}{x^2\left(4-3x\right)}-\dfrac{x^3\left(4-3x\right)}{x^2\left(4-3x\right)}\ge0\)
\(\Leftrightarrow\dfrac{1-3x^4+4x^3}{x^2\left(4-3x\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2\left(3x^2+2x+1\right)}{x^2\left(4-3x\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2\left(3\left(x+\dfrac{1}{3}\right)^2+\dfrac{2}{3}\right)}{x^2\left(4-3x\right)}\ge0\forall0< x< \dfrac{4}{3}\)