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

Giải:

\(\sqrt{6-2\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)

\(=\sqrt{5-2\sqrt{5}+1}+\sqrt{9-6\sqrt{5}+5}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)

\(=\left|\sqrt{5}-1\right|+\left|3-\sqrt{5}\right|\)

\(=\sqrt{5}-1+3-\sqrt{5}\)

\(=2\)

Áp dụng bđt Bunhiacopxki :

\(\sqrt{c}\cdot\sqrt{a-c}+\sqrt{c}\cdot\sqrt{b-c}\le\sqrt{\left[\left(\sqrt{c}\right)^2+\left(\sqrt{a-c}\right)^2\right]\left[\left(\sqrt{c}\right)^2+\left(\sqrt{b-c}\right)^2\right]}\)

\(=\sqrt{\left(c+a-c\right)\left(c+b-c\right)}=\sqrt{ab}\) ( đpcm )

Dấu "=" xảy ra \(\Leftrightarrow\frac{c}{a-c}=\frac{c}{b-c}\Leftrightarrow a-c=b-c\Leftrightarrow a=b\)

Lời giải:

Ta thấy:

\(x^2+xy+y^2=\frac{3}{4}(x^2+2xy+y^2)+\frac{1}{4}(x^2-2xy+y^2)=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\)

\(\geq \frac{3}{4}(x+y)^2\) với mọi $x,y>0$
\(\Rightarrow \sqrt{x^2+xy+y^2}\geq \frac{\sqrt{3}}{2}(x+y)\)

Hoàn toàn tương tự:

\(\sqrt{y^2+yz+z^2}\geq \frac{\sqrt{3}}{2}(y+z); \sqrt{z^2+zx+x^2}\geq \frac{\sqrt{3}}{2}(x+z)\)

Cộng theo vế các BĐT trên và rút gọn:

\(\Rightarrow \sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\geq \sqrt{3}(x+y+z)\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z$

Xét số hạng tổng quát:\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

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

\(\Rightarrow A=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}-\dfrac{1}{\sqrt{121}}\)

\(A=1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(\left(x^2+y\right)\left(x+y^2\right)=\left(x-y\right)^3\)

\(\Leftrightarrow x^3+\left(xy\right)^2+xy+y^3=x^3-xy\left(x-y\right)+y^3\)

\(\Leftrightarrow xy\left(xy+1\right)=xy\left(y-x\right)\)

Xét hai TH:

+ TH1: xy = 0: Khi đó \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\).

+ TH2: xy \(\ne\) 0: Ta được xy + 1 = y - x

\(\Leftrightarrow xy+x-y+1=0\)

\(\Leftrightarrow x\left(y+1\right)-y-1+2=0\)

\(\Leftrightarrow\left(x-1\right)\left(y+1\right)=-2\). Ở đây dễ rồi

a) ta có : \(x^2+x+6=y^2\) \(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=y^2-\dfrac{23}{4}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}y\ge\dfrac{\sqrt{23}}{2}\\y\le\dfrac{-\sqrt{23}}{2}\end{matrix}\right.\) \(\Rightarrow\) \(y=???\) thế vào tìm x

b) tương tự

c) \(x^3-y^3=3xy+1\Leftrightarrow x^3-1=y^3+3xy\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=y\left(y^2+3x\right)\)

\(\Leftrightarrow x-1\) và \(y\left(y^2+3x\right)\) cùng dấu \(\Rightarrow\) ...

d) ai gỏi lm giùm nha

Mysterious Person góp ý vsssssss

\(P=\left(\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}\right):\left(\dfrac{x+y+2xy}{1-xy}+1\right)\)

Điều kiện : \(xy\ge0\) hoặc \(xy\le0\) ; \(xy\ne1\); \(x\ge0\);\(y\ge0\)

\(P=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\right):\left(\dfrac{x+2xy+y+1-xy}{1-xy}\right)\)

\(P=\left(\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}\right):\left(\dfrac{x+xy+y+1}{1-xy}\right)\)

\(P=\left(\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\right):\left(\dfrac{x\left(1+y\right)+\left(y+1\right)}{1-xy}\right)\)

\(P=\left(\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}\right):\left(\dfrac{\left(1+y\right)\left(x+1\right)}{1-xy}\right)\)

\(P=\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}.\dfrac{1-xy}{\left(1+y\right)\left(x+1\right)}\)

\(P=\dfrac{2\sqrt{x}}{x+1}\)

b) ta có :\(x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{4-3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

thay \(x=\left(\sqrt{3}-1\right)^2\) vào biểu thức P
ta được : \(P=\dfrac{2\sqrt{\left(\sqrt{3}-1\right)^2}}{\left(\sqrt{3}-1\right)^2+1}\)

\(P=\dfrac{2\left|\sqrt{3}-1\right|}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}\)

\(P=\dfrac{\left(2\sqrt{3}-2\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}=\dfrac{10\sqrt{3}+12-10-4\sqrt{3}}{25-12}\)

\(P=\dfrac{6\sqrt{3}+2}{13}\)

c) để P\(\le\)1 thì \(\dfrac{2\sqrt{x}}{x+1}\le1\)

\(\Leftrightarrow\dfrac{2\sqrt{x}}{x+1}-1\le0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-x-1}{x+1}\le0\)

\(\Leftrightarrow\dfrac{-\left(x-2\sqrt{x}+1\right)}{x+1}\le0\)

\(\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x+1}\le0\)

Vì \(-\left(x-1\right)^2\le0\) nên x + 1 \(\ge\) 0

\(\Leftrightarrow\) x \(\ge\) -1
đúng thì cho xin 1 like nha

đk: x≥1

\(pt\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}+3\sqrt{x-1}=18\)

\(\Leftrightarrow6\sqrt{x-1}=18\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x-1=9\Leftrightarrow x=10\left(tm\right)\)

Vậy pt có nghiệm x = 10