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2b)
Áp dụng BĐT bunhiacopxki có:
\(\left(1+1\right)\left(x^4+y^4\right)\ge\left(x^2+y^2\right)^2\)
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2\)\(\Leftrightarrow x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Rightarrow2\left(x^4+y^4\right)\ge\dfrac{\left(x+y\right)^4}{4}\Leftrightarrow x^4+y^4\ge\dfrac{1}{8}.\left(x+y\right)^4\)
Dấu "=" xảy ra khi x=y
3)
Áp dụng bđt Holder có:
\(\left(x^3+y^3+z^3\right)\left(1+1+1\right)\left(1+1+1\right)\ge\left(x+y+z\right)^3\)
\(\Leftrightarrow x^3+y^3+z^3\ge\dfrac{1}{9}\left(x+y+z\right)^3\)
Dấu "=" xảy ra khi x=y=z
3)(Nếu không dùng Holder)
Với x,y,z >0, ta có bđt sau:\(2x^3+2y^3+2z^3\ge xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\) (1)
Thật vậy (1)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)+\left(y+z\right)\left(y^2-yz+z^2\right)-yz\left(y+z\right)+\left(z+x\right)\left(z^2-zx+x^2\right)-zx\left(x+z\right)\ge0\)
\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^2+\left(y+z\right)\left(y-z\right)^2+\left(z+x\right)\left(z-x\right)^2\ge0\) (lđ)
Áp dụng AM-GM có:
\(x^3+y^3+z^3\ge3xyz\)
\(\Leftrightarrow\dfrac{2\left(x^3+y^3+z^3\right)}{3}\ge2xyz\) (2)
Từ (1) và (2), cộng vế với vế \(\Rightarrow\dfrac{8}{3}\left(x^3+y^3+z^3\right)\ge xy\left(x+y\right)+yz\left(x+z\right)+xz\left(x+z\right)+2xyz\)
\(\Leftrightarrow\dfrac{8}{3}\left(x^3+y^3+z^3\right)\ge\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(\Leftrightarrow9\left(x^3+y^3+z^3\right)\ge x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)^3\)
\(\Rightarrow x^3+y^3+z^3\ge\dfrac{1}{9}\left(x+y+z\right)^3\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3,\\ a,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=12\\ \Leftrightarrow\sqrt{x+5}=4\Leftrightarrow x+5=16\Leftrightarrow x=11\left(tm\right)\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|x-5\right|=6\Leftrightarrow\left[{}\begin{matrix}x-5=6\\5-x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)
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\(1,\\ a,M=\sqrt{3}-1-6\sqrt{3}+\sqrt{3}+1=-4\sqrt{3}\\ b,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}-\sqrt{x-1}=1\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\\ \Leftrightarrow x-1=\dfrac{1}{4}\Leftrightarrow x=\dfrac{5}{4}\left(tm\right)\\ 2,\\ a,ĐK:x>0;x\ne1\\ P=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\\ b,P< 0\Leftrightarrow x-1< 0\left(\sqrt{x}>0\right)\\ \Leftrightarrow0< x< 1\\ c,P\sqrt{x}=m-\sqrt{x}\\ \Leftrightarrow x-1=m-\sqrt{x}\\ \Leftrightarrow x+\sqrt{x}-m-1=0\\ \text{PT có nghiệm nên }\Delta=1+4\left(m+1\right)\ge0\\ \Leftrightarrow4m+5\ge0\Leftrightarrow m\ge-\dfrac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
b: Phương trình hoành độ giao điểm là: \(\dfrac{1}{2}x^2-\left(m-1\right)x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x^2-x\left(2m-2\right)+1=0\)
\(\text{Δ}=\left(2m-2\right)^2-4=\left(2m-6\right)\left(2m+2\right)\)
Để (P) cắt (d) tại hai điểm phân biệt thì(m-3)(m+1)>0
=>m>3 hoặc m<-1
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3.1
\(\frac{3+\sqrt{3}}{1+\sqrt{3}}=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1}=\sqrt{3}\)
3.2
\(=\frac{4(\sqrt{5}-1)-4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}:\sqrt{(\sqrt{2}+1)^2}\)
\(=\frac{-8}{4}:(\sqrt{2}+1)=\frac{-2}{\sqrt{2}+1}=\frac{-2(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{-2(\sqrt{2}-1)}{1}=2(1-\sqrt{2})\)
3.3.
\(=\frac{\sqrt{2}+1-(\sqrt{2}-1)}{(\sqrt{2}-1)(\sqrt{2}+1)}:\sqrt{(\sqrt{2}-1)^2}=\frac{2}{1}:(\sqrt{2}-1)=\frac{2}{\sqrt{2}-1}=\frac{2(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=2(\sqrt{2}+1)\)
3.4 Cái này đã gọn lắm rồi không cần rút gọn nữa.
3.5
\(=\frac{(\sqrt{5}+\sqrt{3})^2+(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{16}{5-3}=8\)
3.6
\(=\frac{(\sqrt{5}-1)^2-(\sqrt{5}+1)^2}{(\sqrt{5}+1)(\sqrt{5}-1)}=\frac{-4\sqrt{5}}{5-1}=-\sqrt{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
cot B = \(\dfrac{5}{13}=>tanB=\dfrac{13}{5}\)
AC=AB.tanB
AC= 15.\(\dfrac{13}{5}\)
AC= 39cm
BC2=AB2+AC2
BC2=225+1521=1746
BC=3 \(\sqrt{194}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a) Ta có: \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\)
\(=\sqrt{a}+2-\left(\sqrt{a}+2\right)\)
=0
b) Ta có: \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{a-6\sqrt{a}+9}{\sqrt{a}-3}\)
\(=3-\sqrt{a}-\sqrt{a}+3\)
\(=6-2\sqrt{a}\)
c) Ta có: \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)\)
=0
d) Ta có: \(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\)
\(=a+\sqrt{ab}+b+\sqrt{ab}\)
\(=a+2\sqrt{ab}+b\)
Bài 1:
a.
\(\frac{4\sqrt{5}+\sqrt{15}}{\sqrt{5}}=\frac{\sqrt{5}(4+\sqrt{3})}{\sqrt{5}}=4+\sqrt{3}\)
$\frac{7-\sqrt{7}}{3\sqrt{7}}=\frac{\sqrt{7}(\sqrt{7}-1)}{3\sqrt{7}}=\frac{\sqrt{7}-1}{3}$
\(\frac{4\sqrt{2}-\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{2}(4-\sqrt{3})}{\sqrt{2}.\sqrt{6}}=\frac{4-\sqrt{3}}{\sqrt{6}}\)
\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\frac{(3\sqrt{2}-2\sqrt{3})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\frac{\sqrt{6}}{3-2}=\sqrt{6}\)
b.
\(\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}(\sqrt{a}-2)}{\sqrt{a}-2}=\sqrt{a}\)
\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}=\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}=1+\sqrt{a}+a\)
\(\frac{a+10\sqrt{a}+25}{\sqrt{a}+5}=\frac{(\sqrt{a}+5)^2}{\sqrt{a}+5}=\sqrt{a}+5\)
\(\frac{a-9}{\sqrt{a}+3}=\frac{(\sqrt{a}-3)(\sqrt{a}+3)}{\sqrt{a}-3}=\sqrt{a}+3\)
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