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Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
Đặt a - 1 = x > 0; b - 1 = y > 0
\(A=\frac{\left(x+1\right)^2}{x}+\frac{\left(y+1\right)^2}{y}\\ A=\frac{x^2+2x+1}{x}+\frac{y^2+2y+1}{y}\\ A=\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)+4\)
Với x > 0; y > 0, theo BĐT AM-GM ta có:
\(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}\Rightarrow x+\frac{1}{x}\ge2\)
\(y+\frac{1}{y}\ge2\sqrt{y.\frac{1}{y}}\Rightarrow y+\frac{1}{y}\ge2\)
\(\Rightarrow A\ge8\)
Dấu "=" xảy ra khi và chỉ khi x = y = 1 => a = b = 2
Vậy...
?Amanda?, Phạm Lan Hương, Phạm Thị Diệu Huyền, Vũ Minh Tuấn, Nguyễn Ngọc Lộc , @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @Trần Thanh Phương
giúp e với ạ! Cần trước 5h chiều nay! Cảm ơn mn nhiều!
Tranh thủ làm 1, 2 bài rồi ăn cơm:
1/ Đặt \(m=n-2008>0\)
\(\Rightarrow2^{2008}\left(369+2^m\right)\) là số chính phương
\(\Rightarrow369+2^m\) là số chính phương
m lẻ thì số trên chia 3 dư 2 nên ko là số chính phương
\(\Rightarrow m=2k\Rightarrow369=x^2-\left(2^k\right)^2=\left(x-2^k\right)\left(x+2^k\right)\)
b/
\(2\left(a^2+b^2\right)\left(a+b-2\right)=a^4+b^4\) \(\left(a+b>2\right)\)
\(\Rightarrow2\left(a^2+b^2\right)\left(a+b-2\right)\ge\frac{1}{2}\left(a^2+b^2\right)^2\)
\(\Rightarrow a^2+b^2\le4\left(a+b-2\right)\)
\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2\le0\Rightarrow a=b=2\)
\(\Rightarrow x=y=4\)
Lời giải:
a)
\(\frac{4}{\sqrt{10}}(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})=\frac{4}{\sqrt{20}}(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}})\)
\(=\frac{4}{2\sqrt{5}}(\sqrt{5+1+2\sqrt{5}}+\sqrt{5+1-2\sqrt{5}})=\frac{2}{\sqrt{5}}[\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}]\)
\(=\frac{2}{\sqrt{5}}(\sqrt{5}+1+\sqrt{5}-1)=\frac{2}{\sqrt{5}}.2\sqrt{5}=4\)
b)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})\)
\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
c)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+8\sqrt{3}+18}=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(3+1+2\sqrt{3})+2}\)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(\sqrt{3}+1)^2+2}\)
\(=\sqrt{(2\sqrt{3}+2)^2+(\sqrt{2})^2+2.(2\sqrt{3}+2).\sqrt{2}}\)
\(=\sqrt{(2\sqrt{3}+2+\sqrt{2})^2}=2\sqrt{3}+2+\sqrt{2}\)
Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
a) \(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(=\sqrt{75}-\sqrt{\frac{16}{3}}+\frac{9}{2}\sqrt{\frac{8}{3}}+2\sqrt{27}\)
\(=5\sqrt{3}-\frac{4}{\sqrt{3}}+3\sqrt{6}+6\sqrt{3}\)
\(=-\frac{4}{\sqrt{3}}+5\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)
\(=-\frac{4}{\sqrt{3}}+11\sqrt{3}+3\sqrt{6}\)
\(=-\frac{4\sqrt{3}}{3}+11\sqrt{3}+3\sqrt{6}\)
b) \(\sqrt{48}-\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48}-\sqrt{\frac{16}{3}}+2\sqrt{75}-5\sqrt{\frac{4}{3}}\)
\(=4\sqrt{3}-\frac{4}{\sqrt{3}}+10\sqrt{3}-\frac{10}{\sqrt{3}}\)
\(=-\frac{4}{\sqrt{3}}-\frac{10}{\sqrt{3}}+4\sqrt{3}+10\sqrt{3}\)
\(=-\frac{14\sqrt{3}}{3}+4\sqrt{3}+10\sqrt{3}\)
\(=-\frac{14\sqrt{3}}{3}+14\sqrt{3}\)
c)\(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(=27+12\sqrt{5}+12\sqrt{5}\)
\(=27+24\sqrt{5}\)
d)\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{6}+2-\sqrt{3}-\sqrt{2}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
\(=4+2\sqrt{3}-2\sqrt{3}+4\)
= 8
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
= 14
a) \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
\(=2\sqrt{2}.\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)
= 9 (đpcm)
b) \(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
\(=\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2^{\frac{1}{2}}\left(\sqrt{2}-1\right)}\)
\(=\sqrt{2\left(\sqrt{2}-1\right)}\) (đpcm)
a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
Áp dụng bđt bunhiacopski có:
\(\left(a^4+1\right)\left(1+4^2\right)\ge\left(a^2+4\right)^2\)
=> \(\sqrt{a^4+1}\ge\sqrt{\frac{\left(a^2+4\right)^2}{1+4^2}}=\frac{a^2+4}{\sqrt{17}}\)(1)
Tương tự cx có: \(\sqrt{b^4+1}\ge\frac{b^2+4}{\sqrt{17}}\) (2)
Từ (1),(2) => \(F\ge\frac{a^2+b^2+8}{\sqrt{17}}\)
Có (a+2)(b+2)=\(\frac{25}{4}\)
=> \(ab+2a+2b+4=\frac{25}{4}\) <=> \(ab+2a+2b=\frac{9}{4}\)
Áp dụng cosi có:
\(ab\le\frac{a^2+b^2}{2}\)
\(2a\le2\left(a^2+\frac{1}{4}\right)\)
\(2b\le2\left(b^2+\frac{1}{4}\right)\)
=> \(\frac{a^2+b^2}{2}+2a^2+\frac{1}{2}+2b^2+\frac{1}{2}\ge ab+2a+2b=\frac{9}{4}\)
<=> \(\frac{a^2+b^2+4a^2+4b^2}{2}\ge\frac{9}{4}-\frac{1}{2}-\frac{1}{2}=\frac{5}{4}\)
<=> \(\frac{5\left(a^2+b^2\right)}{2}\ge\frac{5}{4}\)
<=> \(a^2+b^2\ge\frac{1}{2}\)
Thay \(a^2+b^2\ge\frac{1}{2}\) vào F có:
\(F\ge\frac{\frac{1}{2}+8}{\sqrt{17}}\)
<=> F \(\ge\frac{\sqrt{17}}{2}\)
Dấu "=" xảy ra <=>\(a=b=\frac{1}{2}\)