Chịu khó tìm kiếm trước khi hỏi nha anh/chị: Câu hỏi của san nguyễn - Toán lớp 9 | Học trực tuyến

hả?

bài để thi hok kì I đó hả? đúng khó *_*

mk sẽ ghi lại để sau này mk hok

câu hỏi tương tự ko có đâu

a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)

\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)

\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)

\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)

b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)

\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)

C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)

\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)

\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)

\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)

d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)

\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)

e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)

Ta có:

\(a+b+c=4\)

\(\Rightarrow\)  \(a< 4\)

\(\Rightarrow\)  \(a^4< 4a^3\)  (do  \(a>0\)  nên  \(a^3>0\)  )

Do đó,  \(a^3>\frac{a^4}{4}\)  hay nói cách khác,  \(\sqrt[4]{a^3}>\sqrt[4]{\frac{a^4}{4}}=\frac{a}{\sqrt[4]{4}}\)  \(\left(1\right)\)

Từ đó, ta cũng tương tự thiết lập được:   \(\sqrt[4]{b^3}>\frac{b}{\sqrt[4]{4}}\)  \(\left(2\right)\)  và   \(\sqrt[4]{c^3}>\frac{c}{\sqrt[4]{4}}\)  \(\left(3\right)\)

Cộng từng vế các bđt   \(\left(1\right);\)  \(\left(2\right);\)  và  \(\left(3\right)\)  ta có:

\(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}>\frac{a+b+c}{\sqrt[4]{4}}=\frac{4}{\sqrt[4]{4}}=2\sqrt{2}\)

Áp dụng  \(x^2+y^2+z^2\ge xy+yz+zx\) và \(x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^2\)

\(N\ge\dfrac{a^2b}{c}+\dfrac{b^2c}{a}+\dfrac{c^2a}{b}\ge\dfrac{1}{3}\left(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}\right)^2=3\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đề: \(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\sqrt{3}\) ???

*Ta chứng minh : \(x^4-x^3+2\ge x+1\forall x>0\)

\(\Leftrightarrow x^4-x^3-x+1\ge0\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)\ge0\) ( đúng )

Do đó: \(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\) \(\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}=\sqrt{3}\)

Dấu "=" \(\Leftrightarrow a=b=c=1\)

\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-8\sqrt{2}}}}}-\sqrt{3}\)\(=\sqrt{6+2.1,4.\sqrt{3-\sqrt{1,4+2.1,7+\sqrt{18-8.1,4\text{​​}}}}}-1,7\)

\(=\sqrt{6+2,8\sqrt{3-\sqrt{1,4+3,4+\sqrt{18-11,2}}}}-1,7\)

\(=\sqrt{8,8\sqrt{3-\sqrt{4,8+\sqrt{6,8}}}}-1,7\)

\(=\sqrt{8,8\sqrt{3-\sqrt{4,8+2,6}}}-1,7\)

\(=\sqrt{8,8\sqrt{3-\sqrt{7,4}}}-1,7\)

\(=\sqrt{8,8\sqrt{3-2,7}}-1,7\)

\(=\sqrt{88\sqrt{0,3}}-1,7\)

\(=\sqrt{88.0,54}-1,7\)

\(=\sqrt{47,52}-1,7\)

\(=6,9-1,7\)

\(=5,2\)

2,Mệt với câu 1 rồi nên câu 2 và câu 3 chịu