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8 tháng 9 2018

\(a.\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{2}\sqrt{2+\sqrt{3}}.\)

\(=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3+1}\right)^2}\)

\(=\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)^2=\left(2-\sqrt{3}\right)\left(4+2\sqrt{3}\right)\)

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

\(1.A=x-3\sqrt{x}+5=\left(\sqrt{x}-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)          Điều kiện: \(x\ge0\)
\(\Rightarrow MinA=\frac{11}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\left(TM\right)\)
\(2.B=\left(x-2015\right)-\sqrt{x-2015}+2015=\left(\sqrt{x-2015}-\frac{1}{2}\right)^2+2015-\frac{1}{4}\)    điều kiện: \(x\ge2015\)
\(B\ge2015-\frac{1}{4}=\frac{8059}{8060}\)
Dấu "=" xảy ra khi \(\sqrt{x-2015}-\frac{1}{2}=0\Leftrightarrow x-2015=\frac{1}{2^2}\Leftrightarrow x=\frac{8061}{8060}\left(TM\right)\)

31 tháng 5 2021

\(=>x^3=(\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)})^3\)

\(x^3=2\left(\sqrt{3}+1\right)-3.\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]^2.\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)

+\(3\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]^2\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]-2\left(\sqrt{3}-1\right)\)

\(x^3=\)

\(4-3\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)

\(x^3=4-3.\left[\sqrt[3]{4\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right].\)\(x\)

\(x^3=4-3\left[\sqrt[3]{4\left(3-1\right)}\right].x\)

\(x^3=4-3.2x\)

\(x^3=4-6x\)

thay \(x^3=4-6x\) vào A=>\(A=\left(4-6x+6x-5\right)^{2009}=\left(-1\right)^{2009}=-1\)

1. a) Tính:\(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\) b)Tính giá trị của biểu thức:M = \(\frac{\left(x-1\right).\sqrt{3}}{\sqrt{x^2}-x+1}\) với x = \(2+\sqrt{3}\)2.CMR nếu: a) \(\sqrt{1+b}+\sqrt{1+c}=2\sqrt{1+a}\) thì \(b+c\ge2a\) b) Nếu a,b >0 thì:\(\sqrt{a}+\sqrt{b}\le\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\)3. a) Giải pt:   1.\(\sqrt{x^2-16x+64}-2\sqrt{x^2-8x+16}+\sqrt{x^2}=0\)   2. \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)b) giải bất...
Đọc tiếp

1. a) Tính:

\(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)

 b)Tính giá trị của biểu thức:

\(\frac{\left(x-1\right).\sqrt{3}}{\sqrt{x^2}-x+1}\) với \(2+\sqrt{3}\)

2.CMR nếu:

 a) \(\sqrt{1+b}+\sqrt{1+c}=2\sqrt{1+a}\) thì \(b+c\ge2a\)

 b) Nếu a,b >0 thì:

\(\sqrt{a}+\sqrt{b}\le\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\)

3. a) Giải pt:

   1.\(\sqrt{x^2-16x+64}-2\sqrt{x^2-8x+16}+\sqrt{x^2}=0\)

   2. \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)

b) giải bất pt

 \(\sqrt{x^2-4x}< \sqrt{5}\)

4*.Chứng minh rằng với mọi số nguyên dương n ta luôn có:

\(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)

5*. Tìm GTNN của hàm số:

\(y=\sqrt{x+2\left(1+\sqrt{x+1}\right)}+\sqrt{x+2\left(1-\sqrt{x+1}\right)}\)

Có ai làm đc bài nào thì làm giúp mình nhé...  1 bài tkoy cũng được ạ. mình cảm ơn.

3
23 tháng 7 2018

Mấy bài này dài vật vã ghê =)))))))))))))

1, a, \(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\) 

\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)

=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-5}\)

=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{8+4\sqrt{3}-5}\)

\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}\)

=\(\sqrt{6}+\sqrt{2}+\sqrt{5}\)

b, M \(\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x^2}-x+1}\)(ĐKXĐ: \(x\ge0\))

\(\frac{\sqrt{3}\left(x-1\right)}{x-x+1}\)

\(\sqrt{3}\left(x-1\right)\)

Thay x = \(2+\sqrt{3}\)(TMĐK) vào M ta có:

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

Vậy với x = \(2+\sqrt{3}\)thì M\(3+\sqrt{3}\)

2, Mình chỉ giải câu a thôi nhé:

\(\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt{1+a}\)

\(\Leftrightarrow\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge\left(2\sqrt{1+a}\right)^2\)

\(\Leftrightarrow1+b+2\sqrt{\left(1+b\right)\left(1+c\right)}+1+c\ge4\left(1+a\right)\)

\(\Leftrightarrow2+b+c+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)\left(1\right)\)

Vì \(\left(\sqrt{1+b}-\sqrt{1+c}\right)^2\ge0\)

\(\Rightarrow2+b+c\ge2\sqrt{\left(1+b\right)\left(1+c\right)}\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow4+2\left(b+c\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\)

\(\Leftrightarrow4+2\left(b+c\right)\ge4\left(1+a\right)\)

\(\Leftrightarrow4+2\left(b+c\right)\ge4+4a\)

\(\Leftrightarrow2\left(b+c\right)\ge4a\)

\(\Leftrightarrow b+c\ge2a\)

4*. Thật ra cái này mình xài làm trội, làm giảm là được mà

Đặt A = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}\)

\(\frac{1}{2}A=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+....+\frac{1}{2\sqrt{n}}\)

\(\frac{1}{2}A=\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+....+\frac{1}{\sqrt{n}+\sqrt{n}}\)

Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2}}>\frac{1}{\sqrt{3}+\sqrt{2}}\)

          \(\frac{1}{\sqrt{3}+\sqrt{3}}>\frac{1}{\sqrt{4}+\sqrt{3}}\)

  +      .........................................................

          \(\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{n+1}+\sqrt{n}}\)  

Cộng tất cả vào

\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{n+1}+\sqrt{n}}\)\(\frac{1}{2}A>\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)

\(\frac{1}{2}A>\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\)

\(\frac{1}{2}A>\sqrt{n+1}-\sqrt{2}\)

\(A>2\sqrt{n+1}-2\sqrt{2}>2\sqrt{n+1}-3\)

\(A+1>2\sqrt{n+1}-3+1\)

\(A+1>2\sqrt{n+1}-2\)

\(A+1>2\left(\sqrt{n+1}-1\right)\)

Vậy ta có điều phải chứng minh.

23 tháng 7 2018

Cảm ơn b Trần Bảo Như nha <3

28 tháng 2 2022

Bo thi:>

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28 tháng 2 2022

+ đk x > 0 , x khác 1

7 tháng 7 2021

\(3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3\)

Xét \(a\ge0\Rightarrow\) biểu thức \(=9a^3-6a^3=3a^3\)

Xét \(a< 0\Rightarrow\) biểu thức \(=-9a^3-6a^3=-15a^3\)

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

\(=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x\)

\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)

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

\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16\)

\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)

\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)

\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)

\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|\)

Xét \(x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow\)biểu thức \(=\sqrt{x-4}+2+\sqrt{x-4}-2\)

\(=2\sqrt{x-4}\)

Xét \(x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow\) biểu thức \(=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)

 

4 tháng 7 2018

\(b1:=\sqrt{2}\left(\sqrt{3}+1\right).\sqrt{2-\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\sqrt{4-2\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)\\ =2\\ \\ b2:a,=\sqrt{\dfrac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}-3\right)}{\left(2\sqrt{5}-3\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{2}}{\sqrt{2}}.\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{54-14\sqrt{5}}}{2\sqrt{10}-3\sqrt{2}} .\left(\sqrt{10}-\sqrt{2}\right)\\ \)\(=\dfrac{\sqrt{\left(7-\sqrt{5}\right)^2}}{2\sqrt{10}-3\sqrt{2}}.\left(\sqrt{10}-\sqrt{2}\right)\)\(\\ =\dfrac{8\sqrt{10}-12\sqrt{2}}{2\sqrt{10}-3\sqrt{2}}\\ =4\)

13 tháng 9 2019

ĐK: \(x\ge-7\)

PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)

\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)

\(\Leftrightarrow x=9\) 

P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((