Giúp tôi giải toán và làm văn

Ta có: \(\sqrt{9}+\sqrt{8}>\sqrt{8}+\sqrt{7}\)

\(\Leftrightarrow\frac{1}{\sqrt{9}+\sqrt{8}}< \frac{1}{\sqrt{8}+\sqrt{7}}\)

\(\Leftrightarrow\frac{\left(\sqrt{9}-\sqrt{8}\right)}{\left(\sqrt{9}+\sqrt{8}\right)\left(\sqrt{9}-\sqrt{8}\right)}< \frac{\left(\sqrt{8}-\sqrt{7}\right)}{\left(\sqrt{8}+\sqrt{7}\right)\left(\sqrt{8}-\sqrt{7}\right)}\)

\(\Leftrightarrow\sqrt{9}-\sqrt{8}< \sqrt{8}-\sqrt{7}\)

\(\Leftrightarrow3-2\sqrt{2}< 2\sqrt{2}-\sqrt{7}\)

\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{23}+\sqrt{25}}\)

\(2A=\frac{2}{\sqrt{3}+\sqrt{1}}+\frac{2}{\sqrt{5}+\sqrt{3}}+...+\frac{2}{\sqrt{25}+\sqrt{23}}\)\(2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{\left(\sqrt{3}+\sqrt{1}\right)\left(\sqrt{3}-\sqrt{1}\right)}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{\left(\sqrt{25}+\sqrt{23}\right)\left(\sqrt{25}-\sqrt{23}\right)}\)

\(2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{2}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{2}\)

\(2A=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{25}-\sqrt{23}\)

\(2A=\sqrt{25}-\sqrt{1}\)

\(2A=4\)

\(A=2\)

1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)

\(\Rightarrow1+2019^2=2020^2-2.2019\)

\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)

\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)

\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)

\(=2020\)

Vậy M=2020.

2) Xét  : \(k\in N;k\ge2\)ta có:

\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)

                                          \(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)

\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)

\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)

Cho \(k=3,4,...,2020.\)Ta có:

\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)

\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)

Vậy \(N=2018\frac{1009}{2020}.\)

Bạn @tam mai ơi, nếu bạn không biết làm thì thôi, mình cần cách làm tử tế chứ không cần bấm máy tính như bạn, nếu vậy ai chả làm được.

uh một cách giải ngoài dự đoán của bạn

Liên quan gì bạn @Tam Mai, chứng minh chứ không phải bấm máy tính

\(\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{9+4\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}\right)\) 

\(=\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{\left(2^2+2.2\sqrt{5}+\sqrt{5^2}\right)}+\sqrt[3]{2+\sqrt{5}}\right)\) 

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

\(=2\sqrt[3]{2-\sqrt{5}}.\sqrt[3]{2+\sqrt{5}}=2\sqrt[3]{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}=2\sqrt[3]{4-5}=2\sqrt[3]{-1}=-1.2=-2\)

Cảm ơn bạn Shitbo nha.

ha dẻ vcayk mafd  bạn lét  123=4=1=5342=6678=+493076

\(\left|x-3\right|=\left|x+5\right|\)

\(\cdot TH1:x-3=x+5\)

\(\Leftrightarrow x-x=5+3\)

\(\Leftrightarrow0=8\)(Vô Lí)

\(\cdot TH2:x-3=-\left(x+5\right)\)

\(\Leftrightarrow x-3=-x-5\)

\(\Leftrightarrow x+x=-5+3\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\)

Vậy \(S=\left\{-1\right\}\)

\(\left|x-3\right|=\left|x+5\right|\)

\(\cdot TH1:x-3=x+5\)

\(\Leftrightarrow x-x=5+3\)

\(\Leftrightarrow0=8\)(Vô Lí)

\(\cdot TH2:x-3=-\left(x+5\right)\)

\(\Leftrightarrow x-3=-x-5\)

\(\Leftrightarrow x+x=-5+3\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\)

Vậy \(S=\left\{-1\right\}\)

\(\left|x-3\right|=\left|x+5\right|\)

\(\Rightarrow\orbr{\begin{cases}x-3=x+5\\x-3+x+5=0\end{cases}}\)

\(\Rightarrow2x+2=0\)

=> 2x = -2

=> x = -1

vậy_

\(A-1=\frac{x-2\sqrt{x}+1}{\sqrt{x}}=\frac{(\sqrt{x}-1)^2}{\sqrt{x}}\ge0\)\(\Rightarrow A\ge1\)

\(x=1-\sqrt{2012}\Leftrightarrow1-x=\sqrt{2012}\)

\(\Leftrightarrow\left(1-x\right)^2=2012\Leftrightarrow x^2-2x-2011=0\)

Ta có: 

\(A=\left(x^5-2x^4-2012x^3+3x^2+2009x-2012\right)^{2012}\)

\(A=\left[\left(x^5-2x^4-2011x^3\right)-\left(x^3-2x^2-2011x\right)+\left(x^2-2x-2011\right)-1\right]^{2012}\)

\(A=\left[\left(x^3-x+1\right)\left(x^2-2x-2011\right)-1\right]^{2012}=1\)

Ta có

\(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left[4\left(4-y-z\right)+yz\right]}\)

                                             \(=\sqrt{x\left(4\left(x+\sqrt{xyz}\right)+yz\right)}\)

                                             \(=\sqrt{4x^2+4x\sqrt{xyz}+xyz}\)

                                              \(=2x+\sqrt{xyz}\)

Khi đó \(T=2\left(x+y+z\right)+3\sqrt{xyz}-\sqrt{xyz}=2.4=8\)

+ \(\frac{\sqrt{47}-\sqrt{48}}{\left(\sqrt{47}-\sqrt{48}\right)\left(\sqrt{47}+\sqrt{48}\right)}\)

= \(\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{47}-\sqrt{48}}{-1}\)

= \(\sqrt{48}-1\left(1\right)\)

Lại có: \(3=4-1=\sqrt{16}-1\left(2\right)\)

Từ (1) và (2) 

=> \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{47}+\sqrt{48}}>3\)

Ta có:

\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{47}+\sqrt{48}}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{47}-\sqrt{48}}{-1}\)

\(=\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{47}-\sqrt{48}}{-1}\)

\(=\frac{\sqrt{1}-\sqrt{48}}{-1}\)

\(=4\sqrt{3}-1\approx5,9>3\left(đpcm\right)\)

Ta có \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{47}+\sqrt{48}}=\frac{1-\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}-\sqrt{4}\right)\left(\sqrt{3}+\sqrt{4}\right)}\)

Sorry

\(\frac{\sqrt{a}-1}{a\sqrt{a}-a+\sqrt{a}}:\frac{1}{a^2+\sqrt{a}}\)

\(=\frac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}.\left(\sqrt{a}\left(\sqrt{a}^3+1\right)\right)\)

\(=\frac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}.\left(\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)\right)\)

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

\(=a-1\)

Bạn Nguyễn Công Tỉnh, cảm ơn nha.

Bạn Nguyễn Công Tỉnh ơi, bạn chép nhần đề bài rồi .... 

Cảm ơn các bạn. Theo như đáp án thì đúng là \(4\sqrt{7}.\)

Mình nhầm 1 đấy là a nhg bây giờ xem lại thì thấy dù sửa lại 1 thành a thì vẫn sai

Cảm ơn bạn đã góp y

Nguyễn Minh Trang ơi \(5a^3-1\ne a\left(5a^2-1\right)\)mà

Với mọi a nguyên dương , 

Ta có: 

\(\frac{1}{\sqrt{a}}=\frac{2}{2\sqrt{a}}>\frac{2}{\sqrt{a}+\sqrt{a+1}}=2\left(\sqrt{a-1}-\sqrt{a}\right)=2\sqrt{a-1}-2\sqrt{a}\)

Biểu thức:

\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}\)

\(>2\sqrt{2}-2\sqrt{1}+2\sqrt{3}-2\sqrt{2}+2\sqrt{4}-2\sqrt{3}+...+2\sqrt{25}-2\sqrt{24}\)

\(=-2\sqrt{1}+2\sqrt{25}=-2+10=8\)

Vậy B>8

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

Đặt \(\hept{\begin{cases}a=\sqrt[3]{2+x}\\b=\sqrt[3]{2-x}\end{cases}\Leftrightarrow}\hept{\begin{cases}a^3=2+x\\b^3=2-x\end{cases}\Rightarrow}a^3+b^3=4\)

Khi đó phương trình đã cho tương đương với:

\(\hept{\begin{cases}a+b=1\\a^3+b^3=4\end{cases}\Leftrightarrow}\hept{\begin{cases}\left(a+b\right)\left(a^2-ab+b^2\right)=4\\a+b=1\end{cases}\Leftrightarrow}\hept{\begin{cases}a^2-ab+b^2=4\\a+b=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(a+b\right)^2-3ab=4\\a+b=1\end{cases}\Leftrightarrow}\hept{\begin{cases}-3ab=3\\a+b=1\end{cases}\Leftrightarrow\hept{\begin{cases}ab=-1\\a+b=1\end{cases}.}}\)

Suy ra a, b là nghiệm của phương trình \(X^2-X-1=0\Leftrightarrow\left(X^2-X+\frac{1}{4}\right)=\frac{5}{4}\)

   \(\Leftrightarrow\left(X-\frac{1}{2}\right)^2=\frac{5}{4}\Leftrightarrow\orbr{\begin{cases}X-\frac{1}{2}=-\frac{\sqrt{5}}{2}\\X-\frac{1}{2}=\frac{\sqrt{5}}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}X=\frac{1-\sqrt{5}}{2}\\X=\frac{1+\sqrt{5}}{2}\end{cases}.}}\)

Suy ra có 2 trường hợp :

 \(\hept{\begin{cases}a=\frac{1+\sqrt{5}}{2}\\b=\frac{1-\sqrt{5}}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt[3]{2+x}=\frac{1+\sqrt{5}}{2}\\\sqrt[3]{2-x}=\frac{1-\sqrt{5}}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}2+x=2+\sqrt{5}\\2-x=2-\sqrt{5}\end{cases}\Leftrightarrow x=\sqrt{5}.}\)

\(\hept{\begin{cases}a=\frac{1-\sqrt{5}}{2}\\b=\frac{1+\sqrt{5}}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt[3]{2+x}=\frac{1-\sqrt{5}}{2}\\\sqrt[3]{2-x}=\frac{1+\sqrt{5}}{2}\end{cases}\Leftrightarrow x=-\sqrt{5}.}\)

Vậy phương trình đã cho có 2 nghiệm phân biêt là \(x_1=\sqrt{5},x_2=-\sqrt{5}.\)

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

=>\(4+3\sqrt[3]{\left(4-x^2\right)}.1=1\)

=>\(\sqrt[3]{\left(4-x^2\right)}=-1\)

=>\(x^2=5\)

=>\(x=\sqrt{5}\)và \(x=-\sqrt{5}\)

=1 nha các bạn

\(A=\left(\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1-a}}\right):\left(\frac{1}{\sqrt{1-a^2}}+1\right)\)

\(A=\left[\frac{\sqrt{1-a}+\sqrt{1+a}}{\sqrt{1+a}\cdot\sqrt{1-a}}\right]:\left(\frac{1+\sqrt{1-a^2}}{\sqrt{1-a^2}}\right)\)

\(A=\frac{\sqrt{1-a}+\sqrt{1+a}}{\sqrt{1-a^2}}\cdot\frac{\sqrt{1-a^2}}{1+\sqrt{1-a^2}}\)

\(A=\frac{\sqrt{1-a}+\sqrt{1+a}}{1+\sqrt{1-a^2}}\)