Giúp tôi giải toán

\(\sqrt{15-2.\sqrt{15}+1}+\sqrt{9+2.3.\sqrt{5}+5}=\sqrt{15}-1+3+\sqrt{5}\)

tự lm nốt

cái này qua học 24 tha hồ surt

Bạn dùng quy nạp nhé, chứng minh bài này nhanh thôi.

Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)

=>\(A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\)\(\sqrt{10+2\sqrt{5}}.\)

=\(8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)\)

=\(8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)

=>\(\sqrt{5}+1\)

(x-√(x^2+2010).(x+√(x^2+2010)).(y+√(y^2+... = 2010.(x-√(x^2+2010) 
<=> -2010.(y+√(y^2+2010) = 2010.(x-√(x^2+2010) 
<=> - (y+√(y^2+2010) = (x-√(x^2+2010) 
<=> (x-√(x^2+2010) = - (y+√(y^2+2010) 
+++ (x+√(x^2+2010)) (y+√(y^2+2010))(y-√(y^2+2010)) = 2010.(y-√(y^2+2010)) 
<=> -2010.(x+√(x^2+2010) = 2010.(y-√(y^2+2010)) 
<=> - (x+√(x^2+2010) = (y-√(y^2+2010) (**) 
...Lấy (*) - (**) vế theo vế,ta có: 
2x = -2y 
<=> x + y = 0 

Cám ơn mn :D

bằng 2010 hay \(\sqrt{2010}\) vậy bạn

Ta có:\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>...>\frac{1}{\sqrt{100}}\)

=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{.1}{\sqrt{100}}>100.\frac{1}{\sqrt{100}}=100\cdot\frac{1}{10}=10\)

Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{.1}{\sqrt{100}}>10\)

mọi người giải giùm đi!

ngu xi dốt nát mà bn

Đề lạ!!!!!!!!!!!! tại sao x = ... = 2  ( mình làm theo đề mình nha )

ĐKXĐ : ( giúp mình )

\(\Leftrightarrow\sqrt{x+\frac{1}{4}+2.\sqrt{x+\frac{1}{4}}\cdot\frac{1}{2}+\frac{1}{4}}=2\)

\(\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)

=>  \(l\sqrt{x+\frac{1}{4}}+\frac{1}{2}l=2\)  ( \(l...l\) là trị tuyệt đối ) 

=>  \(\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\) ( vì \(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\ge0\) với mọi x ) 

=> \(\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)

\(\Rightarrow x+\frac{1}{4}=\frac{9}{4}\)

 => x = 2

moi tay

Ta có:

\(\sqrt{50}+\sqrt{26}+1>\sqrt{49}+\sqrt{25}+1\)

\(=7+5+1=13\)

Mà \(\sqrt{168}< \sqrt{169}=13\)

Vì \(\sqrt{50}+\sqrt{26}+1>13>\sqrt{168}\)

Nên \(\sqrt{50}+\sqrt{26}+1>\sqrt{168}\)

Đúng rồi bạn ơi

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Đúng rồi: \(\sqrt{100}-1=9\) khử căn ở mẫu là ra

hello

mk ko biết mk mới học lớp nhỏ thôi . Đó là lớp này nè bn...... tự vào trang của mk coi đi nhé

a)\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=5\Leftrightarrow5\left(\sqrt{x}-1\right)=\sqrt{x}+1\)

\(\Leftrightarrow5\sqrt{x}-5=\sqrt{x}+1\)\(\Leftrightarrow4\sqrt{x}=6\)\(\Leftrightarrow\sqrt{x}=\frac{6}{4}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)

Vậy A=5 khi x=9/4

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

A nguyên <=> \(\frac{2}{\sqrt{x}-1}\) nguyên

<=>2 chia hết cho \(\sqrt{x}-1\)

<=>\(\sqrt{x}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

<=>\(\sqrt{x}\in\left\{-1;0;2;3\right\}\)

Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}\in\left\{0;2;3\right\}\)

<=>\(x\in\left\{0;4;9\right\}\)

Vậy A nguyên khi \(x\in\left\{0;4;9\right\}\)

\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

\(\Rightarrow\sqrt{2}.C=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)

\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

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

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

\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\)

\(=\sqrt{7}-1-\sqrt{7}-1=-2\)

\(\Rightarrow C=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

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

\(\Rightarrow\sqrt{2}.D=\sqrt{2}\sqrt{4-\sqrt{7}}+\sqrt{2}\sqrt{2-\sqrt{3}}\)

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

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

\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)

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

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

\(\Rightarrow D=\frac{\sqrt{7}+\sqrt{3}-2}{\sqrt{2}}=\frac{\sqrt{14}+\sqrt{6}-2\sqrt{2}}{2}\)

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

=> \(C=-\sqrt{2}\)

\(D.\sqrt{2}=\sqrt{8-2\sqrt{7}}+\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{7}-1+\sqrt{3}-1\)

=> \(D=\frac{\sqrt{7}+\sqrt{3}-2}{\sqrt{2}}\)

\(\frac{1}{\sqrt{1}}=1>\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(...\)

\(\frac{1}{\sqrt{100}}=\frac{1}{10}\)

Mà dựa vào mẫu của các số hạng mà ta biết tổng trên có \(\frac{100-1}{1}+1=100\)( số hạng )

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\) ( 100 số \(\frac{1}{10}\))

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}.100\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>10\)

Vậy \(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>10.\)

Cảm ơn Trần Thuỳ Dung nhiều ạ!

(x-2).(y-3)=-4

=> (x-2).(y-3)=1.(-4)=(-4).1

+) x-2=1 ; y-3=-4

=> x=3;y=-1

+)x-2=-4; y-3=1

=> x=-2; y=4

vậy...

\(\frac{1}{2}\)

do (x-2)mux2.(y-3)=-4

suy ra xmux2 - 2mux2 = -4

          y-3 = -4

lưu ý trước cả hai biểu thức trên la dau ngoac vuong cho minh nhe (-_-)

suy ra xmux2 - 4 = -4

          y = -4 + 3

suy ra xmux2 = -4 + 4

          y = -1

suy ra xmux2 = 0

          y = -1

suy ra x = 0 chia x

          y = -1

ma cac ban biet roi day 0 chia cho so nao cung = 0 suy ra x = 0 va y = -1

\(VT=\frac{2015-1}{\sqrt{2015}}+\frac{2014+1}{\sqrt{2014}}=\sqrt{2015}-\frac{1}{\sqrt{2015}}+\sqrt{2014}+\frac{1}{\sqrt{2014}}\)

        \(>\sqrt{2014}+\sqrt{2015}\)(do \(\frac{1}{\sqrt{2014}}-\frac{1}{\sqrt{2015}}>0\))

\(ĐKXĐ:x\ne2009;y\ne2010;z\ne2011;x,y,z\in R\)

\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{x-2009}-\frac{\sqrt{x-2009}}{x-2009}+\frac{1}{y-2010}-\frac{\sqrt{y-2011}}{y-2011}+\frac{1}{z-2011}-\frac{\sqrt{z-2011}}{z-2011}=\frac{-3}{4}\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}^2}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{y-2010}^2}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{z-2011}^2}+\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^{^2}+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)

• \(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}=0\)

• \(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}=0\)
• \(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{1}{\sqrt{x-2009}}=\frac{1}{2};\frac{1}{\sqrt{y-2010}}=\frac{1}{2};\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\)

\(\Leftrightarrow x=2013;y=2014;z=2015\inĐKXĐ\)

  VẬY       \(x=2013;y=2014;z=2015\)

\(\sqrt{3^2-2.3.\sqrt{6}+6}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2.\sqrt{6}\right)^2}\)

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