Điều kiện: \(x\ge2\)

\(\left(\sqrt{x+1}-2\right)+\left(\sqrt{x-2}-1\right)+\left(3-\sqrt{2x+3}\right)+\left(\sqrt{5x+1}-4\right)=0\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{x+1}+2}+\dfrac{x-3}{\sqrt{x-2}+1}-\dfrac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\dfrac{5\left(x-3\right)}{\sqrt{5x+1}+4}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{x+1}+2}+\dfrac{1}{\sqrt{x-2}+1}-\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{5}{\sqrt{5x+1}+4}\right)=0\)

\(\Leftrightarrow x=3\)

PS: Phần trong ngoặc tự chứng minh nha.

Buoc 2 dau ra the phan dua ve phan so

a) Ta có: \(\widehat{BIM}\) + \(\widehat{MIA}\) = 180 - (\(\widehat{\dfrac{A}{2}}\) + \(\widehat{\dfrac{B}{2}}\))

=> \(\widehat{BIM}\) = 90 - (\(\widehat{\dfrac{A}{2}}\) + \(\widehat{\dfrac{B}{2}}\))

Và \(\widehat{BCI}\) = 90 - (\(\widehat{\dfrac{A}{2}}\) + \(\widehat{\dfrac{B}{2}}\))

=> \(\widehat{BIM}\) = \(\widehat{BCI}\)

=> \(\Delta\)BIM \(\sim\)\(\Delta\)BCI (g.g)

=> \(\overset{ }{\dfrac{BI}{BM}}\) = \(\overset{ }{\dfrac{BC}{BI}}\) => BI² = BM.BC (1)

C/m tương tự ta có \(\Delta\)ICN \(\sim\)\(\Delta\)BCI (g.g)

=> \(\overset{ }{\dfrac{CI}{CN}}\) = \(\overset{ }{\dfrac{BC}{CI}}\) => CI² = CN.BC (2)

Từ (1) và (2) => \(\overset{ }{\dfrac{BI^2}{CI^2}}\) = \(\overset{ }{\dfrac{BM}{CN}}\) (đpcm)

b) Tam giác MIB đồng dạng với tam giác NIC, viết ra tỉ số rồi thay vào VT là ra

\(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)

\(=\sqrt{xy}+\sqrt{x}-\sqrt{y}\)

Ta có: \(P=\sqrt{xy}+\sqrt{x}-\sqrt{y}=2\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{y}+1\right)=3\)

\(\Rightarrow\left(\sqrt{x}-1,\sqrt{y}+1\right)=\left(1,3;3,1\right)\)

\(\Rightarrow\left(x,y\right)=\left(4,4;16,0\right)\)

Phần đầu lm kiểu gì để ra được ạ ?

\(\dfrac{B}{2}=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{2014.2015.2016.2017}\)

\(\Leftrightarrow\dfrac{3B}{2}=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{2014.2015.2016.2017}\)

\(=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2014.2015.2016}-\dfrac{1}{2015.2016.2017}\)

\(=\dfrac{1}{1.2.3}-\dfrac{1}{2015.2016.2017}\)

Tự làm nốt nhé

Còn lại thì trời làm ak

Thay \(a=b=c=0,25\)thì ta có:

\(\dfrac{1}{\sqrt{0,25}}+\dfrac{1}{\sqrt{0,25}}+\dfrac{2\sqrt{2}}{\sqrt{0,25}}\approx9,657\)

\(\dfrac{8}{0,25+0,25+0,25}\approx10,667\)

Vậy đề sai