rút gọn bt:
\(\frac{\sqrt{2}-\sqrt{6}}{1-\sqrt{3}}-\frac{3+\sqrt{27}}{1+\sqrt{3}}\)
Giải pt:
x=\(\frac{1}{\sqrt{2019}-\sqrt{2018}}\)và y=\(\frac{1}{\sqrt{2018}-\sqrt{2017}}\)
b,So sánh
(giúp mk vs huhu...)
Rút gọn \(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{4}+\sqrt{5}}{1+\sqrt{4}+\sqrt{5}}+...+\frac{1-\sqrt{2018}+\sqrt{2019}}{1+\sqrt{2018}+\sqrt{2019}}\)
Rút gọn : \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{4}-\sqrt{3}\right)}+...+\frac{\sqrt{2018}-\sqrt{2017}}{\left(\sqrt{2017}+\sqrt{2018}\right)\left(\sqrt{2018}-\sqrt{2017}\right)}\)
\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{2018}-\sqrt{2017}}{2018-2017}\)
\(=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}+...+\frac{\sqrt{2018}-\sqrt{2017}}{1}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2018}-\sqrt{2017}=\sqrt{2018}-1\)
\(=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2018}\)
\(=-\left(\sqrt{1}+\sqrt{2018}\right)\)
Ta có :
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\)\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2018}-\sqrt{2017}\)
\(=\)\(\sqrt{2018}-1\)
Chúc bạn học tốt ~
Rút gọn: \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.........+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{2-1}{\sqrt{1}+\sqrt{2}}+\frac{3-2}{\sqrt{2}+\sqrt{3}}+........+\frac{2018-2017}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}+......+\)
\(\frac{\left(\sqrt{2018}-\sqrt{2017}\right)\left(\sqrt{2018}+\sqrt{2017}\right)}{\sqrt{2017}+\sqrt{2018}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+........+\left(\sqrt{2018}-\sqrt{2017}\right)\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+......+\sqrt{2018}-\sqrt{2017}\)
\(=-\sqrt{1}+\sqrt{2018}=\sqrt{2018}-\sqrt{1}\)
Rút gọn biểu thức S = \(\frac{2019}{2\sqrt{1}+1\sqrt{2}}+\frac{2019}{3\sqrt{2}+2\sqrt{3}}+\frac{2019}{4\sqrt{3}+3\sqrt{4}}+...+\frac{2019}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Mk chỉ cần kết quả thôi , cảm ơn nhiều ạ
a) Cho a,b,c là các số thực thỏa mãn a+b+c=2018 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2018}\) . Tính giá trị của biểu thức \(A=\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}\)
b) Rút gọn biểu thức : \(\frac{\sqrt{\sqrt{5}+2}\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
Nhờ các bn giải dùm !!!
Giải phương trình:
x=\(\frac{1}{\sqrt{2019}-\sqrt{2018}}\)và y=\(\frac{1}{\sqrt{2018}-\sqrt{2017}}\)
b,So sánh
a, x=\(\frac{1\left(\sqrt{2019}+\sqrt{2018}\right)}{2019-2018}\) và y=\(\frac{1\left(\sqrt{2018}+\sqrt{2017}\right)}{2018-2017}\) (Trục căn thức ở mẫu)
\(\Leftrightarrow\) x=\(\sqrt{2019}+\sqrt{2018}\) và y=\(\sqrt{2018}+\sqrt{2017}\)
b, Ta có : x - y = (\(\sqrt{2019}+\sqrt{2018}\) ) - ( \(\sqrt{2018}+\sqrt{2017}\) )
= \(\sqrt{2019}-\sqrt{2017}\) > 0
\(\Rightarrow\) x - y > 0 \(\Leftrightarrow\) x > y
Giải phương trình:
a) \(2\sqrt{x^2-4}-3=6\sqrt{x-2}-\sqrt{x+2}\)
b) \(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2017}-1}{y-2017}+\frac{\sqrt{z-2018}-1}{z-2018}=\frac{3}{4}\)
c) \(\sqrt{3+\sqrt{3+x}}=x\)
d) \(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
e) \(\sqrt{x^2+3x+5}+\sqrt{x^2-2x+5}=5\sqrt{x}\)
f) \(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}+5}\)
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
e/ ĐKXĐ: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+3x+5}=a>0\\\sqrt{x^2-2x+5}=b>0\\\sqrt{x}=c\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=5c^2\)
Ta được hệ: \(\left\{{}\begin{matrix}a^2-b^2=5c^2\\a+b=5c\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=5c^2\\a+b=5c\end{matrix}\right.\)
\(\Rightarrow5c\left(a-b\right)=5c^2\)
\(\Leftrightarrow\left[{}\begin{matrix}c=0\\a-b=c\end{matrix}\right.\)
f/ ĐKXĐ: \(x>0\)
\(\Leftrightarrow\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}\)
\(\Leftrightarrow\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}-2\sqrt{x+2}+2x-2\sqrt{x\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{\frac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)-2\sqrt{x}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{\frac{x+2}{x}}-2\sqrt{x}\right)\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x}=4x\\x+3=4x\end{matrix}\right.\)
Rút gọn biểu thức: A= \(\frac{\sqrt{x-2017-2\sqrt{x-2018}}}{\sqrt{x-2018}-1}\)Với x > 2019
Bài 1: Rút gọn biểu thức:
\(A=\frac{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}-2}{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}+2}\left(a>2\right)\)
\(B=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\left(ab\ne0\right)\)
Bài 2: Tính giá trị của biểu thức:
\(E=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyên
\(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)