\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}\)
M.n ai bik thì giúp mình với ạ
Trục căn thức ở mẫu các biểu thức sau:
A=\(\frac{\sqrt{2}+1}{2\sqrt{3}+\sqrt{2}}\); B=\(\frac{1}{\sqrt{2}+\sqrt{2-\sqrt{2}}}\); C=\(\frac{1}{\sqrt{3}-\sqrt{2}+1}\)
GIÚP MÌNH VỚI M.N (T_T) ~ MÌNH CẦN GẤP LẮM
bạn hãy nhân ở mẫu với biểu thức tương ướng để tạo ra biểu thức liên hợp , là HĐT số 3 ạ
bài 1: thực hiện phép tính
1) 2\(\sqrt{5}\)\(-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
3) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
4) \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}-\sqrt{27}}{\sqrt{30}-\sqrt{162}}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
6) \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}-\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}}\)
GIÚP MÌNH VỚI M.N~~~!!!!!!!!! MÌNH ĐANG CẦN GẤP LẮM (T_T)
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)
\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
\(=4\sqrt{5}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-\sqrt{6^2.6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-6\sqrt{6}+3^2}+\sqrt{\left(2\sqrt{6}\right)^2-12\sqrt{6}+3^2}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-3\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vi \(\sqrt{6}-3< 0\))
\(=\sqrt{6}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)
\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)
\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)
\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)
Báo cáo sai phạm
1) 2√5−√125−√80+√605
=2√5−√52.5−√42.5+√112.5
=2√5−5√5−4√5+11√5
=4√5
2) √15−√216+√33−12√6
=√15−√62.6+√33−12√6
=√15−6√6+√33−12√6
=√(√6)2−6√6+32+√(2√6)2−12√6+32
=√(√6−3)2+√(2√6−3)2
=|√6−3|+|2√6−3|
=3−√6+2√6−3 ( vi √6−3<0)
=√6
5) 2√163 −3√127 −6√475
=24√3 −3.13 −6√223.52
=8√33 −1−6.25 .√13
=8√33 −1−125 .√33
=285 .√33 −1
Bài 1:
a) \(\sqrt{8}+\sqrt{18}-\sqrt{\frac{1}{2}}\)
b) \(\frac{3-\sqrt{3}}{\sqrt{3}}+\frac{2\cdot\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)
c) Gải phương trình: \(\sqrt{x^2-2x+1=3}\)
M.n giải nhanh hộ mình nhé, mình cần gấp lắm. Cảm ơn ạ !!! <3
a, \(\sqrt{8}+\sqrt{18}-\sqrt{\frac{1}{2}}=2\sqrt{2}+3\sqrt{2}-\frac{1}{2}\sqrt{2}\)
\(=\frac{9}{2}\sqrt{2}\)
b, \(\frac{3-\sqrt{3}}{\sqrt{3}}+\frac{2\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}+\frac{2\sqrt{2}}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)
\(=\sqrt{3}-1+\frac{2\sqrt{2}}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)
\(=\frac{2\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+1\right)\) \(=\frac{2\sqrt{2}-\left(\sqrt{2}+1\right)^2}{\sqrt{2}+1}\)
\(=\frac{2\sqrt{2}-2-2\sqrt{2}-1}{\sqrt{2}+1}=-\frac{2+1}{\sqrt{2}+1}\)
c, PT xác định với mọi x nha!
\(\sqrt{x^2-2x+1}=3\) \(\Rightarrow x^2-2x+1=9\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(2x-8\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy...
bạn tự kl
A
√8+√18-√1/√2
=> 2√2+3√2+√2/2
=> 5√2-√2/2
=>10√2/2-√2/2
=>9√2/2
C
=> √(x-1)2=√9 đk x\(\ge\)
= > |x-1|=9
<=> x-1=9<=>x=10 tm
x-1=-9<=>x=-8 loại
\(B=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
Giúp mình với mai mình nộp bài rồi
\(\frac{B}{\sqrt{2}}=\frac{\frac{2+\sqrt{3}}{2}}{\sqrt{2}+\sqrt{\frac{4+2\sqrt{3}}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\sqrt{2}-\sqrt{\frac{4-2\sqrt{3}}{2}}}\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}+\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}-\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}}\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}-\frac{\sqrt{3}-1}{\sqrt{2}}}=\frac{\frac{2+\sqrt{3}}{2}}{\frac{\sqrt{3}+3}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{3-\sqrt{3}}{\sqrt{2}}}\)
\(=\frac{\left(2+\sqrt{3}\right).\sqrt{2}}{2\cdot\left(3+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right).\sqrt{2}}{2.\left(3-\sqrt{3}\right)}\)
=> \(B=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(B=\frac{3+\sqrt{3}}{6}+\frac{3-\sqrt{3}}{6}=1\)
----
Vài chỗ mình làm vắn tắt không hiểu cứ hỏi nhé, còn kết quả mình ấn máy tính ra chính xác rùi :)
P=\(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)
Rút gọn P
AI GIÚP MÌNH VỚI...GIÚP MÌNH MÌNH SẼ TÍCH CHO BẠN GIÚP MÌNH T^T <3
Ta có \(P=\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)
\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{2\left(1-\sqrt{5}\right)}\right).\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}\right).\left(\sqrt{7}-\sqrt{3}\right)=\frac{\sqrt{7}+\sqrt{3}}{2}.\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\frac{7-3}{2}=2\)
Vậy \(P=2\)
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\)\(\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
Rút gọn biểu thức A
\(B=x^3-3x+2000\). Rút gọn B biết \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
Mong mọi người giúp đỡ mình ạ , mình rất cần ạ
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
\(M=\frac{x-\sqrt[3]{x}}{x-1}+\frac{1}{\sqrt[3]{x}-1}+\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\)
Rút gọn ạ...giúp mình:))
3 tick ạ
ĐKXĐ: x khác 1
\(M=\frac{x-\sqrt[3]{x}}{x-1}+\frac{1}{\sqrt[3]{x}-1}+\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\)
\(=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1+\sqrt[3]{x}-1}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+2\sqrt[3]{x}}{\left(\sqrt[3]{x}\right)^3-1}\)
\(=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+2\sqrt[3]{x}}{x-1}=\frac{x+\sqrt[3]{x^2}+\sqrt[3]{x}}{x-1}=\frac{\sqrt[3]{x}\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)
\(=\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\)
bạn nhớ kiểm tra lại nhé
Tính: \(\frac{\sqrt{1+\frac{2\sqrt{2}}{3}}+\sqrt{1-\frac{2\sqrt{2}}{3}}}{\sqrt{1+\frac{2\sqrt{2}}{3}}-\sqrt{1-\frac{2\sqrt{2}}{3}}}\)
Làm ơn giúp mình với T^T
biểu thức\(=\frac{2+\frac{2}{9}}{1-\frac{8}{9}}=\frac{\frac{20}{9}}{\frac{1}{9}}=20\)
bt = \(\frac{2+\frac{2}{9}}{\frac{4\sqrt{2}}{3}}=\frac{\frac{20}{9}}{\frac{4\sqrt{2}}{3}}=\frac{5\sqrt{2}}{6}\)
còn sai đâu ko?
1\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{25}}\)
Giải gấp giúp mình với ạ. Cảm ơn rất nhiều💖
Em muốn giải cái gì? Kiểm tra lại đề bài nhé!