Tính GTBT chứa căn:
a,(\(\sqrt{28}\)-\(2\sqrt{14}\)+\(\sqrt{7}\)).\(\sqrt{7}\)+\(7\sqrt{8}\)
Tính GTBT chứa căn:
a,\(\dfrac{2}{4-3\sqrt{2}}\)-\(\dfrac{2}{4+3\sqrt{2}}\)
b,\(\dfrac{2}{1+\sqrt{2}}\)+\(\dfrac{2}{1-\sqrt{2}}\)
c,\(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
d,\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
e,\(\left(\sqrt{6}-\sqrt{5}\right)^2-2\sqrt{120}\)
\(a.\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=\dfrac{2\left(4+3\sqrt{2}\right)-2\left(4-3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{8+6\sqrt{2}-8+6\sqrt{2}}{16-18}=\dfrac{12\sqrt{2}}{-2}=-6\sqrt{2}\)\(b.\dfrac{2}{1+\sqrt{2}}+\dfrac{2}{1-\sqrt{2}}=\dfrac{2\left(1-\sqrt{2}\right)+2\left(1+\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}=\dfrac{2-2\sqrt{2}+2+2\sqrt{2}}{1-2}=-4\)\(c.\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}=14-12\sqrt{7}+18+12\sqrt{7}=14+18=32\)\(d.\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}=14-14\sqrt{2}+7+14\sqrt{2}=14+7=21\)\(e.\left(\sqrt{6}-\sqrt{5}\right)^2-2\sqrt{120}=6-2\sqrt{30}+5-4\sqrt{30}=11-6\sqrt{30}\)
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
Tính
1.\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
2. \(\left(\sqrt{8}-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{18}-\sqrt{8}+\sqrt{5}\right)\)
1) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
Xin lỗi xin lỗi :v
1)\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
= \(\sqrt{7}.\left(3\sqrt{7}-2\sqrt{14}\right)+14\sqrt{2}\)
= 21 - \(14\sqrt{2}+14\sqrt{2}\)
= 21
2) \(\left(\sqrt{8}-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{18}-\sqrt{8}+\sqrt{5}\right)\)
= \(\left(2\sqrt{2}-\sqrt{2}-\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{5}-2\sqrt{2}\right)\)
= \(\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)\)
=\(\left(\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2\)
= -3
Bài 1:Tìn ĐKXĐ
a.\(\sqrt{\dfrac{2}{^{^{^{ }}}x^2}}\)
b.\(\sqrt{\dfrac{-3}{3x+5}}\)
Bài 2:
a.\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)
b.\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
c,\(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
Trả lời giúp mình với ạ!Mình cảm ơn nhiều!
Bài 1:
a. Ta có \(\sqrt{\dfrac{2}{x^2}}=\dfrac{\sqrt{2}}{\left|x\right|}=\dfrac{\sqrt{2}}{x}\) ,để biểu thức có nghĩa thì \(x>0\)
b. Để biểu thức \(\sqrt{\dfrac{-3}{3x+5}}\) có nghĩa thì \(\dfrac{-3}{3x+5}\ge0\)
mà \(-3< 0\Rightarrow3x+5< 0\) \(\Rightarrow x< \dfrac{-5}{3}\)
Bài 2:
a. \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1-2}=\dfrac{-\sqrt{2}}{-1}=\sqrt{2}\)
b. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}\)
\(=21\)
c. \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
\(=14-6\sqrt{28}+18+6\sqrt{28}\)
\(=32\)
rút gọn
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
giúp mink vs ạ
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)
\(=21-14\sqrt{2}+14\sqrt{2}\)
=21
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
Bạn ơi ! Cho mình hỏi,đề bài yêu cầu gì vậy bạn ?
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+7\sqrt{8}\)
\(=21-14\sqrt{2}+14\sqrt{2}=21\)
Thực hiện phép tính:
a)\(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
b) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
d) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=-\frac{3}{2}\)
\(b,\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(7-5\right)\)
\(=-2\)
rút gọn
A=\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
B=\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
C=\(\left(\sqrt{7}-\sqrt{10}\right)^2+\sqrt{280}\)
D=\(\dfrac{\sqrt{99}}{\sqrt{11}}+\sqrt{7}\cdot\sqrt{63}-\sqrt{\sqrt{81}}\)
E=\(\sqrt{27}\left(s-\sqrt{5}\right)^2\cdot\left(3\sqrt{48}\right)\)
giải chi tiết ra giúp mik nha,cảm ơn nhiều
Tính GTBT:
\(B=\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{8+2\sqrt{7}}{6+\sqrt{8+2\sqrt{7}}}+\frac{8-2\sqrt{7}}{6-\sqrt{8-2\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{6+\sqrt{7}+1}+\frac{\left(\sqrt{7}-1\right)^2}{6-\sqrt{7}+1}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{\sqrt{7}\left(\sqrt{7}+1\right)}+\frac{\left(\sqrt{7}-1\right)^2}{\sqrt{7}\left(\sqrt{7}-1\right)}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\sqrt{7}+1}{\sqrt{7}}+\frac{\sqrt{7}-1}{\sqrt{7}}=\frac{2\sqrt{7}}{\sqrt{7}}=2\)
=> B = \(\sqrt{2}\)