Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(5\sqrt{\left(-2\right)^4}=5\sqrt{2^4}=5.2^2=5.4=20\)
b, \(-4\sqrt{\left(-3\right)^6}=-4\sqrt{3^6}=-4.3^3=-4.27=-108\)
c,\(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{5^8}}=\sqrt{5^4}=5^2=25\)
d ,\(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\)
\(=2\sqrt{5^6}+3\sqrt{2^8}\)
=\(2.5^3+3.2^4=2.125+3.16=298\)
a) \(5\sqrt{\left(-2\right)^4}\) \(=5\left|\left(-2\right)^2\right|=5.4=20\)
b) \(-4\sqrt{\left(-3\right)^6}=-4\left|\left(-3\right)^3\right|=-4.27=-108\)
c) \(\sqrt{\sqrt{\left(-5\right)^8}}=\left|\left(-5\right)^4\right|=5^4=625\)
d) \(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\) \(=2\left|\left(-5\right)^3\right|+3\left|\left(-2\right)^4\right|\)
\(=-2.\left(-125\right)+3.16\)
\(= 250 + 48 = 298\)
a) \(5\sqrt{\left(-2\right)^4}=5\sqrt{\left(2^2\right)^2}=5\left|4\right|=5.4=20\)
b)\(-4\sqrt{\left(-3\right)^6}=-4\sqrt{\left(3^3\right)^2}=-4\left|27\right|=-4.27=-108\)
c) \(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{\left(5^4\right)^2}}=\sqrt{\left(5^2\right)^2}=25\)
d)
a) \(5\sqrt{\left(-2\right)^4}=5\sqrt{\left(\left(-2\right)^2\right)^2}\) = \(5\left|\left(-2\right)^2\right|=5.4=20\)
b) \(-4\sqrt{\left(-3\right)^6}=-4\sqrt{\left(\left(-3\right)^3\right)^2}=-4\left|\left(-3\right)^3\right|\) = -4.27 = -108
\(c,2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\) = \(2\sqrt{\left(\left(-5\right)^3\right)^2}+3\sqrt{\left(\left(-2\right)^4\right)^2}=2\left|\left(-5\right)^3\right|+3\left|\left(-2\right)^4\right|=2.125+3.16=298\)
a) \(5\sqrt{\left(-2\right)^4}=5\sqrt{\left(\left(-2\right)^2\right)^2}=5\sqrt{4^2}=5\left|4\right|=5.4=20\)
b) \(-4\sqrt{\left(-3\right)^6}=-4\sqrt{\left(\left(-3\right)^3\right)^2}=-4\sqrt{\left(-27\right)^2}=-4\left|-27\right|=-4.27=-108\)
c) \(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}=2\sqrt{\left(\left(-5\right)^3\right)^2}+3\sqrt{\left(\left(-2\right)^4\right)^2}\)
\(=2\sqrt{\left(-125\right)^2}+3\sqrt{16^2}=2\left|-125\right|+3\left|16\right|=2.125+3.16=250+48=298\)
1) không có gt nào của x để căn thức trên có nghĩa
2) Câu hỏi của Phuong Nguyen dang - Toán lớp 9 | Học trực tuyến
mình đã trả lời trước đó
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
a) \(\sqrt{\left(4+\sqrt{2}\right)^2}=4+\sqrt{2}\)
b) \(-4\sqrt{\left(-3\right)^6}=-4\left|\left(-3\right)^3\right|=-4\cdot27=-108\)
c) \(\sqrt{\left(4-\sqrt{17}\right)^2}=\sqrt{17}-4\)
d) \(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(x-2\right)^8}=2\cdot\left|\left(-5\right)^3\right|+3\left(x-2\right)^4=250+3\left(x-2\right)^4\)
\(5\sqrt{\left(-2\right)^4}=5\sqrt{4^2}=5.4=20\)
\(-4\sqrt{\left(-3\right)^6}=-4\sqrt{27^2}=-4.27=-108\)
\(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{\left(5^4\right)^2}}=\sqrt{5^4}=\sqrt{25^2}=25\)
cảm ơn thầy ạ