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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) \(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}=2\sqrt{\left(\left(-5\right)^3\right)^2}+3\sqrt{\left(\left(-2\right)^4\right)^2}\)
\(=2\sqrt{\left(-125\right)^2}+3\sqrt{16^2}=2\left|-125\right|+3\left|16\right|=2.125+3.16=250+48=298\)
a) \(VT=2\sqrt{6}-4\sqrt{2}+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)
\(=-4\sqrt{2}+1+4\sqrt{2}+8\)
\(=1+8\)
\(=9\)
\(\Rightarrow VT=VP\) (đpcm).
b) \(VT=\left(3\sqrt{10}-3\sqrt{2}+\sqrt{50}-\sqrt{10}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(3\sqrt{10}-3\sqrt{2}+5\sqrt{2}-\sqrt{10}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{10}+2\sqrt{2}\right)^2\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(40+8\sqrt{20}+8\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(48+16\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{16\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{16\left(9-5\right)}\)
\(=\sqrt{64}\)
\(=8\)
\(\Rightarrow VT=VP\) (đpcm).
c) \(VT=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{2+\sqrt{5}}\)
\(=2\left(\sqrt{5}+2\right)-\dfrac{2\left(2-\sqrt{5}\right)}{-1}\)
\(=2\sqrt{5}+4+2\left(2-\sqrt{5}\right)\)
\(=2\sqrt{5}+4+4-2\sqrt{5}\)
\(=4+4\)
\(=8\)
\(\Rightarrow VT=VP\) (đpcm).
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
\(5\sqrt{\left(-2\right)^4}=5\sqrt{4^2}=5.4=20\)
\(-4\sqrt{\left(-3\right)^6}=-4\sqrt{27^2}=-4.27=-108\)
\(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{\left(5^4\right)^2}}=\sqrt{5^4}=\sqrt{25^2}=25\)
cảm ơn thầy ạ