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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)
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
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).
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) \(\sqrt{\left(4+\sqrt{2}\right)^2}=4+\sqrt{2}\)
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) \(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(x-2\right)^8}=2\cdot\left|\left(-5\right)^3\right|+3\left(x-2\right)^4=250+3\left(x-2\right)^4\)
a) Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{4+2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)(Vì \(\sqrt{3}>1>0\))
\(=\left(4+2\sqrt{3}\right)\cdot\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(3-4\right)\)
\(=-2\)
b) Ta có: \(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
c) Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=8-2\sqrt{15}\)
d) Ta có: \(\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)\)
\(=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=-\sqrt{3}\cdot\left|\sqrt{3}+\sqrt{2}\right|\)
\(=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)(Vì \(\sqrt{3}>\sqrt{2}>0\))
\(=-3-\sqrt{6}\)
e) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)\)(Vì \(\sqrt{3}>1\))
\(=\frac{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{2}\)
\(=\frac{16-12}{2}=\frac{4}{2}=2\)
f) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)(Vì \(2>\sqrt{3}>0\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)(Vì \(5>\sqrt{3}\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=\sqrt{9}=3\)
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) \(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\) \(=2\left|\left(-5\right)^3\right|+3\left|\left(-2\right)^4\right|\)
\(=-2.\left(-125\right)+3.16\)
\(= 250 + 48 = 298\)