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a) Ta có:
\(5\sqrt{a}-4b\sqrt{25a^3}+5a\sqrt{16ab^2}-2\sqrt{9a}\)
\(=5\sqrt{a}-4b.5a\sqrt{a}+5a.4b\sqrt{a}-2.3\sqrt{a}\)
\(=5\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-6\sqrt{a}\) \(=-\sqrt{a}\)
b) Ta có:
\(5a\sqrt{64ab^3}-\sqrt{3}.\sqrt{12a^3b^3}+2ab\sqrt{9ab}\) \(-5b\sqrt{81a^3b}\)
\(=5a.8b\sqrt{ab}-\sqrt{3.12a^3b^3}+2ab.3\sqrt{ab}\) \(-5b.9a\sqrt{ab}\)
\(=40ab\sqrt{ab}-6ab\sqrt{ab}+6ab\sqrt{ab}-45ab\)\(\sqrt{ab}\)
\(=-5ab\sqrt{ab}\)
a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
a) Ta có: \(\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)
\(=5\sqrt{5}-4.3\sqrt{5}+3.2\sqrt{5}-4\sqrt{5}\)
\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)
\(=-5\sqrt{5}\)
\(\approx-11,18033989\)
câu g
(câu cuối) đề nhiều trôi hết nhìn thấy mỗi câu (g)
\(G=0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)
\(G=0,1.10\sqrt{2}+\dfrac{2.2}{10}\sqrt{2}+0,4.5\sqrt{2}\)
\(G=\sqrt{2}\left(1+\dfrac{2}{5}+2\right)=\dfrac{\sqrt{2}\left(5+2+10\right)}{5}=\dfrac{17\sqrt{2}}{5}\)
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
= \(2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
= \(-\sqrt{5}+15\sqrt{2}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
= \(2.7-2\sqrt{21}+7+2\sqrt{21}=14+7=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{6}.\sqrt{5}+5-2\sqrt{30}\)
= \(11+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
= \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
= \(4-4\sqrt{2}-12\sqrt{2}+64\sqrt{2}=4+48\sqrt{2}\)
Bài này dễ ẹc ( đâu có khó đâu :)) )
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=\sqrt{2^2.5}-\sqrt{3^2.5}+3\sqrt{3^2.2}+\sqrt{6^2.2}\)
\(=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=\left(2-3\right)\sqrt{5}+\left(9+6\right)\sqrt{2}\)
\(=15\sqrt{2}-\sqrt{5}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\sqrt{2^2.7}.\sqrt{7}-2\sqrt{3}.\sqrt{7}+\sqrt{7}.\sqrt{7}+\sqrt{2^2.21}\)
\(=2.7-2\sqrt{21}+7+2\sqrt{21}\)
\(=14+7+\left(2-2\right)\sqrt{21}=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
\(=6+2\sqrt{30}+5-\sqrt{2^2.30}\)
\(=6+5+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{2^2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{10^2.2}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
\(=2\sqrt{2}-12\sqrt{2}+64\sqrt{2}=54\sqrt{2}\)
Hok tốt
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
Bài 2:
a: \(=\sqrt{5}-2\)
b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)
d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
1.8:
2:
a: \(\frac{\sqrt{15}-\sqrt5}{\sqrt3-1}-\frac{5-2\sqrt5}{2\sqrt5-4}\)
\(=\frac{\sqrt5\left(\sqrt3-1\right)}{\sqrt3-1}-\frac{\sqrt5\left(\sqrt5-2\right)}{2\left(\sqrt5-2\right)}=\sqrt5-\frac{\sqrt5}{2}=\frac{\sqrt5}{2}\)
b: \(\frac{\sqrt3+1}{\sqrt3-1}+\frac{\sqrt3-1}{\sqrt3+1}\)
\(=\frac{\left(\sqrt3+1\right)^2+\left(\sqrt3-1\right)^2}{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}\)
\(=\frac{4+2\sqrt3+4-2\sqrt3}{3-1}=\frac82=4\)
c: \(\frac{3-\sqrt3}{2\sqrt3-1}+\frac{3+\sqrt3}{2\sqrt3-1}\)
\(=\frac{3-\sqrt3+3+\sqrt3}{2\sqrt3-1}=\frac{6}{2\sqrt3-1}=\frac{6\left(2\sqrt3+1\right)}{12-1}=\frac{12\sqrt3+6}{11}\)
d: \(\frac{2}{\sqrt3-1}-\frac{2}{\sqrt3+1}\)
\(=\frac{2\left(\sqrt3+1\right)-2\left(\sqrt3-1\right)}{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}\)
\(=\frac{2\sqrt3+2-2\sqrt3+2}{3-1}=\frac42=2\)
1:
a: \(\frac{\sqrt{3-2\sqrt2}}{\sqrt{17-12\sqrt2}}\)
\(=\frac{\sqrt{\left(\sqrt2-1\right)^2}}{\sqrt{\left(3-2\sqrt2\right)^2}}=\frac{\sqrt2-1}{3-2\sqrt2}\)
\(=\frac{\sqrt2-1}{\left(\sqrt2-1\right)^2}=\frac{1}{\sqrt2-1}=\sqrt2+1\)
b: \(B=\frac{\sqrt{6+\sqrt{35}}}{\sqrt2}\)
\(=\frac{\sqrt{12+2\sqrt{35}}}{2}\)
\(=\frac{\sqrt{\left(\sqrt7+\sqrt5\right)^2}}{2}=\frac{\sqrt7+\sqrt5}{2}\)
c: \(\frac{\sqrt{45}-\sqrt2}{\sqrt5-\sqrt2}\)
\(=\frac{3\sqrt5-\sqrt2}{\sqrt5-\sqrt2}\)
\(=\frac{\left(3\sqrt5-\sqrt2\right)\left(\sqrt5+\sqrt2\right)}{5-2}=\frac{15+3\sqrt{10}-\sqrt{10}-2}{3}=\frac{13+2\sqrt{10}}{3}\)
d: \(A=\frac{\sqrt{2-\sqrt3}}{\sqrt2}\)
\(=\frac{\sqrt{4-2\sqrt3}}{2}=\frac{\sqrt{\left(\sqrt3-1\right)^2}}{2}=\frac{\sqrt3-1}{2}\)
1.7:
a: \(5\sqrt{a}-3\cdot\sqrt{25a^3}+2\cdot\sqrt{36ab^2}-2\sqrt{9a}\)
\(=5\sqrt{a}-3\cdot5\cdot a\cdot\sqrt{a}+2\cdot6\cdot\sqrt{a}\cdot b-2\cdot3\sqrt{a}\)
\(=5\sqrt{a}-6\sqrt{a}-15a\sqrt{a}+12b\sqrt{a}=-\sqrt{a}-15a\cdot\sqrt{a}+12b\cdot\sqrt{a}\)
b: \(\sqrt{64ab^3}-3\sqrt{12a^3b^3}+2ab\cdot\sqrt{9ab}-5b\cdot\sqrt{81a^3b}\)
\(=8b\cdot\sqrt{ab}-3\cdot2\cdot ab\cdot\sqrt{3ab}+2ab\cdot\sqrt{ab}\cdot3-5b\cdot9a\cdot\sqrt{ab}\)
\(=\sqrt{ab}\left(8b-6ab\sqrt3+6ab-45ab\right)=\sqrt{ab}\left(8b-6ab\sqrt3-39ab\right)\)
c: \(2\sqrt{3a}-\sqrt{75a}+a\cdot\sqrt{\frac{13.5}{2a}}-\frac25\cdot\sqrt{300a^3}\)
\(=2\sqrt{3a}-5\cdot\sqrt{3a}+a\cdot\frac{3\sqrt3}{2\cdot\sqrt{a}}-\frac25\cdot10a\cdot\sqrt{3a}\)
\(=-1,5\cdot\sqrt{3a}-4a\cdot\sqrt{3a}\)
1.6 Rút gọn các biểu thức sau:
a) \(5 \sqrt{1 / 5} + \frac{1}{2} \sqrt{20} + \sqrt{5}\)
Ta có:
👉 Vậy:
\(5 \sqrt{1 / 5} + \frac{1}{2} \sqrt{20} + \sqrt{5} = 5 + \sqrt{5} + \sqrt{5} = 5 + 2 \sqrt{5}\)
b) \(1 / \sqrt{2} + \sqrt{4.5} + \sqrt{12.5}\)
Ta có:
👉 Cộng lại:
\(\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{2}} + \frac{5}{\sqrt{2}} = \frac{9}{\sqrt{2}} \Rightarrow \frac{9 \sqrt{2}}{2}\)
c) \(\sqrt{20} - \sqrt{45} + 3 \sqrt{18} + \sqrt{72}\)
Ta rút gọn từng căn:
👉 Vậy:
\(2 \sqrt{5} - 3 \sqrt{5} + 9 \sqrt{2} + 6 \sqrt{2} = - \sqrt{5} + 15 \sqrt{2}\)
d) \(\sqrt{20} - \sqrt{45} + 3 \sqrt{18} + \sqrt{72}\)
(giống y câu c, kết quả vẫn là: \(- \sqrt{5} + 15 \sqrt{2}\))
e) \(\left(\right. \sqrt{6} + \sqrt{5} \left.\right)^{2} - \sqrt{120}\)
Ta khai triển:
👉 Vậy:
\(11 + 2 \sqrt{30} - 2 \sqrt{30} = 11\)