a)\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)
b)\(\sqrt{27\left(1-\sqrt{3}\right)^4}:3\sqrt{15}\)
c) \(\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}\)
Những câu hỏi liên quan
Bài 1: Rút gọn biểu thức1) sqrt{12}-sqrt{27}+sqrt{48} 2) left(sqrt{25}+sqrt{20}-sqrt{80}right):sqrt{5}3) 2sqrt{27}-sqrt{frac{16}{3}}-sqrt{48}-sqrt{8frac{1}{3}} 4) frac{1}{sqrt{5}-sqrt{3}}-frac{1}{sqrt{5}+sqrt{3}}5) left(sqrt{125}-sqrt{12}-2sqrt{5}right)left(3sqrt{5}-sqrt{3}+sqrt{27}right) 6) left(3sqrt{20}-sqrt{125}-15sqrt{frac{1}{5}}right).sqrt{5}7) left(6sqrt{128}-frac{3}{5}sqrt{50}+7sqrt{8}right):3sqrt{2} 8) left(2sqrt{48}-frac{3}{2}sqrt{frac{4}{3}}+sqrt{27}right).2sqrt{3...
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Bài 1: Rút gọn biểu thức
1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\) 2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)
3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\) 4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) 6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)
7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\) 8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)
9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\) 10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\) 12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
13) \(\sqrt{15-6\sqrt{6}}\) 14) \(\sqrt{8-2\sqrt{15}}\) 15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)
Tính giá trị các biểu thức:
a.\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\sqrt{3}\)
b.\(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
c.\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
d.\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
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a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
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d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)
\(=19\sqrt{3\sqrt{2}}\)
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Rút gọn bt a,Afrac{left(5+2sqrt{6}right)left(49-20sqrt{6}right)left(sqrt{5-2sqrt{6}}right)}{9sqrt{3}-11sqrt{2}}b,Csqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}c,left(sqrt{3}-1right)sqrt{6+2sqrt{2}sqrt{3-sqrt{sqrt{2}+sqrt{12}+sqrt{18-sqrt{128}}}}}d,sqrt[3]{3+sqrt{9+frac{125}{27}}}-sqrt[3]{-3+sqrt{9+frac{125}{27}}}
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Rút gọn bt
\(a,A=\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(b,C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(c,\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(d,\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
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c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
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b/ \(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=3\)
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Xem thêm câu trả lời
Câu 1: Thực hiện phép tính
a,left(sqrt{12}+3sqrt{15}-4sqrt{135}right)cdotsqrt{3}
b,sqrt{252}-sqrt{700}+sqrt{1008}-sqrt{448}
c,2sqrt{40sqrt{12}}-2sqrt{sqrt{75}}-3sqrt{5sqrt{48}}
Câu 2: Rút gọn
a,frac{9sqrt{5}+3sqrt{27}}{sqrt{5}+sqrt{3}}
b,frac{3sqrt{8}+2sqrt{12}+sqrt{20}}{3sqrt{18}-2sqrt{27}+sqrt{45}}
c,left(4+sqrt{15}right)cdotleft(sqrt{10}-sqrt{6}right)cdotsqrt{4-sqrt{15}}
Câu 3:So sánh
a,3+sqrt{5}và2sqrt{2}+sqrt{6}
b,2sqrt{3}+4và3sqrt{2}+sqrt{10}
c,18vàsqrt{15}cdotsqrt{17}
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Câu 1: Thực hiện phép tính
\(a,\left(\sqrt{12}+3\sqrt{15}-4\sqrt{135}\right)\cdot\sqrt{3}\\ b,\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\\ c,2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
Câu 2: Rút gọn
\(a,\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\\ b,\frac{3\sqrt{8}+2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\\ c,\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
Câu 3:So sánh
\(a,3+\sqrt{5}và2\sqrt{2}+\sqrt{6}\\ b,2\sqrt{3}+4và3\sqrt{2}+\sqrt{10}\\ c,18và\sqrt{15}\cdot\sqrt{17}\)
Tinh
a,sqrt{75}-sqrt{5frac{1}{3}}+frac{9}{2}sqrt{2frac{2}{3}}+2sqrt{27}
b,sqrt{48}+sqrt{5frac{1}{3}}+2sqrt{75}-5sqrt{1frac{1}{3}}
c,left(sqrt{15}+2sqrt{3}right)^2+12sqrt{5}
d,left(sqrt{6}+2right)left(sqrt{3}-sqrt{2}right)
e,left(sqrt{3}+1right)^2-2sqrt{3}+4
f,frac{1}{7+4sqrt{3}}+frac{1}{7-4sqrt{3}}
g,left(frac{1}{sqrt{5}-sqrt{2}}-frac{1}{sqrt{5}+sqrt{2}}+1right)frac{1}{left(sqrt{2}+1right)^2}
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Tinh
\(a,\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(b,\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
\(c,\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(d,\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(e,\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
\(f,\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(g,\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}\)
a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)
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Rút Gọna,sqrt{75}-sqrt{5frac{1}{3}}+frac{9}{2}sqrt{2frac{2}{3}}+2sqrt{27}b,sqrt{48}+sqrt{5frac{1}{3}}+2sqrt{75}-5sqrt{1frac{1}{3}}c,left(sqrt{12}+2sqrt{27}right)frac{sqrt{3}}{2}-sqrt{150}d,left(sqrt{18}+sqrt{0,5}-3sqrt{frac{1}{3}}right)-left(sqrt{frac{1}{8}-sqrt{75}}right)e,6sqrt{frac{8}{9}}-5sqrt{frac{32}{25}}+14sqrt{frac{18}{49}}f,2sqrt{frac{16}{3}}-3sqrt{frac{1}{27}}-6sqrt{frac{4}{75}}g,left(2sqrt{frac{16}{3}}-3sqrt{frac{1}{27}}-6sqrt{frac{4}{75}}right)sqrt{3}h,left(6sqrt{frac{8}{9}}-5sqrt{fr...
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Rút Gọn
a,\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
b,\(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
c,\(\left(\sqrt{12}+2\sqrt{27}\right)\frac{\sqrt{3}}{2}-\sqrt{150}\)
d,\(\left(\sqrt{18}+\sqrt{0,5}-3\sqrt{\frac{1}{3}}\right)-\left(\sqrt{\frac{1}{8}-\sqrt{75}}\right)\)
e,\(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\)
f,\(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
g,\(\left(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\right)\sqrt{3}\)
h,\(\left(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\right)\sqrt{\frac{1}{2}}\)
i,\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
j,\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+7\right):\sqrt{7}\)
Tính
\(a,\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(b,\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
\(c,\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)
\(d,\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(e,\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
Bài 2. Tính giá trị biểu thứca/ 2sqrt{27}-sqrt{frac{16}{3}}-sqrt{48}-sqrt{8frac{1}{3}}b/ left(3sqrt{20}-sqrt{125}-15sqrt{frac{1}{5}}right)sqrt{5}c/left(2sqrt{48}-frac{3}{2}sqrt{frac{4}{3}}+sqrt{27}right).2sqrt{3}d/ sqrt{left(3-2sqrt{2}right)^2}-sqrt{left(sqrt{8}-4right)^2}e/ sqrt{left(4-sqrt{15}right)^2}+sqrt{left(3-sqrt{15}right)^2}
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Bài 2. Tính giá trị biểu thức
a/ \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)
b/ \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right)\sqrt{5}\)
c/\(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)
d/ \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\)
e/ \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
a) A=\(\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{\sqrt{15}-\sqrt{35}}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)
b) B=\(\dfrac{12}{3+\sqrt{3}}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{27}-3\sqrt{2}}{\sqrt{3}.\sqrt{2}}\)
c)C=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)(x>0,x≠1,x≠4)
\(A=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{7}\right)}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)
\(=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)=2-5=-3\)
\(B=\dfrac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{2\sqrt{3}.\sqrt{3}}{\sqrt{3}}+\dfrac{3}{\sqrt{2}}-\dfrac{3}{\sqrt{3}}\)
\(=\dfrac{12\left(3-\sqrt{3}\right)}{6}-2\sqrt{3}+\dfrac{3\sqrt{2}}{2}-\sqrt{3}\)
\(=2\left(3-\sqrt{3}\right)-3\sqrt{3}+\dfrac{3\sqrt{2}}{2}=6-5\sqrt{3}+\dfrac{3\sqrt{2}}{2}\) (câu này khả năng đề sai, dấu \(\sqrt{3}.\sqrt{2}\) ở mẫu cuối cùng là dấu trừ mới hợp lý)
\(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
Dấu giữa 2 dấu ngoặc là dấu chia sẽ hợp lý hơn
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