RUT GON :
\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}\)
Những câu hỏi liên quan
1) Rut gon bieu thuc A = \(\left(4+\sqrt{7}\right).\left(\sqrt{2}-\sqrt{14}\right).\sqrt{4-\sqrt{7}}\)
rut gon
A=\(\sqrt{\left(\sqrt{6}-2\sqrt{2}\right)^2}-\sqrt{24-12\sqrt{3}}\)
Ta có: A = \(\sqrt{\left(\sqrt{6}-2\sqrt{2}\right)^2}-\sqrt{24-12\sqrt{3}}\)
= \(\left|\sqrt{6}-2\sqrt{2}\right|\) \(-\sqrt{18-2.6\sqrt{3}+6}\)
= \(2\sqrt{2}-\sqrt{6}-\sqrt{\left(\sqrt{18}-\sqrt{6}\right)^2}\)
= \(2\sqrt{2}-\sqrt{6}-\sqrt{18}+\sqrt{6}\)
= \(2\sqrt{2}-3\sqrt{2}=-\sqrt{2}\)
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rut gon bieu thuc
A=\(\left(\dfrac{\sqrt{X}+2}{X-9}-\dfrac{\sqrt{X}-2}{X+6\sqrt{X}+9}\right)\left(\sqrt{X}\dfrac{9}{\sqrt{X}}\right)\)
XEM CÓ SAI ĐỀ BÀI KHÔNG, MK RÚT GỌN RA TO LẮM
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\(=\dfrac{x+5\sqrt{x}+6-x+5\sqrt{x}-6}{\left(\sqrt{x}+3\right)^2\cdot\left(\sqrt{x}-3\right)}\cdot\dfrac{x-9}{\sqrt{x}}\)
\(=\dfrac{10\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{10}{\sqrt{x}+3}\)
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21 tháng 7 2023 lúc 9:12
Rút gọn: (Giải chi tiết từng bước)
7) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
8) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.
`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.
`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.
`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.
`= 7`.
`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`
`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`
`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`
`= 33 - 11 = 22`.
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Rut gon bieu thuc \(Q=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{9-x}\right).\frac{x-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{21}\)
\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
= \(14-2\sqrt{21}-7+2\sqrt{21}\) = \(7\)
\(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{21}\)
= \(33-3\sqrt{22}-11+3\sqrt{21}\) = \(22-3\sqrt{22}+3\sqrt{21}\)
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Mình sẽ làm cụ thể một tí nhé:
a) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=\left(\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}\)
\(=7\)
b) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{21}\)
\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{21}\)
\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{21}\)
\(=22-3\sqrt{22}+3\sqrt{21}\)
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Rut gon \(B=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{2}+3}-\frac{9-x}{x+\sqrt{x-6}}\right)\)
Rut gon:
\(\left(\sqrt{3-\sqrt{5}}\right).\left(\sqrt{10}-\sqrt{2}\right).\left(3+\sqrt{5}\right)\)
Mình rút gọn như sau:
\(\left(\sqrt{3-\sqrt{5}}\right).\left(\sqrt{10}-\sqrt{2}\right).\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}\right)^2}.\left(3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\)
\(=\left(\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}\right).\left(2\sqrt{10}+2\sqrt{2}\right)\)
\(=10+2\sqrt{5}-2\sqrt{5}-2\)
\(=8\)
(Chúc bạn học giỏi và tíck cho mìk vs nhá!)
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rut gon \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)