rút gọnA=(2+1)*(2^2+1)(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)
Những câu hỏi liên quan
bài tập, rút gọn
a, √8-2√7 + √16-6√7
b, √(√7-1)2 - √11+4√7
c.ơn các bạn trước
`a)sqrt{8-2sqrt7}+sqrt{16-6sqrt7}`
`=sqrt{(sqrt7-1)^2}+sqrt{(3-sqrt7)^2}`
`=sqrt7-1+3-sqrt7=2`
`b)sqrt{(sqrt7-1)^2}-sqrt{11+4sqrt7}`
`=sqrt7-1-sqrt{(2+sqrt7)^2}`
`=sqrt7-1-2-sqrt7=-3`
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a, \(=\sqrt{7-2\sqrt{7}+1}+\sqrt{7-2.3\sqrt{7}+9}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\left|\sqrt{7}-1\right|+\left|3-\sqrt{7}\right|\)
\(=\sqrt{7}-1+3-\sqrt{7}=2\)
\(b,=\left|\sqrt{7}-1\right|-\sqrt{7+2.2\sqrt{7}+4}\)
\(=\left|\sqrt{7}-1\right|-\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+2\right|\)
\(=\sqrt{7}-1-\sqrt{7}-2=-3\)
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a \(\Rightarrow\sqrt{7-2\sqrt{7}+1}+\sqrt{9-6\sqrt{7}+7}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\sqrt{7}-1+3-\sqrt{7}=2\) b \(\Rightarrow\sqrt{7}-1-\sqrt{7-4\sqrt{7}+4}=\sqrt{7}-1-\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-1-\sqrt{7}+2=1\)
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Bài 1. Rút gọna. 2sqrt{8}-3sqrt{18}+sqrt{32}b. sqrt{left(1-sqrt{2}right)^2}+sqrt{left(1+sqrt{2}right)^2}c. sqrt{left(2-sqrt{3}right)^2}+sqrt{4-2sqrt{3}}d. sqrt{2-sqrt{3}+sqrt{2+sqrt{3}}}Bài 2. Giải phương trìnha. xsqrt{8}-6sqrt{2}0b. sqrt{2x+1}-30c. sqrt{x^2-4x+4}-30d. sqrt{x-1}+sqrt{4x-4}-sqrt{25x-25+2}0
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Bài 1. Rút gọn
a. \(2\sqrt{8}-3\sqrt{18}+\sqrt{32}\)
b. \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(1+\sqrt{2}\right)^2}\)
c. \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
d. \(\sqrt{2-\sqrt{3}+\sqrt{2+\sqrt{3}}}\)
Bài 2. Giải phương trình
a. \(x\sqrt{8}-6\sqrt{2}=0\)
b. \(\sqrt{2x+1}-3=0\)
c. \(\sqrt{x^2-4x+4}-3=0\)
d. \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25+2}=0\)
a) Ta có: \(2\sqrt{8}-3\sqrt{18}+\sqrt{32}\)
\(=4\sqrt{2}-6\sqrt{2}+4\sqrt{2}\)
\(=2\sqrt{2}\)
b) Ta có: \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(1+\sqrt{2}\right)^2}\)
\(=\sqrt{2}-1+\sqrt{2}+1\)
\(=2\sqrt{2}\)
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c) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
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a) Ta có: 2√8−3√18+√3228−318+32
=4√2−6√2+4√2=42−62+42
=2√2=22
b) Ta có: √(1−√2)2+√(1+√2)2(1−2)2+(1+2)2
=√2−1+√2+1=2−1+2+1
=2√2
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(x^2+x+1)(x^4+x^2+1)(x^8+x^4+1)(x^16+x^8+1)(x^32+x^16+1) rút gọn zùm mình với
https://www.youtube.com/watch?v=cFZDEMTQQCs
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8 tháng 4 2021 lúc 20:38
1.Rút gọn
a) C = 1/2 + 1/2^2 + 1/2^3 + ....+ 1/2^2020
\(C=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\\ 2C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\\ 2C-C=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\right)\\ C=1-\dfrac{1}{2^{2020}}=\dfrac{2^{2020}-1}{2^{2020}}\)
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8 tháng 4 2021 lúc 20:38
1.Rút gọn
a) C = 1/2 + 1/2^2 + 1/2^3 + ....+ 1/2^2020
Giải:
C=1/2 + 1/2^2 + 1/2^3 + ... + 1/2^2020
2C=1 + 1/2 + 1/2^2 + ... +1/2^2019
2C-C=(1+1/2+1/2^2+...+1/2^2019)-(1/2+1/2^2+1/2^3+...+1/2^2020)
C=1-1/2^2020
Chúc bạn học tốt!
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Rút gọn biểu thức sau:
3.(22+1)(24+1)(28+1)(216+1)(232+1)
( bài này áp dụng hằng đẳng thức \(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
Ta có
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(=2^{64}-1\)
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3.(22+1)(24+1)(28+1)(216+1)(232+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)
=(24-1)(24+1)(28+1)(216+1)(232+1)
=(28-1)(28+1)(216+1)(232+1)
=(216-1)(216+1)(232+1)
=(232-1)(232+1)
=264-1
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1. Rút Gọn
a)√6-2√5
b)√8+2√7
2 Tính
a) √(√10-3)2 -√10
b)√(5+√7)2 - √8-2√7
\(1,\)
\(a,\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2.\sqrt{5}.1+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
\(b,\sqrt{8+2\sqrt{7}}=\sqrt{\sqrt{7^2}+2.\sqrt{7}.1+1}=\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}+1\right|=\sqrt{7}+1\)
\(2,\)
\(a,\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{10}\)
\(=\left|\sqrt{10}-3\right|-\sqrt{10}\)
\(=\sqrt{10}-\sqrt{10}-3\)
\(=-3\)
\(b,\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\)
\(=\left|5+\sqrt{7}\right|-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=5+\sqrt{7}-\left|\sqrt{7}-1\right|\)
\(=5+\sqrt{7}-\sqrt{7}+1\)
\(=6\)
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RÚT GỌN BIỂU THỨC SAU
3(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
3(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(24-1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(28-1)(28+1)(216+1)(232+1)(264+1)
=(216-1)(216+1)(232+1)(264+1)
=(232-1)(232+1)(264+1)
=(264-1)(264+1)
=(2128-1)
Nếu thấy đúng thì thích cho mình nha
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Rút gọn biểu thức:
a) (x-2)(x^2-2x+4)(x+2)(x^2+2x+4)
b)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)