Rút gọn các biểu thức:
M=√3−2√2−√6+4√2M=3−22−6+42
N=√2+√3+√2−√3
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a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
a) \(M=\sqrt{3-2\sqrt{2}}+\sqrt{6+4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{2+2.\sqrt{2}.2+4}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+2\right)^2}=\left|\sqrt{2}-1\right|+\sqrt{2}+2=\sqrt{2}-1+\sqrt{2}+2=2\sqrt{2}+1\)
b) \(N=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\left|\sqrt{3}-1\right|}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{2}.\sqrt{3}=\sqrt{6}\)
Hướng dẫn trả lời:
M=√3−2√2−√6+4√2=√(√2)2−2√2.1+12−√(2)2+2√2+(√2)2=√(√2−1)3−√(2+√2)2=∣∣√2−1∣∣−∣∣2+√2∣∣=√2−1−2−√2=−3M=3−22−6+42=(2)2−22.1+12−(2)2+22+(2)2=(2−1)3−(2+2)2=|2−1|−|2+2|=2−1−2−2=−3
N=√2+√3+√2−√3⇒N2=(√2+√3+√2−√3)2=2+√3+2√(2+√3)(2−√3)+2−√3=4+2√4−3=6N=2+3+2−3⇒N2=(2+3+2−3)2=2+3+2(2+3)(2−3)+2−3=4+24−3=6
Vì N > 0 nên N2 = 6 ⇒ N = √6. Vậy
a: \(=\dfrac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)
\(=\dfrac{6+4\sqrt{2}}{2+2\sqrt{2}}+\dfrac{6-4\sqrt{2}}{2\sqrt{2}-2}\)
\(=\dfrac{3+2\sqrt{2}}{\sqrt{2}+1}+\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}\)
=căn 2+1+căn 2-1=2căn 2
b: \(=\dfrac{\sqrt{3}+\sqrt{3+\sqrt{3}}+\sqrt{3}-\sqrt{3+\sqrt{3}}}{1-\sqrt{3}-1}=\dfrac{-2\sqrt{3}}{\sqrt{3}}=-2\)
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