giải hộ mình bài 15 với
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14a) \(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{2}.2+2^2}-\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{2}.2+2^2}\)
\(=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\)
\(=\sqrt{5}+2-\sqrt{5}+2=4\)
b) \(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}-\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
15a) \(P=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{3^2+2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}=\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\)
\(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
b) \(Q=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}+\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}=\left|3+2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}=6\)
= 12/11
Câu 1:
0,9 x 218 x 2 + 0,18 x 4290 + 0,6 x 353 x 3
= 9/10 x 436 + 9/50 x 4290 + 6/10 x 1059
= 9 x 43,6 + 9 x 85,8 + 6 x 105,9
= 3 x 130,8 + 3 x 257,4 + 3 x 211,8
= 3 x ( 130,8 + 257,4 + 211,8 )
= 3 x 600
= 1800
Câu 2:
3/4 x X + 1/2 x X - 15 = 35
X x ( 3/4 + 1/2 ) - 15 = 35
X x ( 3/4 + 1/2 ) = 50
X x 5/4 = 50
X = 40
VẬy X = 40
A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ..+ \(\dfrac{1}{55}\)+ \(\dfrac{1}{66}\)
A = 2 \(\times\) ( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) +...+ \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\))
A = 2 \(\times\) ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\) +...+ \(\dfrac{1}{10.11}\)+ \(\dfrac{1}{11.12}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +...+ \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))
A = 1 - \(\dfrac{1}{6}\) < 1
Vậy A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{55}\)+ \(\dfrac{1}{66}\) < 1
ta có: x = 2018 => 2019 = x + 1. Do đó:
\(C=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-1.\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-1.\)
\(=x-1=2019-1=2018\)
Vậy C = 2018 với x = 2018.
Học tốt nhé ^3^
\(Ta \) \(có :\)
\(x = 2018\)\(\Leftrightarrow\)\(x + 1 = 2019\)
\(Thay \) \(x + 1 = 2019\)\(vào \) \(C , ta \) \(được :\)
\(C = x\)\(15\)\(- ( x + 1 ).x\)\(14\)\(+ ( x + 1 ).x\)\(13\) \(- ( x + 1 ).x\)\(12\) \(+ ...+ ( x + 1 ).x - 1\)
\(C = x\)\(15\)\(- x\)\(15\)\(- x\)\(14\) \(+ x\)\(14\) \(+ x\)\(13\)\(- x\)\(13\)\(- x\)\(12\)\(+ ... + x^2 + x - 1\)
\(C = x - 1\)
\(Thay \) \(x = 2018\) \(vào \) \(C\) \(, ta \) \(được :\)
\(C = 2018 - 1 = 2017\)
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Ko thấy đề sao giải??