\(\frac{a^5+a^6+a^7+a^8}{a^{-5}+a^{-6}+a^{-7}+a^{-8}}\)
Rut gon bieu thuc tren.
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
H24
Những câu hỏi liên quan
rut gon bieu thuc
C=\(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\)= ?
G=\(\sqrt{\frac{7}{5}}+\sqrt{\frac{5}{7}}-\frac{12}{35}\sqrt{35}\)= ?
P=\(\frac{a\sqrt{a}-1}{\sqrt{a}-1}\)=?
cho bieu thuc A = /x-1/ + 3x-7
a. rut gon bieu thuc
b. tinh A khi x=3;x=-5
A = \(\frac{a^5+a^6+a^7+a^8}{a^{-5}+a^{-6}+a^{-7}+a^{-8}}\)
Tinh bieu thuc A tai a=2015
Ta có:
\(\frac{a^5+a^6+a^7+a^8}{a^{-5}+a^{-6}+a^{-7}+a^{-8}}\)
\(=\frac{a^5+a^6+a^7+a^8}{\frac{1}{a^5}+\frac{1}{a^6}+\frac{1}{a^7}+\frac{1}{a^8}}\)
\(=a^{5+6+7+8}=a^{26}\)
Thay vào sẽ là:
\(2015^{26}=8.149881843.10^{85}\)
Đúng 0
Bình luận (0)
\(A=\frac{a^3\left(a^3+a^2+a+1\right).a^8}{a^3+a^2+a^1+a}=a^{24}\)
Đúng 0
Bình luận (0)
Bo dau ngoac roi rut gon bieu thuc :
a) - ( - 12 + 25 - 36 ) - ( 12 - 36 + 57 )
b) - ( 23 + 5 - 6 + 17 ) + ( 23 - 5 - 83 - 6 )
c) 12 . ( 6 - 7 - 8 ) - 10 . ( 11 + 12 - 13 )
Giup minh voi minh dang can rat gap
a. = 12 - 25 + 36 - 12 + 36 - 57
= (12 - 12) + (36 + 36) - (25 + 57)
= 0 + 72 - 82
= -10
b. = -23 - 5 + 6 - 17 + 23 - 5 - 83 - 6
= (-23 + 23) + (-5 - 5) + (6 - 6) - (17 + 83)
= 0 + (-10) + 0 - 100
= -110
c.?
Đúng 0
Bình luận (0)
Cho bieu thuc:
P=\(\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
a. Rut gon bieu thuc P
b.Tim GTLN cua P sau khi rut gon
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
Đúng 0
Bình luận (0)
1/ Rut gon bieu thuc sau:
a) \(\sqrt{12-2\sqrt{35}}+\sqrt{7-2\sqrt{10}}-\sqrt{\sqrt{49}}\)
b) \(\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{3}{5+2\sqrt{7}}\)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) rut gon bieu thuc gium em a thanks
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Đúng 0
Bình luận (0)
rut gon bieu thuc
A=\(\frac{2}{5}.\sqrt{50x}-\frac{3}{4}.\sqrt{8}\left(x\ge0\right)\)
Bo dau ngoac roi rut gon bieu thuc :
A= a .( 5-b) + b . ( a- 5) - (5a + 3b )
B = ( x-1 ) . (x-2) - ( x2 + x + 8 )
A=a.5-ab+ba-b.5-5a-3b=-8b
Đúng 0
Bình luận (0)