a^4+a^2+1
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3 tháng 8 2023
a,a+1/4=2 3/4-1 1/2
a+1/2=5/4
a=5/4-1/2
a=3/4
b,a-7/4=13/4-7/9
a-7/4=89/36
a= 89/36+7/4
a=152/36
c,3/2-a=17/6-1/6
3/2-a=8/3
a= 3/2-8/3
a= -7/6
25 tháng 8 2023
a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)
\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)
\(3-a=\dfrac{3}{2}\)
\(a=3-\dfrac{3}{2}\)
\(a=\dfrac{6}{2}-\dfrac{3}{2}\)
\(a=\dfrac{3}{2}\)
b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)
\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)
\(0-a=\dfrac{7}{8}\)
\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)
c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)
\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)
\(\dfrac{8}{6}-a=\dfrac{1}{6}\)
\(a=\dfrac{8}{6}-\dfrac{1}{6}\)
\(a=\dfrac{7}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 8 2023
a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)
a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)
a = 2 + 1 - 1 - \(\dfrac{1}{2}\)
a = 2 - \(\dfrac{1}{2}\)
a = \(\dfrac{3}{2}\)
b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
(3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)
a = - \(\dfrac{7}{8}\)
c, 2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a = \(\dfrac{1}{6}\)
a = 2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)
a = (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)
a = 1 + \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)
a = \(\dfrac{7}{6}\)
10 tháng 5 2016
\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)
\(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)
\(=\frac{a-b-a+a^{\frac{1}{2}}.b^{\frac{1}{2}}}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}.\frac{1}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}=\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}\frac{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}\sqrt{\frac{a}{b}}.\sqrt{\frac{a}{b}}=1\)