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1) \(A=\left(2x^2+1\right)^4-3\ge0-3=-3\) (do \(\left(2x^2+1\right)^4\ge0\forall x\))
Dấu "=" xảy ra \(\Leftrightarrow\left(2x^2+1\right)=0\Leftrightarrow2x^2=-1\Leftrightarrow x^2=-\frac{1}{2}\) (vô lí)
Vậy đề sai ~v (hay là tui làm sai ta)
a, 7/12 - 3/4 . 5/6
= 7/12 - 5/8
= 14/24- 15/24
= -1/24
b,( 2/1/3 + 1/3/4 ) . 12/13
= ( 6/3 + 7/3 ) . 12/13
= 13/3 . 12/13
=4
c, 12 : ( 3/4 -5/6 ) . 2
= 12 : ( -1/12 ) .2
= 12 . -12 . 2
= -228
d, 7/22 : 3/11 + 7/22 : 4/11
= 7/22 . 11/3 + 7/22 . 11/4
= 7/22 . ( 11/3 + 11/4 )
....
tiếp theo bạn tự làm nhé!
tính nhanh :
a) ( 3 - 1/4 + 2/3 ) - ( 5 + 1/3 - 6/5 ) - ( 6 - 7/4 + 3/2 )
giúp mk vs nha mai nộp bài r
a) \(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=3-\frac{1}{4}+\frac{7}{4}-\frac{3}{2}+\frac{2}{3}-\frac{1}{3}-5+\frac{6}{5}-6\)
\(=3+\frac{3}{2}-\frac{3}{2}+\frac{1}{3}-11+\frac{6}{5}\)
\(=3+0+\frac{23}{15}-11\)
\(=\frac{68}{15}-\frac{165}{15}=\frac{-97}{15}.\)
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2019}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2018}{2019}\)
\(A=\frac{1.2.3.....2018}{2.3.4....2019}\)
\(A=\frac{1}{2019}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2018}{2019}\)
\(A=\frac{1}{2019}\)
Bn ơi tk cho mk nha!!!!!!!!!!!
