Tính giá trị của các biểu thức sau:
1. \(2,5.\left( {4,1 - 3 - 2,5 + 2.7,2} \right) + 4,2:2\)
2. \(2,86.4 + 3,14.4 - 6,01.5 + {3^2}\)
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\(A=11,44+12,56-30,05+1024\)
\(=24-30,56+1024\)
\(=-6,05+1024\)
\(=1017,95\)
a) \(0,2 + 2,5:\frac{7}{2} = \frac{2}{{10}} + \frac{25}{10}:\frac{7}{2} = \frac{1}{5} + \frac{25}{10}.\frac{2}{7} \\= \frac{1}{5} + \frac{5}{7} = \frac{7}{{35}} + \frac{{25}}{{35}} = \frac{{32}}{{35}}\)
b)
\(\begin{array}{l}9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\\ = 9.\frac{1}{9} - {\left( {\frac{{ - 1}}{{10}}} \right)^3}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}.\frac{{15}}{2}\\ = 1 + \frac{3}{{400}}\\=\frac{400}{400}+\frac{3}{400}\\ = \frac{{403}}{{400}}\end{array}\)
A= (3,1 - 2,5) - (-2,5 + 3,1)
=3.1-2.5+2.5-3.1
=0
B= (5,3 - 2,8) - (4 + 5,3)
=5.3 - 2.8-4-5.3
= -6.8
C= -(251.3 + 281) + 3.251 - (1-281)
=-251.3-281+3.251-1+281
=-1
D= -(3/5+3/4)−(−3/4+2/5)
= -3/5-3/4+3/4-2/5
= -5/5
=-1
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)=3,1-2,5+2,5-3,1=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)=5,3-2,8-4-5,3=-6,8\)
\(C=-\left(215\cdot3+281\right)+3\cdot215-\left(1-281\right)=-215\cdot3-281+3\cdot215-1+281=1\)
\(D=-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(A=3,1-2,5+2,5-3,1\)
\(A=\left(3,1-3,1\right)-\left(2,5-2,5\right)\)
\(A=0-0\)
\(A=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(B=5,3-2,8-4-5,3\)
\(B=\left(5,3-5,3\right)-\left(2,8+4\right)\\ B=0-6,8\\ B=-6,8\)
Ta có\(\hept{\begin{cases}\left(x-\frac{2}{5}\right)^2\ge0\\\left|2y+1\right|\ge0\end{cases}}\)
\(\Rightarrow\frac{1}{3}\left(x-\frac{2}{5}\right)^2+\left|2y+1\right|-2,5\)có GTNN khi \(\hept{\begin{cases}\left(x-\frac{2}{5}\right)^2=0\\\left|2y+1\right|=0\end{cases}}\)
\(\Rightarrow\frac{1}{3}\left(x-\frac{2}{5}\right)^2-\left|2y+1\right|-2,5\)có GTNN là \(\frac{1}{3}\cdot0+0-2,5=-2,5\)
Vậy GTNN của biểu thức trên là -2,5
Vì: \(\begin{cases}\frac{1}{3}\left(x-\frac{2}{5}\right)^2\ge0\\\left|2y+2\right|\ge0\end{cases}\)
\(\Rightarrow\frac{1}{3}\left(x-\frac{2}{5}\right)^2+\left|2y+1\right|\ge0\)
\(\Rightarrow\frac{1}{3}\left(x-\frac{2}{5}\right)^2+\left|2y+1\right|-2,5\ge-2,5\)
Vậy GTNN của C là -2,5 khi \(\begin{cases}x-\frac{2}{5}=0\\2y+1=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=\frac{2}{5}\\y=-\frac{1}{2}\end{cases}\)
1,
\(\begin{array}{l}2,5.\left( {4,1 - 3 - 2,5 + 2.7,2} \right) + 4,2:2\\ = 2,5.\left( {4,1 - 3 - 2,5 + 14,4} \right) + 4,2:2\\ = 2,5.\left( {1,1 - 2,5 + 14,4} \right) + 2,1\\ = 2,5.\left( { - 1,4 + 14,4} \right) + 2,1\\ = 2,5.13 + 2,1\\ = 32,5 + 2,1\\ = 34,6\end{array}\)
2,
Cách 1:
\(\begin{array}{l}2,86.4 + 3,14.4 - 6,01.5 + {3^2}\\ = 11,44 + 12,56 - 30,05 + 9\\ = \left( {11,44 + 12,56} \right) + \left( { - 30,05 + 9} \right)\\ = 24 + \left( { - 21,05} \right)\\ = 24 - 21,05\\ = 2,95\end{array}\)
Cách 2:
\(\begin{array}{l}2,86.4 + 3,14.4 - 6,01.5 + {3^2}\\ = 4.(2,86+3,14) - 30,05 + 9\\ = 4.6 + \left( { - 30,05 + 9} \right)\\ = 24 + \left( { - 21,05} \right)\\ = 24 - 21,05\\ = 2,95\end{array}\)