cho x=3.7 so sánh A= [x]+[x+1/5]+[x+2/5]+[x+3/5]+[x+4/5] và B=[5x]
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+x+\frac{1}{5}+x+\frac{2}{5}+x+\frac{3}{5}+x+\frac{4}{5}=5x+\frac{10}{5}\)
\(5x+2>5x\)
\(A>B\)
\(x+x+\frac{1}{5}+x+\frac{2}{5}+x+\frac{3}{5}+x+\frac{5}{5}=5x+\frac{10}{5}\)
\(5x+2>5x\)
\(A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x\right)+\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{4}{5}\right)\)
\(A=\left(x+x+x+x+x\right)+\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)\)
\(A=5x+2>5x\Rightarrow A>B\)
\(A=x+x+\frac{1}{5}+x+\frac{2}{5}+x+\frac{3}{5}+x+\frac{4}{5}\)
\(=x+x+x+x+x+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\)
\(=5x+\frac{10}{5}\)
\(=5x+2\)
Vì 5x + 2 > 5x
Vậy A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x+\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{4}{5}\right)\)
\(=5x+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\)
\(=5x+2\)
\(B=5x\)
\(\Rightarrow A>B\)Với \(\forall\)\(x\)
#)Giải :
\(A=\left[x\right]+\left[1+\frac{1}{5}\right]+\left[x+\frac{2}{5}\right]+\left[x+\frac{3}{5}\right]+\left[x+\frac{4}{5}\right]\)
Thay x = 3,7 vào biểu thức, ta có :
\(A=\left[3,7\right]+\left[3,7+\frac{1}{5}\right]+\left[3,7+\frac{2}{5}\right]+\left[3,7+\frac{3}{5}\right]+\left[3,7+\frac{4}{5}\right]\)
\(A=\left[3,7+3,7+3,7+3,7+3,7\right]+\left[1+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right]\)
\(A=18,5+3\)
\(A=21,5\)
\(B=\left[5x\right]=\left[5\times3,7\right]=18,5\)
Vì 21,5 > 18,5 \(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải
xét vế A :
thay x=3,7 vào biểu thức ta có:
A=[ 3,7 ]+[ 3,7+1/5]+[3,7+2/5]+[3,7+3/5]+[ 3,7+4/5]
=(3.7*5)+(1/5+2/5+3/5+4/5)=20,5
xét vế B
thay x=3,7 vào biểu thức ta có
B=[5x]
=>b=[5*3.7]=5.3,7=18,5
+, ta có A=20,5 ; B=18,5
=>A>B
\(A=x+\left(x+\dfrac{1}{5}\right)+\left(x+\dfrac{2}{5}\right)+\left(x+\dfrac{3}{5}\right)+\left(x+\dfrac{4}{5}\right)\)
\(A=x+x+\dfrac{1}{5}+x+\dfrac{2}{5}+x+\dfrac{3}{5}+x+\dfrac{4}{5}=5x+2\)
\(5x+2>5x\forall x\Rightarrow A>B\) (đương nhiên)
=)) đề có gì đó nhầm lẫn