a,|2x|.|3.5|=|-2.8| b,|x-1/2|+|x+y|=0
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(A\le0,5\)
Dấu "=" xảy ra khi x-3,5 = 0
<=> x = 3,5
Vậy max A = 0,5 khi x = 3,5
\(B\le-2\)
Dấu "=" xảy ra khi 1,4 -x =0
<=> x = 1,4
Vậy max B = -2 khi x =1,4
1.
A nhỏ hơn hoặc bằng 0,5 suy ra GTLN của A là 0,5.
B sẽ nhơ hơn hoặc bằng 2 suy ra GTLN
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,A=0,5-\left|x-3,5\right|\)
Có \(\left|x-3,5\right|\ge0\)
\(\Rightarrow A\le0,5+0=0,5\)
Dấu "=" xảy ra khi \(\left|x-3,5\right|=0\Leftrightarrow x=3,5\)
Vậy \(A_{max}=0,5\Leftrightarrow x=3,5\)
\(B=-\left|1,4-x\right|-2\)
Có \(-\left|1,4-x\right|\le0\)
\(\Rightarrow B\le0-2=-2\)
Dấu "=" xảy ra khi \(1,4-x=0\Leftrightarrow x=1,4\)
Vậy \(B_{max}=-2\Leftrightarrow x=1,4\)
\(2,C=1,7+\left|3,4-x\right|\)
Có \(\left|3,4-x\right|\ge0\)
\(\Rightarrow C\ge1,7+0=1,7\)
Dấu "=" xảy ra khi \(3,4-x=0\Leftrightarrow x=3,4\)
Vậy \(C_{min}=1,7\Leftrightarrow x=3,4\)
\(D=\left|x+2,8\right|-3,5\)
Có \(\left|x+2,8\right|\ge0\)
\(\Rightarrow D\ge0-3,5=-3,5\)
Dấu "=" xảy ra khi \(x+2,8=0\Leftrightarrow x=-2,8\)
Vậy \(D_{min}=-3,5\Leftrightarrow x=-2,8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/\(240\times20-\left(846+202\times2,5\right)\)
\(=4800-\left(846+505\right)\)
\(=4800-1351\)
\(=3449\)
b/\(2000+\left(32,4:3-2,8\right)\)
\(=2000+\left(10,8-2,8\right)\)
\(=2000+8\)
\(=2008\)
c/\(320\times1,25-\left(933+302\times3,5\right)\)
\(=400-\left(933+1057\right)\)
\(=400-1990\)
\(=-1590\)
![](https://rs.olm.vn/images/avt/0.png?1311)
0,16 : ( x : 3,5) = 2,8
x : 3,5 = 0,16 : 2,8
x = 0,16 : 2,8 x 3,5
x = 0,2
Thử lại: 0,16 : (0,2 : 3,5) = 2,8 (đúng)
![](https://rs.olm.vn/images/avt/0.png?1311)
0,16 : ( x : 3,5 ) = 2,8
x : 3,5 = 2/35
x = 2/35 * 3,5
x = 0,2
mình làm đầu tiên!
0,16 : ( x : 3,5 ) = 2,8
x : 3,5 = 0,16 : 2,8
x : 3,5 = 0,057142857
x = 0,057142857 x 3,5
x = 0,2
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
C = 2-3 + (52)3.5-3 + 4-3.16 - 2.32 - 105.(\(\dfrac{24}{51}\))0
C = \(\dfrac{1}{8}\) + 56.5-3 + 4-3.42 - 2.9 - 105.1
C = \(\dfrac{1}{8}\) + 53 + \(\dfrac{1}{4}\) - 18 - 105
C = (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\)) - (105 - 125 + 18)
C = \(\dfrac{3}{8}\) - (-20 + 18)
C = \(\dfrac{3}{8}\) + 2
C = \(\dfrac{19}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,A=\(\frac{1}{2}.\left(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2016.2016}{2015.2017}\right)=\frac{1}{2}.\left(\frac{2.3.4...2016}{1.2....2015}.\frac{2.3.4...2016}{3.4....2017}\right)=\frac{1}{2}.\left(\frac{2016.2}{2017}\right)=\frac{4032}{4034}=\frac{2016}{2017}\)
Hok tốt
\(\left|x\right|=\frac{1}{2}\Rightarrow x=\orbr{\begin{cases}\frac{1}{2}\\-\frac{1}{2}\end{cases}}\)
TH1:\(x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{3}{2}+5=4\)
TH2:\(x=\frac{-1}{2}\)
\(\Rightarrow\frac{1}{2}+\frac{3}{2}+5=7\)
Vậy
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1,7+\left|3,4-x\right|\)
\(C\text{ó};\left|3,4-x\right|\)lớn hơn or =0
\(\Rightarrow\)\(1,7+\left|3,4-x\right|=1,7+0\)
\(\Rightarrow1,7+\left|3,4-x\right|=1,7\)
\(\Rightarrow\left|3,4-x\right|=0\)
\(\Rightarrow x=3,4\)
\(B=\left|x+2,8\right|-3,5\)
\(C\text{ó}:\left|x+2,8\right|\)Lớn hơn or = 0
\(\Rightarrow\left|x+2,8\right|-3,5=0-3,5\)
\(\Rightarrow\left|x+2,8\right|-3,5=-3,5\)
\(\Rightarrow\left|x+2,8\right|=0\)
\(\Rightarrow x=-2,8\)
\(C=\left|4,3-x\right|+3,7\)
\(C\text{ó}:\left|4,3-x\right|\) Lớn hơn or = 0
\(\Rightarrow\left|4,3-x\right|+3,7=0+3,7\)
\(\Rightarrow\left|4,3-x\right|+3,7=3,7\)
\(\Rightarrow\left|4,3-x\right|=0\)
\(\Rightarrow x=4,3\)
a) \(|2x|.|3.5|=|-2.8|\)
\(\Rightarrow|2x|.|15|=|-16|\)
\(\Rightarrow|2x|.15=16\)
\(\Rightarrow|2x|=\dfrac{16}{15}\)
Ta có hai trường hợp:
1) \(2x\ge0\)
\(\Rightarrow|2x|=2x=\dfrac{16}{15}\Rightarrow x=\dfrac{8}{15}\)
2) \(2x< 0\)
\(\Rightarrow|2x|=-2x=\dfrac{16}{15}\Rightarrow-x=\dfrac{8}{15}\Rightarrow x=-\dfrac{8}{15}\)
Vậy \(x=\dfrac{8}{15}\) hoặc \(x=-\dfrac{8}{15}\).
b) \(|x-\dfrac{1}{2}|+|x+y|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\x+y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)