GTNN của K = \(\left|1,5+3x\right|+\left|3x-7,5\right|-1,7\)là 1,7 khi x bao nhiu?
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\(a,A=\left|3,4-x\right|+1,7\ge1,7\)
Dấu \("="\Leftrightarrow3,4-x=0\Leftrightarrow x=3,4\)
\(c,C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}4x-3=0\\5y+7,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-\dfrac{3}{2}\end{matrix}\right.\)
a: |3x-1|>=0
=>2|3x-1|>=0
=>2|3x-1|-4>=-4
Dấu = xảy ra khi x=1/3
b: |2-x|>=0
=>|2-x|+1,5>=1,5
Dấu = xảy ra khi x=2
a) \(\left|x+1,1\right|\ge0\Leftrightarrow-\left|x+1,1\right|\le0\Leftrightarrow1,5-\left|x+1,1\right|\le1,5\)
\(\Leftrightarrow A_{Max}=1,5\)
\("="\Leftrightarrow x=-1,1\)
b) \(\left|1,7-x\right|\ge0\Leftrightarrow-\left|1,7-x\right|\le0\Leftrightarrow-3,7-\left|1,7-x\right|\le-3,7\)
\(\Leftrightarrow B_{Max}=-3,7\)
\("="\Leftrightarrow x=1,7\)
Bài 1:
a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)
\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)
\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)
\(=-10\cdot1.7=-17\)
b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)
\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)
\(=-\dfrac{539}{60}\)
c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
\(=10\)
\(\begin{array}{l}K = \left( {{x^2}y + 2x{y^3}} \right) - \left( {7,5{x^3}{y^2} - {x^3}} \right) + \left( {3x{y^3} - {x^2}y + 7,5{x^3}{y^2}} \right)\\ = {x^2}y + 2x{y^3} - 7,5{x^3}{y^2} + {x^3} + 3x{y^3} - {x^2}y + 7,5{x^3}{y^2}\\ = \left( {{x^2}y - {x^2}y} \right) + \left( {2x{y^3} + 3x{y^3}} \right) + \left( { - 7,5{x^3}{y^2} + 7,5{x^3}{y^2}} \right) + {x^3}\\ = 5x{y^3} + {x^3}\end{array}\)
Thay x=2, y=-1 vào K ta được \(K = 5.2.{\left( { - 1} \right)^3} + {2^3} = - 10 + 8 = - 2.\)
Bài 1:
a, \(A=3,7+\left|4,3-x\right|\ge3,7\)
Dấu " = " khi \(\left|4,3-x\right|=0\Rightarrow x=4,3\)
Vậy \(MIN_A=3,7\) khi x = 4,3
b, \(B=\left|3x+\dfrac{41}{5}\right|-14,2\ge-14,2\)
Dấu " = " khi \(\left|3x+\dfrac{41}{5}\right|=0\Rightarrow x=\dfrac{-41}{15}\)
Vậy \(MIN_B=-14,2\) khi \(x=\dfrac{-41}{15}\)
c, \(C=\left|4x-3y\right|+\left|5y+7,5\right|\ge17,5\)
( do \(\left|4x-3y\right|+\left|5y+7,5\right|\ge0\) )
Dấu " = " khi \(\left\{{}\begin{matrix}\left|4x-3y\right|=0\\\left|5y+7,5\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{8}\\y=-1,5\end{matrix}\right.\)
Vậy \(MIN_C=17,5\) khi \(\left\{{}\begin{matrix}x=\dfrac{-9}{8}\\y=-1,5\end{matrix}\right.\)
Bài 2:
a, \(A=5,5-\left|2x-1,5\right|\le5,5\)
Dấu " = " khi \(\left|2x-1,5\right|=0\Rightarrow x=0,75\)
Vậy \(MIN_A=5,5\) khi x = 0,75
b, c tương tự