a, A = 3,5 + |x - 2017| - 9
= -5,5 + |x - 2017|
Ta có : |x - 2017| \(\ge0\Rightarrow-5,5+\left|x-2017\right|\ge-5,5\)
Dấu ''='' xảy ra <=> x - 2017 = 0 <=> x = 2017
Vậy GTNN của A = -5,5 <=> x = 2017
@Cô Bé Dễ Thương

|2x-1|=1,5

TH(1)2x-1=1,5

2x =1,5+1

2x =2,5

x =2,5 :2

x =1,25

TH(2) 2x-1=-1,5

2x =-1,5+1

2x =-0,5

x =-0,5:2

x =-0,25

các câu khác cứ tương tự bạn nhé

b) \(7,5-\left|5-2x\right|=-4,5\)

\(\left|5-2x\right|=7,5+4,7\)

\(\left|5-2x\right|=12\)

th1 :\(5-2x=12\)

\(2x=5-12\)

\(2x=-7\)

\(x=-7:2\)

\(x=-3,5\)

th2: \(5-2x=-12\)

\(2x=5+12\)

\(2x=17\)

\(x=17:2\)

\(x=8,5\)

c) \(-3+\left|x\right|=-1\)

\(\left|x\right|=-1+3\)

\(\left|x\right|=2\)

th1: \(x=-2\)

th2 : \(x=2\)

d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)

\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)

th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)

\(x=\dfrac{7}{3}-\dfrac{1}{2}\)

\(x=\dfrac{11}{6}\)

th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)

\(x=\dfrac{7}{3}+\dfrac{1}{6}\)

\(x=\dfrac{-5}{2}\)

e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)

\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)

\(\left|x+1\right|=\dfrac{9}{14}\)

th1 :\(x+1=\dfrac{9}{14}\)

\(x=\dfrac{9}{14}-1\)

\(x=\dfrac{-5}{14}\)

th2 : \(x+1=\dfrac{-9}{14}\)

\(x=\dfrac{-9}{14}-1\)

\(x=\dfrac{-5}{14}\)

\(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\forall x>0\)

\(\Rightarrow2x-2,5=7,5\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

\(|2x|-|-2,5|=|-7,5|\)

\(\Rightarrow|2x|-2=7,5\)

\(\Rightarrow|2x|=10\)

\(\Rightarrow\orbr{\begin{cases}2x=10\\2x=-10\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}}\)

\(KL:x\in\left\{\pm5\right\}\)

\(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)

\(\Rightarrow\left|2x\right|-2,5=7,5\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\) 

\(8 - x + 15 = 6 - 4x\)

\( - x + 4x = 6 - 8 - 15\)

\(3x =  - 17\)

\(x = \left( { - 17} \right):3\)

\(x = \dfrac{{ - 17}}{3}\)

Vậy nghiệm của phương trình là \(x = \dfrac{{ - 17}}{3}\).

b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\)

\( - 9 + 12u =  - 45 + 6u\)

\(12u - 6u =  - 45 + 9\)

\(u = \left( { - 36} \right):6\)

\(6u =  - 36\)

\(u =  - 6\)

Vậy nghiệm của phương trình là \(u =  - 6\).

c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\)

\(\left( {{x^2} + 6x + 9} \right) - \left( {{x^2} + 4x} \right) = 13\)

\({x^2} + 6x + 9 - {x^2} - 4x = 13\)

\(\left( {{x^2} - {x^2}} \right) + \left( {6x - 4x} \right) = 13 - 9\)

\(2x = 4\)

\(x = 4:2\)

\(x = 2\)

Vậy nghiệm của phương trình là \(x = 2\).

d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\)

\(\left( {{y^2} - 25} \right) - \left( {{y^2} - 4y + 4} \right) = 5\)

\({y^2} - 25 - {y^2} + 4y - 4 = 5\)

\(\left( {{y^2} - {y^2}} \right) + 4y = 5 + 4 + 25\)

\(4y = 34\)

\(y = 34:4\)

\(y = \dfrac{{17}}{2}\)

Vậy nghiệm của phương trình là \(y = \dfrac{{17}}{2}\).

\(\begin{array}{l}K = \left( {{x^2}y + 2x{y^3}} \right) - \left( {7,5{x^3}{y^2} - {x^3}} \right) + \left( {3x{y^3} - {x^2}y + 7,5{x^3}{y^2}} \right)\\ = {x^2}y + 2x{y^3} - 7,5{x^3}{y^2} + {x^3} + 3x{y^3} - {x^2}y + 7,5{x^3}{y^2}\\ = \left( {{x^2}y - {x^2}y} \right) + \left( {2x{y^3} + 3x{y^3}} \right) + \left( { - 7,5{x^3}{y^2} + 7,5{x^3}{y^2}} \right) + {x^3}\\ = 5x{y^3} + {x^3}\end{array}\)

Thay x=2, y=-1 vào K ta được \(K = 5.2.{\left( { - 1} \right)^3} + {2^3} =  - 10 + 8 =  - 2.\)

\(a,A=\left|3,4-x\right|+1,7\ge1,7\)

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)\(2\left|2x-3\right|=\frac{1}{2}\)

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{8}\\x=\frac{11}{8}\end{matrix}\right.\)

Vậy....

b)\(7,5-3\left|5-2x\right|=-4,5\)

\(\Leftrightarrow\left|5-2x\right|=4\)

\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)

VẬy...

c)\(\left|3x-4\right|+\left|5-2x\right|=0\)

Có: \(\left|3x-4\right|\ge0với\forall x\\ \left|5-2x\right|\ge0với\forall x\)

\(\Rightarrow\left[{}\begin{matrix}3x-4=0\\5-2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=\frac{5}{2}\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

\(a,B=4,2+\left|x+1,5\right|\ge4,2\\ B_{min}=4,2\Leftrightarrow x+1,5=0\Leftrightarrow x=-1,5\\ b,C=\dfrac{4}{5}-\left|2x+1\right|\le\dfrac{4}{5}\\ C_{max}=\dfrac{4}{5}\Leftrightarrow2x+1=0\Leftrightarrow x=-\dfrac{1}{2}\)

a, Do |x +1,5| ≥ 0 ⇒ 4,2 + |x + 1,5| ≥ 4,2

Dấu "=" xảy ra ⇔ x + 1,5 = 0 ⇔  x = - 1,5

Vậy B_min=  4,2 ⇔ x= -1,5

b, Do |2x + 1| ≥ 0 ⇒ \(\dfrac{4}{5}-\left|2x+1\right|\le\dfrac{4}{5}\)

Dấu "=" xảy ra ⇔ 2x + 1 = 0 ⇔ 2x = -1 ⇔ \(x=-\dfrac{1}{2}\)

Vậy C_max = \(\dfrac{4}{5}\Leftrightarrow x=-\dfrac{1}{2}\)