Gửi em

a) \(\left|x-1,7\right|+\dfrac{1}{2}=23\)

\(\left|x-1,7\right|=22,5\)

\(\Rightarrow x-1,7=\pm22,5\)

\(\Rightarrow x=\left\{24,2;-20,8\right\}\)

b) \(\left|x+\dfrac{4}{15}\right|-\left|3,75\right|=-\left|2,15\right|\)

\(\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75\)

\(\left|x+\dfrac{4}{15}\right|=1,6\)

\(\Rightarrow x+\dfrac{4}{15}=\pm1,6\)

\(\Rightarrow x=\left\{\dfrac{4}{3};-\dfrac{28}{15}\right\}\)

a)\(\left|x-1,7\right|+\dfrac{1}{2}=23\)

<=>\(\left|x-1,7\right|=22,5\)

<=>\(-\left(x-1,7\right)=-22,5\) hoặc \(x-1,7=22,5\)

<=>\(-x+1,7=-22,5\) hoặc \(x-1,7=22,5\)

<=>\(-x=-22,5-1,7\) hoặc \(x=22,5+1,7\)

<=>\(x=24,2\)

vậy \(x=24,2\)

1)

25+27+x=21+l-22l

=>25+27+x=21+22

=>25+27+x=43

=>52+x=43

=>x=43-52

=>x=-9

2)

l-5l+l-7l=x+3

=>5+7=x+3

=>12=x+3

=>x=12-3

=>x=9

3)

8+lxl=l-8l+15

=>8+lxl=8+15

=>8+lxl=23

=>lxl=23-8

=>lxl=15

=>x=15 hoặc x=-15

4)

lxl+15=-7

=>lxl=-7-15

=>lxl=-22

=>x ko tồn tại

1)25+27+x=21+22                                      3)8+x=8+15

52+x=43                                                     8+x=23

x=52-43=9                                                 x=23-8=15

2)5+7=x+3                                                 x=15 hoặc x=-15

12=x+3                                                    4) x+15=-7

x=12-3=9                                                     x=(-7)-15=-22

                                                                   x=22 hoặc x=-22

1. -9

2. 9

3.15

4.ko có

Bài 1:

a)
\(|x+\frac{4}{15}|-|-3,75|=-|-2,15|\)

\(\Leftrightarrow |x+\frac{4}{15}|-3,75=-2,15\)

\(\Leftrightarrow |x+\frac{4}{15}|=-2,15+3,75=\frac{8}{5}\)

\(\Rightarrow \left[\begin{matrix} x+\frac{4}{15}=\frac{8}{5}\\ x+\frac{4}{15}=-\frac{8}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{3}\\ x=\frac{-28}{15}\end{matrix}\right.\)

b )

\(|\frac{5}{3}x|=|-\frac{1}{6}|=\frac{1}{6}\)

\(\Rightarrow \left[\begin{matrix} \frac{5}{3}x=\frac{1}{6}\\ \frac{5}{3}x=-\frac{1}{6}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{10}\\ x=-\frac{1}{10}\end{matrix}\right.\)

c)

\(|\frac{3}{4}x-\frac{3}{4}|-\frac{3}{4}=|-\frac{3}{4}|=\frac{3}{4}\)

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

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

Bài 3:

a) Ta thấy:

\(|x+\frac{15}{19}|\geq 0, \forall x\Rightarrow A\ge 0-1=-1\)

Vậy GTNN của $A$ là $-1$ khi \(x+\frac{15}{19}=0\Leftrightarrow x=-\frac{15}{19}\)

b)Vì \(|x-\frac{4}{7}|\geq 0, \forall x\Rightarrow B\geq \frac{1}{2}+0=\frac{1}{2}\)

Vậy GTNN của $B$ là $\frac{1}{2}$ khi \(x-\frac{4}{7}=0\Leftrightarrow x=\frac{4}{7}\)

Bài 4:

a)

\(|\frac{5}{3}-x|\geq 0, \forall x\)

\(\Rightarrow 5=A+|\frac{5}{3}-x|\geq A+0=A\)

Vậy GTLN của $A$ là $5$ khi \(\frac{5}{3}-x=0\Leftrightarrow x=\frac{5}{3}\)

b)

\(|x-\frac{1}{10}|\geq 0, \forall x\Rightarrow 9=B+|x-\frac{1}{10}|\geq B+0=B\)

Vậy GTLN của $B$ là $9$ khi \(x-\frac{1}{10}=0\Leftrightarrow x=\frac{1}{10}\)

Bài 1: 

a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)

=>x+4/15=8/5 hoặc x+4/15=-8/5

=>x=4/3 hoặc x=-28/15

b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)

c: \(\Leftrightarrow\left|x-1\right|-1=1\)

=>|x-1|=2

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

Bài 2: 

b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)

Bài 3: 

a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)

Dấu '=' xảy ra khi x=-15/19

b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)

Dấu '=' xảy ra khi x=4/7