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

bai 1

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

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

\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=1,6\\x+\dfrac{4}{15}=-1,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)

Vậy ....

b) \(\left|\dfrac{5}{3}x\right|=\left|-\dfrac{1}{6}\right|\)

\(\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)

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

c) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)

\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)

\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{2}\)

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

bai 2

a) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\dfrac{1}{4}-\left|y\right|\)

\(\left|\dfrac{1}{6}+x\right|=\dfrac{1}{4}-\left|y\right|\) (*)

với mọi x ta luôn có \(\left|\dfrac{1}{6}+x\right|\ge0\)

\(\Rightarrow\dfrac{1}{4}-\left|y\right|\ge0\)

\(\Rightarrow\left|y\right|\le\dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{4}-\left|y\right|=\left|\dfrac{1}{4}-y\right|\)

Nên từ * \(\Rightarrow\left|\dfrac{1}{6}+x\right|=\left|\dfrac{1}{4}-y\right|\)

\(\Rightarrow\left|\dfrac{1}{6}+x\right|-\left|\dfrac{1}{4}-y\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{6}+x=0\\\dfrac{1}{4}-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{matrix}\right.\)

b) \(\left|x-y\right|+\left|y+25\right|=0\)

với mọi x, y tao luôn có \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+25\right|\ge0\end{matrix}\right.\)

mà \(\left|x-y\right|+\left|y+25\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+25\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\y=-25\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-25\\y=-25\end{matrix}\right.\)

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

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

bài 1)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{15}\right) \)
\(\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}+\dfrac{15}{28}-\dfrac{11}{15}\)
\(x=\dfrac{5}{42}-\dfrac{3541}{5460}=-\dfrac{413}{780}\)
b) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|=-\left|2,15\right|+\left|3,75\right|=1,6\)
\(\Rightarrow x+\dfrac{4}{15}=1,6\) hoặc \(x+\dfrac{4}{15}=-1,6\)
\(\Rightarrow x=\dfrac{4}{3}\) hoặc \(x=-\dfrac{28}{15}\)
c) \(\dfrac{5}{3}-\left|x-\dfrac{3}{2}\right|=-\dfrac{1}{2}\)
\(\Rightarrow\left|x-\dfrac{3}{2}\right|=\dfrac{5}{3}+\dfrac{1}{2}=\dfrac{13}{6}\)
\(\Rightarrow x-\dfrac{3}{2}=\dfrac{13}{6}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{11}{3}\) hoặc \(x=-\dfrac{2}{3}\)
d)\(\left(x-\dfrac{2}{3}\right).\left(2x-\dfrac{3}{2}\right)=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(2x-\dfrac{3}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)
3) a) \(\left(x^{^2}-4\right)^{^2}+\left(x+2\right)^{^2}=0\)
Vì \(\left(x^{^2}-4\right)^{^2}\ge0,\left(x+2\right)^{^2}\ge0\) nên :
\(\left\{{}\begin{matrix}x^{^2}-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=\pm2\)

b) \(\left(x-y\right)^{^2}+\left|y+2\right|=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^{^2}\ge0\\\left|y+2\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\end{matrix}\right.\Rightarrow x=-2;y=-2\)
c) \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
Vì \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow y=-\dfrac{9}{25};x=-\dfrac{9}{25}\)
d) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\left(-\dfrac{1}{4}\right)-\left|y\right|\)
\(\Rightarrow\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\)
Vì \(\left\{{}\begin{matrix}\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|\ge0\\\left|y\right|\ge0\end{matrix}\right.\) mà \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) nên không tồn tại x,y thỏa mãn đề bài .

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

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

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=\dfrac{8}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=-\dfrac{8}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\\ \Rightarrow\left|x+\dfrac{4}{15}\right|=1,6=\dfrac{8}{5}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=-\dfrac{8}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)

d)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)

=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)