\(a,\Leftrightarrow x-28=-45\\ \Leftrightarrow x=-27\\ b,\Leftrightarrow3+x=0\\ \Leftrightarrow x=-3\\ c,\Leftrightarrow\left[{}\begin{matrix}7-x=0\\-x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow16\left(x^2-4\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

a)-28+x=-34+(-11)             b)-12(3+x)=0

<=>-28+x=-45                   <=>-36-12x=0

<=>x=-17                            <=>-12x=36

Vậy x=-17                             <=>x=-3

                                         Vậy x=-3

c)(7-x)(-x+2)=0

<=>7-x=0 hoặc -x+2=0

Th1:7-x=0                    Th2:-x+2=0

 <=>x=7                             <=>x=2

           Vậy xϵ{7;2}

d)16x2-64=0

<=>32x=64

<=>x=2

Vậy x=2

`#3107.101107`

\(\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{11}\times\dfrac{1}{7}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{7}\times\dfrac{1}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)

a)\(x=-\frac{4}{7}\Rightarrow\left|x\right|=\left|-\frac{4}{7}\right|=\frac{4}{7}\)

b)\(x=\frac{-3}{-11}=\frac{3}{11}\Rightarrow\left|x\right|=\left|\frac{3}{11}\right|=\frac{3}{11}\)

c)\(x=-0,749\Rightarrow\left|x\right|=\left|-0,749\right|=0,749\)

d)\(x=-34\Rightarrow\left|x\right|=\left|-34\right|=34\)

\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\left(1-\frac{3}{100}\right)\)

\(=\frac{1}{4}\cdot\frac{4}{7}\cdot\frac{7}{10}\cdot\cdot\cdot\frac{97}{100}\)

\(=\frac{1.4.7.10...97}{4.7.10.13...100}\)

\(=\frac{1}{100}\)

Còn có cách khác không bạn

a: -28+x=-34+(-11)

=>x-28=-34-11=-45

=>x=-45+28

=>x=-17

b: -12(x+3)=0

=>x+3=0

=>x=0-3

=>x=-3

c: \(\left(7-x\right)\left(-x+2\right)=0\)

=>(x-7)(x-2)=0

=>\(\left[\begin{array}{l}x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=2\end{array}\right.\)

d: \(16x^2-64=0\)

=>\(16x^2=64\)

=>\(x^2=4\)

=>\(\left[\begin{array}{l}x=2\\ x=-2\end{array}\right.\)

\(x\in\left\{0;7;14;21;28;35;42;49\right\}\)

49

\(\left(6:3,5-1\dfrac{1}{6}\times\dfrac{6}{7}\right):\left(4,2\times\dfrac{10}{11}+5\dfrac{2}{11}\right)\\ =\left(\dfrac{12}{7}-\dfrac{7}{6}\times\dfrac{6}{7}\right):\left(\dfrac{21}{5}\times\dfrac{10}{11}+\dfrac{57}{11}\right)\\ =\left(\dfrac{17}{7}-1\right):\left(\dfrac{42}{11}+\dfrac{57}{11}\right)\\ =\dfrac{10}{7}:9\\ =\dfrac{10}{63}\)

\(\left(6:\dfrac{3}{5}-1\dfrac{1}{6}\times\dfrac{6}{7}\right):\left(4,2\times\dfrac{10}{11}+5\dfrac{2}{11}\right)\)

\(=\left(6\times\dfrac{5}{3}-\dfrac{7}{6}\times\dfrac{6}{7}\right):\left(\dfrac{21}{5}\times\dfrac{10}{11}+\dfrac{57}{11}\right)\)

\(=\left(10-1\right):\left(\dfrac{42}{11}+\dfrac{57}{11}\right)=9:9=1\)