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4 CÁI LÀ A,B,C,D ĐÓ BẠN

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\(=\frac{7}{5}-\frac{27}{12}+\frac{3}{5}+\frac{2}{3}+\frac{1}{3}-\frac{3}{4}\) 

\(=\frac{7}{5}+\frac{3}{5}+\frac{2}{3}+\frac{1}{3}-\frac{27}{12}-\frac{9}{12}\)          

\(=2+1-3=0\)        

\(\frac{14}{10}\)\(-\frac{27}{12}\)\(+\frac{30}{50}\)\(+\frac{10}{15}\)\(+\frac{16}{48}\)\(-\frac{75}{100}\)

\(=\frac{7}{5}\)\(+\frac{27}{12}\)\(+\frac{3}{5}\)\(+\frac{2}{3}\)\(+\frac{1}{3}\)\(-\frac{3}{4}\)

\(=\frac{7}{5}\)\(+\frac{3}{5}\)\(+\frac{2}{3}\)\(+\frac{1}{3}\)\(-\frac{27}{12}\)\(-\frac{9}{12}\)

\(=2+1-3\)

\(=0\)

a: \(12:\dfrac{3}{4}+75\%\cdot\dfrac{1}{2}-16\cdot50\%\)

\(=12\cdot\dfrac{4}{3}+\dfrac{3}{4}\cdot\dfrac{1}{2}-16\cdot0,5\)

\(=16-8+\dfrac{3}{8}\)

\(=8,375\)

b: \(75\%\cdot4+22,5:3-1\dfrac{3}{5}\)

\(=0,75\cdot4+7,5-1,6\)

\(=3+7,5-1,6\)

=10,5-1,6

=8,9

Bài 1 bạn tự là nhé dễ lắm :))

Bài 2 :

Ta có :

\(4x-15=-75-x\)
\(\Leftrightarrow\)\(4x+x=-75+15\)

\(\Leftrightarrow\)\(5x=-60\)

\(\Leftrightarrow\)\(x=\frac{-60}{5}\)

\(\Leftrightarrow\)\(x=-12\)

Bài 3 :

Ta có :

\(12⋮\left(x-3\right)\)\(\Rightarrow\)\(\left(x-3\right)\inƯ\left(12\right)\)

Mà \(Ư\left(12\right)=\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

Suy ra : ( lập bảng )

Vậy \(x\in\left\{4;2;5;1;6;0;7;-1;9;-3;15;-9\right\}\)

Chúc bạn học tốt 

\(\left\{{}\begin{matrix}x+y=1500\\x+\dfrac{75}{100}x+y+\dfrac{68}{100}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\left(x+y\right)+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\1500+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\dfrac{3}{4}x+\dfrac{17}{25}y=1083\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=900\\y=600\end{matrix}\right.\)

ĐKXĐ: \(x\ge-15\)

\(\sqrt{16\left(x-1\right)^2}=2x+\sqrt{12}\cdot\sqrt{75}\)

\(\Leftrightarrow\sqrt{16\left(x-1\right)^2}=2x+30\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-4=2x+30\\4x-4=-2x-30\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-\frac{13}{3}\end{matrix}\right.\)(Thỏa mãn)

Vậy phương trình có tập nghiệm \(S=\left\{17;-\frac{13}{3}\right\}\)

ĐKXĐ: \(x\ge-15\)

Ta có: \(\sqrt{16\left(x-1\right)^2}=2x+\sqrt{12}\cdot\sqrt{75}\)

\(\Leftrightarrow\sqrt{\left[4\left(x-1\right)\right]^2}=2x+30\)

\(\Leftrightarrow\left|4x-4\right|=2x+30\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-4=2x+30\\4x-4=-2x-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-4-2x-30=0\\4x-4+2x+30=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-34=0\\6x+26=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=34\\6x=-26\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\left(nhận\right)\\x=-\frac{26}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{17;\frac{-26}{6}\right\}\)

\(\sqrt{16\left(x-1\right)^2}=2x+\sqrt{12}.\sqrt{75}\)

\(\Leftrightarrow\)\(4\sqrt{\left(x-1\right)^2}=2x+\sqrt{12.75}\)

\(\Leftrightarrow4x-4=2x+\sqrt{900}\)

\(\Leftrightarrow4x-2x=4+30\) (chuyển vế đổi dấu)

\(\Leftrightarrow2x=34\)

\(\Leftrightarrow x=17\)

\(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{2\times8\times3}=2.\\ \dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{5\times2\times3\times5\times5\times7\times8}{3\times5\times5\times8\times2\times2\times3\times2\times7}=\dfrac{5}{12}.\)