Bài 6:

a: \(\left|x+\dfrac{1}{2}\right|>=0\forall x;\left|y-\dfrac{3}{4}\right|>=0\forall y;\left|z-1\right|>=0\forall z\)

Do đó: \(\left|x+\dfrac{1}{2}\right|+\left|y-\dfrac{3}{4}\right|+\left|z-1\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\y-\dfrac{3}{4}=0\\z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{3}{4}\\z=1\end{matrix}\right.\)

b: \(\left|x-\dfrac{3}{4}\right|>=0\forall x;\left|\dfrac{2}{5}-y\right|>=0\forall y;\left|x-y+z\right|>=0\forall x,y,z\)

Do đó: \(\left|x-\dfrac{3}{4}\right|+\left|\dfrac{2}{5}-y\right|+\left|x-y+z\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{3}{4}=0\\\dfrac{2}{5}-y=0\\x-y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{2}{5}\\z=-x+y=-\dfrac{3}{4}+\dfrac{2}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{2}{5}\\z=-\dfrac{7}{20}\end{matrix}\right.\)

c: \(\left|x-\dfrac{2}{3}\right|>=0\forall x;\left|x+y+\dfrac{3}{4}\right|>=0\forall x,y;\left|y-z-\dfrac{5}{6}\right|>=0\forall y,z\)

Do đó: \(\left|x-\dfrac{2}{3}\right|+\left|x+y+\dfrac{3}{4}\right|+\left|y-z-\dfrac{5}{6}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\x+y+\dfrac{3}{4}=0\\y-z-\dfrac{5}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-x-\dfrac{3}{4}=-\dfrac{2}{3}-\dfrac{3}{4}\\z=y-\dfrac{5}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{17}{12}\\z=-\dfrac{17}{12}-\dfrac{10}{12}=-\dfrac{27}{12}=-\dfrac{9}{4}\end{matrix}\right.\)

Bài 5:

a: \(\left|-\dfrac{3}{5}+\dfrac{1}{2}\right|-\left(\dfrac{3}{4}-\dfrac{5}{8}\right)+\left|-\dfrac{3}{2}\right|\)

\(=\left|-\dfrac{6}{10}+\dfrac{5}{10}\right|-\dfrac{1}{8}+\dfrac{3}{2}\)

\(=\dfrac{1}{10}-\dfrac{1}{8}+\dfrac{3}{2}=\dfrac{4}{40}-\dfrac{5}{40}+\dfrac{60}{40}=\dfrac{59}{40}\)

b: \(\dfrac{2}{3}-\left|-\dfrac{7}{3}+\dfrac{3}{4}\right|-\left|-\dfrac{5}{2}+1\right|\)

\(=\dfrac{2}{3}-\left|-\dfrac{28}{12}+\dfrac{9}{12}\right|-\left|-\dfrac{5}{2}+\dfrac{2}{2}\right|\)

\(=\dfrac{2}{3}-\dfrac{19}{12}-\dfrac{3}{2}=\dfrac{8}{12}-\dfrac{19}{12}-\dfrac{18}{12}\)

\(=-\dfrac{29}{12}\)

c: \(\dfrac{1}{5}-\left(\dfrac{3}{10}-\dfrac{-3}{5}\right)-\left|\dfrac{1}{4}-\dfrac{2}{5}\right|\)

\(=\dfrac{1}{5}-\dfrac{3}{10}-\dfrac{3}{5}-\left|\dfrac{5}{20}-\dfrac{8}{20}\right|\)

\(=-\dfrac{7}{10}-\left|\dfrac{-3}{20}\right|=-\dfrac{7}{10}-\dfrac{3}{20}=-\dfrac{17}{20}\)

d: \(\left|-\dfrac{5}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right|-\left(-\dfrac{3}{4}+\dfrac{-5}{3}\right)\)

\(=\left|-\dfrac{30}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right|+\dfrac{3}{4}+\dfrac{5}{3}\)

\(=\dfrac{25}{12}+\dfrac{9}{12}+\dfrac{20}{12}=\dfrac{54}{12}=\dfrac{9}{2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)

Để A là số nguyên thì \(\sqrt{x}+2⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3+5⋮\sqrt{x}-3\)

=>\(5⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\{1;-1;5;-5\right\}\)

=>\(\sqrt{x}\in\left\{4;2;8\right\}\)

=>\(x\in\left\{16;4;64\right\}\)

Thể tích phần bể chứa nước ban đầu là:

\(80\cdot50\cdot35=140000\left(cm^3\right)\)

Thể tích phần bể chứa nước lúc này sau khi thêm hòn đá là:

\(140000+20000=160000\left(cm^3\right)\)

Mực nước trong bể lúc này cao là:

\(160000:80:50=40\left(cm\right)\)

Thể tích ban đầu: 80 x 50 x 35 = 140.000 cm3

Sau khi thêm hòn đá: 140.000+20.000 =  160.000 cm3

=> Chiều cao mực nước = 160.000 / (80x50) = 40 cm

Hình đâu em?

Độ dài cạnh huyền là:

\(\sqrt{3^2+7^2}=\sqrt{9+49}=\sqrt{58}\left(cm\right)\)

Bình phương cạnh huyền là:

 3² + 7² =  58(cm²)

Cạnh huyền là:  \(\sqrt{58}\) m

1: \(\left(x^2+2xy-3\right)\left(-xy^2\right)\)

\(=-xy^2\cdot x^2-xy^2\cdot2xy+3\cdot xy^2\)

\(=-x^3y^2-2x^2y^3+3xy^2\)

2: \(3x\left(x+2\right)-3x^2-12=0\)

=>\(3x^2+6x-3x^2-12=0\)

=>6x-12=0

=>6x=12

=>x=2

3: \(\left(2x^3-\dfrac{9}{2}x^2+\dfrac{1}{xy}\right)\cdot x^2y^3\)

\(=2x^3\cdot x^2y^3-\dfrac{9}{2}x^2\cdot x^2y^3+\dfrac{x^2y^3}{xy}\)

\(=2x^5y^3-\dfrac{9}{2}x^4y^3+xy^2\)

2; 3\(x\)(\(x+2\)) - 3\(x^2\) - 12 = 0

    3\(x^2\) + 6\(x\) - 3\(x^2\) - 12 = 0

  (3\(x^2\) - 3\(x^2\)) + 6\(x\) - 12 = 0

   0 + 6\(x\) - 12 = 0

          6\(x\)  = 12

           \(x\) = 12 : 6

           \(x=2\) 

Vậy \(x=2\)

Lấy mẫu số, chứ không phải lấy tử số em nhé!

tất cả các số âm hay dương gì đều lấy mẫu số ạ?

a: \(-\dfrac{15}{19}=-1+\dfrac{4}{19}\)

\(-\dfrac{37}{41}=-1+\dfrac{4}{41}\)

\(-\dfrac{5}{9}=-1+\dfrac{4}{9}\)

\(\dfrac{23}{-27}=-\dfrac{23}{27}=-1+\dfrac{4}{27}\)

\(-\dfrac{7}{11}=-1+\dfrac{4}{11}\)

mà \(\dfrac{4}{41}< \dfrac{4}{27}< \dfrac{4}{19}< \dfrac{4}{11}< \dfrac{4}{9}\)

nên \(-\dfrac{37}{41}< -\dfrac{23}{27}< -\dfrac{15}{19}< -\dfrac{7}{11}< -\dfrac{5}{9}\)

mà \(-\dfrac{37}{41}< -\dfrac{76}{89}< -\dfrac{23}{27}\)

nên  \(-\dfrac{37}{41}< -\dfrac{76}{89}< -\dfrac{23}{27}< -\dfrac{15}{19}< -\dfrac{7}{11}< -\dfrac{5}{9}\)