a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)

<=> \(25x+10-80x+10=24x+12-30\)

<=> \(25x-80x-24x=12-30-10-10\)

<=> \(-79x=-38\)

<=> \(x=\dfrac{-38}{-79}\)

\(x=\dfrac{38}{79}\)

b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)

<=> \(30x-12x+30+5x+40=210+10x-10\)

<=> \(30x-12x+5x-10x=210-10-30-40\)

<=> \(13x=130\)

<=> \(x=\dfrac{130}{13}\)

\(x=10\)

c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)

<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)

<=> \(28x+28+60x+120+105x+420+2520=0\)

<=> \(28x+60x+105x=-28-120-420-2520\)

<=> \(193x=-3088\)

<=> \(x=\dfrac{-3088}{193}\)

\(x=-16\)

d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)

<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)

<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)

<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)

<=> \(22968x=8199576\)

<=> \(x=\dfrac{8199576}{22968}\)

\(x=357\)

Đề là giải PT nha các bn

A -\(\dfrac{24}{25}\)

B -\(\dfrac{5}{21}\)

C -\(\dfrac{24}{47}\)

D -\(\dfrac{19}{42}\)

tick cho mk

a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)

\(\Rightarrow-4< x< \dfrac{-3}{10}\)

\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)

\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)

b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)

\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)

\(\Rightarrow x=\varnothing\)

c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)

\(\Rightarrow x\in\left\{1;2\right\}\)

+) Với \(x=1\)

\(\Rightarrow y\in\left\{1;2\right\}\)

+) Với \(x=2\)

\(\Rightarrow y=2\)

Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)

\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)

\(D=1+-1+\dfrac{1}{41}\)

\(D=0+\dfrac{1}{41}\)

\(D=\dfrac{1}{41}\)

\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)

\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)

=1/57

\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)

giúp mk với

mai mk đi học rùi

Mình chỉ giải câu a thôi,mấy câu còn lại dễ.

a)Ta có:\(\dfrac{x}{27}=\dfrac{-3}{x}\)

=>\(x^2=-3\cdot27=-81\)(Nhân chéo)

Mà x²>0 với mọi x nên :

Không có giá trị nào thỏa mãn điều kiện của x

Tìm x biết :

a) \(\dfrac{x}{27}=-\dfrac{3}{x}\) \(\Rightarrow2x=-3.27\Rightarrow2x=-81\Rightarrow x=-40,5\)

b) \(-\dfrac{9}{x}=-\dfrac{x}{\dfrac{4}{49}}\Rightarrow2x=-9.\left(-\dfrac{4}{9}\right)\Rightarrow2x=4\Rightarrow x=2\)

c) \(\left|7x-\dfrac{5}{3}\right|+\dfrac{7}{19}=-\dfrac{8}{15}\) ( mk nghĩ bn chép sai đề bài câu này )

\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{8}{15}-\dfrac{7}{19}\)

\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{257}{285}\)

\(\Rightarrow\left[{}\begin{matrix}7x-\dfrac{5}{3}=-\dfrac{257}{285}\\7x-\dfrac{5}{3}=\dfrac{257}{285}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{218}{1995}\\x=\dfrac{244.}{665}\end{matrix}\right.\)

d) \(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=\dfrac{18}{19}-1\dfrac{2}{5}\)

\(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=-\dfrac{43}{95}\)

\(\left|\dfrac{1}{23}x\right|=-\dfrac{43}{95}-\dfrac{18}{90}\)

\(\left|\dfrac{1}{23}x\right|=-\dfrac{62}{95}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{23}x=\dfrac{62}{95}\\\dfrac{1}{23}x=-\dfrac{62}{95}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\dfrac{1}{95}\\x=-15\dfrac{1}{95}\end{matrix}\right.\)

a)

\(\dfrac{x}{27}=\dfrac{-3}{x}\\ \Leftrightarrow x^2=-81\\ \Leftrightarrow x\in\varnothing\)

b)

\(-\dfrac{9}{x}=\dfrac{-x}{\dfrac{4}{49}}\\ \Leftrightarrow-9.\dfrac{4}{49}=-x^2\\ \Leftrightarrow-x^2=\dfrac{-36}{49}\\ \Leftrightarrow x^2=\dfrac{36}{49}\\ \Leftrightarrow x=\pm\dfrac{6}{7}\)

c)

\(\left|7x-\dfrac{5}{3}\right|+\dfrac{7}{19}=-\dfrac{8}{15}\\ \Rightarrow\left|7x-\dfrac{5}{3}\right|=\dfrac{-257}{285}\\ \)

Mà \(\left|7x-\dfrac{5}{3}\right|\ge0\Rightarrow x\in\varnothing\)

d) \(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=\dfrac{18}{19}-1\dfrac{2}{5}\\ \Rightarrow\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=-\dfrac{43}{95}\\ \Rightarrow\left|\dfrac{1}{23}x\right|=-\dfrac{62}{95}\\ \Rightarrow x\in\varnothing\)