a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)

\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)

\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)

a,

\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)

b,

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

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

c,

\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)

\(\Rightarrow5\left|x+1\right|^2=180\)

\(\Rightarrow\left|x+1\right|^2=36\)

Mà \(\left|x+1\right|\ge0\)

=> x + 1 = 6 <=> x = 7

Bài 1:

\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)

\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)

\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)

\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)

\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)

\(=\dfrac{168}{89}\)

\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{4}{3}-\dfrac{5}{4}x+\dfrac{5}{4}=\dfrac{15}{2}-\dfrac{3}{2}x-\dfrac{3}{2}\left(2x+3\right)\)

\(\Leftrightarrow x\cdot\dfrac{-7}{12}+\dfrac{31}{12}=\dfrac{-15}{2}x+3\)

=>83/12x=5/12

hay x=5/83

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

\(a,2\left(5x+1\right)-7\left(3x-2\right)=4\left(2x-1\right)+3\left(2-x\right)\)

\(\Leftrightarrow10x+2-21x+14=8x-4+6-3x\)

\(\Leftrightarrow-16x=-14\)

\(\Rightarrow x=\dfrac{7}{8}\)

\(b,-4\left(\dfrac{1}{2}x-3\right)+\dfrac{7}{2}\left(2x-1\right)+x=5x\left(1-x\right)\)

\(\Leftrightarrow-2x+12+7x-\dfrac{7}{2}+x=5x-5x^2\)

\(\Leftrightarrow5x^2+x+\dfrac{17}{2}=0\)

Cái này không biết tách kiểu gì cho vừa nên bạn nhấn máy tính nhé

Mode 5 3 rồi lần lượt điền vào theo thứ tự trên thì

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}+\dfrac{13i}{10}\\x=-\dfrac{1}{10}-\dfrac{13i}{10}\end{matrix}\right.\)

an thế nào hả bạn mk ko có bt an hộ mk đi

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a)$\left(\genfrac{}{}{0ex}{}{1}{2}+1,5\right)\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$2\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$x=\genfrac{}{}{0ex}{}{1}{5}:2$

$x=\genfrac{}{}{0ex}{}{1}{10}$
b) $\left(-1\genfrac{}{}{0ex}{}{3}{5}+x\right):\genfrac{}{}{0ex}{}{12}{13}=2\genfrac{}{}{0ex}{}{1}{6}$

$-1\genfrac{}{}{0ex}{}{3}{5}+x=\genfrac{}{}{0ex}{}{13}{6}\cdot \genfrac{}{}{0ex}{}{12}{13}$
$x=2+1\genfrac{}{}{0ex}{}{3}{5}$

$x=3\genfrac{}{}{0ex}{}{3}{5}$
c) $\left(x:2\genfrac{}{}{0ex}{}{1}{3}\right)\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x\cdot \genfrac{}{}{0ex}{}{3}{7}\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x=\genfrac{}{}{0ex}{}{-3}{8}:\genfrac{}{}{0ex}{}{3}{49}$
$x=\genfrac{}{}{0ex}{}{-49}{8}=-6\genfrac{}{}{0ex}{}{1}{8}$
d) $\genfrac{}{}{0ex}{}{-4}{7}\cdot x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}:\left(-1\genfrac{}{}{0ex}{}{2}{3}\right)$

$\genfrac{}{}{0ex}{}{-4}{7}x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{-3}{5}$
$-\genfrac{}{}{0ex}{}{4}{7}x=\genfrac{}{}{0ex}{}{-3}{40}-\genfrac{}{}{0ex}{}{7}{5}x=\genfrac{}{}{0ex}{}{-59}{40}:\genfrac{}{}{0ex}{}{-4}{7}=\genfrac{}{}{0ex}{}{413}{160}=2\genfrac{}{}{0ex}{}{93}{160}$
 

b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}