`#3107.101107`

`(x + 1)^2 = 121`

`=> (x + 1)^2 = (+-11)^2`

\(\Rightarrow\left[{}\begin{matrix}x+1=11\\x+1=-11\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=11-1\\x=-11-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=10\\x=-12\end{matrix}\right.\)

Vậy, `x \in {10; -12}.`

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`10.`

`a)`

\(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=\left(\dfrac{7}{5}+\dfrac{7}{10}x\right)\)

\(\Rightarrow\dfrac{x}{2}-\left(\dfrac{3x-13}{5}\right)=\dfrac{7}{5}+\dfrac{7}{10}x\)

\(\Rightarrow\dfrac{x}{2}-\dfrac{3x-13}{5}-\dfrac{7}{10}x=\dfrac{7}{5}\\ \Rightarrow\dfrac{5x}{10}-\dfrac{2\left(3x-13\right)}{10}-\dfrac{7x}{10}=\dfrac{14}{10}\\ \Rightarrow5x-6x+26-7x=14\\ \Rightarrow-8x=14-26\\ \Rightarrow-8x=-12\\ \Rightarrow x=\dfrac{3}{2}\)

Vậy, `x = 3/2`

`b)`

\(\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=\dfrac{3}{2}-\dfrac{1}{3}\\ \Rightarrow\dfrac{2\left(2x-3\right)}{6}-\dfrac{5-3x}{6}=\dfrac{7}{6}\\ \Rightarrow4x-6-5+3x=7\\ \Rightarrow7x=7+5+6\\ \Rightarrow7x=18\\ \Rightarrow x=\dfrac{18}{7}\)

Vậy, `x = 18/7`

`c)`

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

\(\Rightarrow\dfrac{1}{x-1}-\dfrac{5}{2-2x}=\dfrac{2}{3}\cdot\left(-\dfrac{9}{20}\right)\\ \Rightarrow\dfrac{2}{2\left(x-1\right)}+\dfrac{5}{2\left(x-1\right)}=\dfrac{3}{10}\\ \Rightarrow\dfrac{2+5}{2\left(x-1\right)}=\dfrac{3}{10}\\ \Rightarrow\dfrac{7}{2\left(x-1\right)}=\dfrac{3}{10}\\ \Rightarrow6\left(x-1\right)=70\\ \Rightarrow6x-6=70\\ \Rightarrow6x=76\\ \Rightarrow x=\dfrac{38}{3}\)

Vậy, `x = 38/3`

`d)`

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

\(\Rightarrow\left[{}\begin{matrix}3\div x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3\div x=1\\-\dfrac{1}{2}x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=10\end{matrix}\right.\)

Vậy, `x \in {3; 10}`

`e)`

\(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}+\dfrac{-3}{7}\div x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{4}{9}=0\\\dfrac{1}{2}+\dfrac{-3}{7}\div x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\-\dfrac{3}{7}\div x=-\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{6}{7}\end{matrix}\right.\)

Vậy, `x \in {2/3; 6/7}.`

`#3107.101107`

`9.`

`a)`

\(2\dfrac{2}{3}\div x-0,98=0,02\\ \Rightarrow\dfrac{8}{3}\div x=0,02+0,98\\ \Rightarrow\dfrac{8}{3}\div x=1\\ \Rightarrow x=\dfrac{8}{3}\div1\\ \Rightarrow x=\dfrac{8}{3}\)

Vậy, \(x=\dfrac{8}{3}\)

`b)`

\(\dfrac{8}{23}\cdot\dfrac{46}{24}-x=\dfrac{1}{3}\\ \Rightarrow\dfrac{1}{2}\cdot\dfrac{2}{3}-x=\dfrac{1}{3}\\ \Rightarrow\dfrac{1}{3}-x=\dfrac{1}{3}\\ \Rightarrow x=0\)

Vậy, `x = 0`

`c)`

\(0,125-3\dfrac{2}{7}x=2\dfrac{3}{4}\\ \Rightarrow3\dfrac{2}{7}x=0,125-2\dfrac{3}{4}\\ \Rightarrow3\dfrac{2}{7}x=-\dfrac{489}{4}\\ \Rightarrow x=-\dfrac{3423}{92}\)

`d)`

\(\dfrac{3}{4}\cdot\left(x-8\right)=\dfrac{5}{7}\cdot\left(14-\dfrac{7}{2}\right)\\ \Rightarrow\dfrac{3}{4}\left(x-8\right)=\dfrac{5}{7}\cdot\dfrac{21}{2}\\ \Rightarrow\dfrac{3}{4}\left(x-8\right)=\dfrac{15}{2}\\ \Rightarrow x-8=\dfrac{15}{2}\div\dfrac{3}{4}\\ \Rightarrow x-8=10\\ \Rightarrow x=18\)

Vậy, `x = 18`

`e)`

\(30\%\cdot x-x+\dfrac{5}{6}=\dfrac{1}{3}\\ \Rightarrow x\left(0,03-1\right)=\dfrac{1}{3}-\dfrac{5}{6}\\ \Rightarrow0,97x=-\dfrac{1}{2}\\ \Rightarrow x=-\dfrac{1}{2}\div0,97\\ \Rightarrow x=-\dfrac{50}{97}\)

Vậy, `x = -50/97`

`f)`

\(1-\left(\dfrac{43}{8}+x-\dfrac{173}{24}\right)\div\dfrac{50}{3}=0\\ \Rightarrow\left(x-\dfrac{11}{6}\right)\div\dfrac{50}{3}=1\\ \Rightarrow x-\dfrac{11}{6}=1\cdot\dfrac{50}{3}\\ \Rightarrow x-\dfrac{11}{6}=\dfrac{50}{3}\\ \Rightarrow x=\dfrac{37}{2}\)

Vậy, `x = 37/2.`

`#3107.101107`

`a)`

\(\left(2x-3\right)\left(\dfrac{3}{4}x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\\dfrac{3}{4}x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy, `x \in {3/2; -4/3}`

`b)`

\(\dfrac{3}{7}x+2\dfrac{3}{5}=1\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{7}x=\dfrac{7}{5}-\dfrac{13}{5}\\ \Rightarrow\dfrac{3}{7}x=-\dfrac{6}{5}\\ \Rightarrow x=-\dfrac{14}{5}\)

Vậy, `x = -14/5`

`c)`

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

Vậy, `x \in {1/5; 1/6}`

`d)`

\(\dfrac{3}{7}+\dfrac{1}{7}\div x=\dfrac{3}{14}\)

$\Rightarrow $ `3/7 + 1/7 * 1/x = 3/14`

$\Rightarrow $ `1/(7x) = 3/14 - 3/7`

$\Rightarrow $ `1/(7x) = -3/14`

$\Rightarrow $ `-3*7x = 14`

$\Rightarrow $ `-21x = 14`

$\Rightarrow $ `x = -2/3`

Vậy, `x = -2/3.`

Đó đâu phải điều kiện mà pt cần phải tuân thủ, quan trọng là bạn làm thế nào cho phù hợp với cấp độ câu hỏi nằm ở đâu. CH này là của lớp 8, sử dụng đúng kiến thức phù hợp với khối lớp chứ tự nhiên nó ở đâu vào, liên quan đến chủ đề nào thì sử dụng chủ đề đấy thôi. Trừ khi pt vô nghiệm, hầu hết đều có cách giải quyết riêng của nó.

`#3107.101107`

`1/3(6xy^2)^2*(-5/4x^4y^3)`

`=(-5/4*1/3)*[(6xy^2)^2*x^4y^3]`

`= -5/12*(36x^2y^4*x^4y^3)`

`=-5/12*36*x^6y^5`

`=-15x^6y^5.`