Bài này bạn lần lượt thay x vào rồi tìm a thôi.

\(x=\dfrac{7}{3a-1}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{3a-1}=-1\\\dfrac{7}{3a-1}=7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3a-1=-7\\3a-1=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3a=-6\\3a=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=-2\\a=\dfrac{2}{3}\end{matrix}\right.\)

a: x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

A=1/7-x-x-3/5+4/5=-2x+12/35

b: B=|x-1/7|+|x+3/5|-1/3

x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

B=1/7-x+3/5+x-1/3=43/105

Bài 1 :

a, \(-1\dfrac{2}{3}\)= \(\dfrac{-5}{3}\)

Dựa vào tính chất của Tỉ lệ thức :

Ta có : \(\dfrac{x}{y}=\dfrac{-5}{3}\rightarrow\dfrac{x}{-5}=\dfrac{y}{3}\)

Dựa vào tính chất của dãy tỉ số = nhau

Ta có : \(\dfrac{x}{-5}=\dfrac{y}{3}=\dfrac{x+y}{\left(-5\right)+3}=\dfrac{18}{-2}=-9\)

\(\rightarrow\dfrac{x}{-5}=-9\rightarrow x=\left(-5\right).\left(-9\right)\Rightarrow x=45\\ \rightarrow\dfrac{y}{3}=-9\rightarrow y=3.\left(-9\right)\Rightarrow y=-27\)b,

Ta có :

( x + 4 ) . 7 = ( y + 7 ) . 4

\(\rightarrow\) 7x + 28 = 4y + 28

\(\rightarrow\) 7x = 4y

Vì 7x = 4y

\(\Rightarrow\) x = 22 / ( 4 + 7 ) . 7 = 14

\(\Rightarrow\) y = 22 - 14 = 8

Đợi mk lm câu 2 nha

hỏi huy dài lắm hôm qua mới nhắn xong ở đây lộ hết

a, \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)

\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{-2}{21}\)

\(\Rightarrow x=\dfrac{-2}{21}:\dfrac{4}{7}\)

\(\Rightarrow x=\dfrac{-1}{6}\)

b, \(25-\left(5-x\right)=-7\)

\(\Rightarrow\) \(5-x=25-\left(-7\right)\)

\(\Rightarrow5-x=32\)

\(\Rightarrow x=5-32\)

\(\Rightarrow x=-27\)

c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(\Rightarrow x=\dfrac{-5}{7}\)

d, \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\) \(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0:2\\x=0+\dfrac{1}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

e, \(\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|-7=-3\)

\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=-3+7\)

\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=4\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4+\dfrac{3}{4}\\\dfrac{1}{2}x=-4+\dfrac{3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{19}{4}\\\dfrac{1}{2}x=\dfrac{-13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{4}:\dfrac{1}{2}\\x=\dfrac{-13}{4}:\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=\dfrac{-13}{2}\end{matrix}\right.\)

a)\(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)

\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\dfrac{4}{7}x=\dfrac{-2}{21}\)

\(x=\dfrac{-2}{21}:\dfrac{4}{7}\)

\(x=\dfrac{-1}{6}\)

b)\(25-\left(5-x\right)=-7\)

\(5-x=25-\left(-7\right)\)

\(5-x=32\)

x= -27

c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{-5}{7}\)

d)\(2x\left(x-\dfrac{1}{7}\right)=0\)

⇒\(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

e)\(|\dfrac{1}{2}x-\dfrac{3}{7}|-7=-3\)

\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=-3+7\)

\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=4\)

⇒\(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4\dfrac{3}{4}\Rightarrow x=9\dfrac{1}{2}=\dfrac{19}{2}\\\dfrac{1}{2}x=-3\dfrac{1}{4}\Rightarrow x=\dfrac{-13}{2}\end{matrix}\right.\)

Nhiều quá, từng bài 1 nhé, bài nào làm được, tớ sẽ cố gắng.

bài 2:

a) \(x>2x\Leftrightarrow x-2x>0\Leftrightarrow-x>0\Leftrightarrow x< 0\)

Kl: x<0

b) \(a+x< a\Leftrightarrow x< 0\)

Kl: x<0

c) \(x^3>x^2\Leftrightarrow x^3-x^2>0\Leftrightarrow x^2\left(x-1\right)>0\) (*)

Mà x^2 > 0 \(\Rightarrow\) (*) \(\Leftrightarrow x-1>0\Leftrightarrow x>1\)

Kl: x>1

Câu 4:

a) \(1-2x< 7\Leftrightarrow2x>-6\Leftrightarrow x>3\)

Kl: x>3

b) \(\left(x-1\right)\left(x-2\right)>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< 2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)

Kl: x>2 hoặc x<1

c) \(\left(x-2\right)^2\left(x+1\right)\left(x+4\right)< 0\Leftrightarrow\left(x+1\right)\left(x+4\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x+4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x+4>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x< -4\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x>-4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-1< x< -4\left(vô-lý\right)\\-4< x< -1\end{matrix}\right.\) \(\Leftrightarrow-4< x< -1\)

Kl: -4<x<-1

d) ĐK: x khác 9\(\dfrac{x^2\left(x+3\right)}{x-9}< 0\Leftrightarrow x^2\left(x+3\right)\left(x-9\right)< 0\Leftrightarrow\left(x+3\right)\left(x-9\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\x-9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\x-9>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x>9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3< x< 9\left(N\right)\\9< x< -3\left(vô-lý\right)\end{matrix}\right.\) \(\Leftrightarrow-3< x< 9\)

Kl: -3<x<9

e) Đk: x khác 0

\(\dfrac{5}{x}< 1\Leftrightarrow\dfrac{5}{x}< \dfrac{5}{5}\Leftrightarrow x>5\left(N\right)\)

KL: x >5

f) ĐK: x khác 1

\(\dfrac{2x-5}{x-1}< 0\Leftrightarrow\left(2x-5\right)\left(x-1\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5>0\\x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5< 0\\x-1>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x>1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{2}< x< 1\left(vô-lý\right)\\1< x< \dfrac{5}{2}\left(N\right)\end{matrix}\right.\)

Kl: 1< x< 5/2

a) \(x+\dfrac{1}{3}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}-\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{12}\) vậy \(x=\dfrac{5}{12}\)

b) \(x-\dfrac{2}{5}=\dfrac{5}{7}\Leftrightarrow x=\dfrac{5}{7}+\dfrac{2}{5}\Leftrightarrow x=\dfrac{39}{35}\) vậy \(x=\dfrac{39}{35}\)

c) \(-x-\dfrac{2}{3}=\dfrac{-6}{7}\Leftrightarrow x=\dfrac{-2}{3}+\dfrac{6}{7}\Leftrightarrow x=\dfrac{4}{21}\) vậy \(x=\dfrac{4}{21}\)

d) \(\dfrac{4}{7}-x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{7}-\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{21}\) vậy \(x=\dfrac{5}{21}\)

a) x + \(\dfrac{1}{3}\) = \(\dfrac{3}{4}\)

x = \(\dfrac{3}{4}\) - \(\dfrac{1}{3}\)

x = \(\dfrac{5}{12}\)

Vậy x = \(\dfrac{5}{12}\)

b) x - \(\dfrac{2}{5}\) = \(\dfrac{5}{7}\)

x = \(\dfrac{5}{7}\) + \(\dfrac{2}{5}\)

x = \(\dfrac{39}{35}\)

Vậy x = \(\dfrac{39}{35}\)

c) -x - \(\dfrac{2}{3}\) = \(-\dfrac{6}{7}\)

- x = \(-\dfrac{6}{7}\) + \(\dfrac{2}{3}\)

- x = \(-\dfrac{4}{21}\)

⇒ x = \(\dfrac{4}{21}\)

Vậy x = \(\dfrac{4}{21}\)

d) \(\dfrac{4}{7}\) - x = \(\dfrac{1}{3}\)

x = \(\dfrac{4}{7}\) - \(\dfrac{1}{3}\)

x = \(\dfrac{5}{21}\)

Vậy x = \(\dfrac{5}{21}\)