1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

a) 1\(\dfrac{2}{3}\).           b)\(\dfrac{1}{7}\).             c) 1               d )0

a: =>x+5>0

hay x>-5

b: =>2x+1<0

hay x<-1/2

c: =>(x-1)(x-4)>0

=>x>4 hoặc x<1

a) x>-5 ĐKXĐ x\(\ne\)-5

b)x<\(-\dfrac{1}{2}\)

c)x>4 hoặc x<1

d)ĐKXĐ x\(\ne\)1, ko tìm đc x

a) Ta có: \(A=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)

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

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

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

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

c) Để A>0 thì 2x-3<0

hay \(x< \dfrac{3}{2}\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x\ne1\end{matrix}\right.\)

Tham khảo:Cho biểu thức P= \((\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}):(3+\frac{2}{1-x})\) a) Rút gọn P b) Tính P với |3x-2|+1=5 c)... - Hoc24

a) ĐKXĐ: $x\notin \left\{1;\frac{3}{2}\right\}$

Ta có: $A=\left(\frac{2x}{2{}^{}-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)$

$=\left(\frac{2x}{\left(x-1\right)\left(2x-3\right)}-\frac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right):(\frac{3\left(x-1\right)\left(2x-3\right)}{\left(x-1\right)(2x-}$

a, \(A=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)ĐK : \(x\ne1;\dfrac{3}{2}\)

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

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

b, \(\left|3x-2\right|+1=5\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Với x = 2 \(A=\dfrac{6-5}{1.3}=\dfrac{1}{3}\)

Với x = -2/3 \(A=\dfrac{3\left(-\dfrac{2}{3}\right)-5}{\left(-\dfrac{2}{3}-3\right)\left(5+\dfrac{2}{3}\right)}=\dfrac{-2-5}{-\dfrac{11}{3}.\dfrac{17}{3}}=-\dfrac{7}{-\dfrac{187}{9}}=\dfrac{63}{187}\)

a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(TH_1:3x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(TH_2:-2x-7=0\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

b) \(2x^3-5x^2+3x=0\)

\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)

\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)

\(TH_1:x=0\)

\(TH_2:x-1=0\)

\(\Leftrightarrow x=1\)

\(TH_3:2x-3=0\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(3x+4\right)\left(2x-4\right)=0\)

\(TH_1:3x+4=0\)

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

\(TH_2:2x-4=0\)

\(\Leftrightarrow x=2\)

Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)

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

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Rightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x-9x=-6-16+12\)

\(\Leftrightarrow11x=-10\)

\(\Leftrightarrow x=-\dfrac{10}{11}\)

Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow3x+1=5x+8\)

\(\Leftrightarrow3x-5x=8-1\)

\(\Leftrightarrow-2x=7\)

\(\Leftrightarrow x=\dfrac{-7}{2}\)

Vậy \(X=\dfrac{-7}{2}\)

b) Ta có: \(2x^3-5x^2+3x=0\)

\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)

\(\Leftrightarrow x\left[\left(2x^2-2x\right)-\left(3x-3\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)

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

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

Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow9x^2-16-3x^2-4x=0\)

\(\Leftrightarrow6x^2-4x-16=0\)

\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)

\(\Leftrightarrow3x^2-6x+4x-8=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)

Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)

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

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Leftrightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x+16-12-9x+6=0\)

\(\Leftrightarrow11x+10=0\)

\(\Leftrightarrow x=\dfrac{-10}{11}\)

Vậy \(x=\dfrac{-10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

a, đk : x khác -2 ; 2 

\(\left(x+2\right)^2-8x=0\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)(ktm) 

pt vô nghiệm 

b, đk : x khác -1 ; 1 

\(x\left(x+1\right)-5x+3=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow x=1\left(ktm\right);x=3\left(tm\right)\)

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

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

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

a. có vấn đề

b.

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

\(\Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-30\)

\(\Leftrightarrow-22x+5x< -30-5\)

\(\Leftrightarrow-17x< -35\)

\(\Leftrightarrow x>\dfrac{35}{17}\)