Ta có : \(\dfrac{3-7x}{1+x}\ge\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{3-7x}{1+x}-\dfrac{1}{2}\ge0\)

\(\Leftrightarrow\dfrac{2\left(3-7x\right)-\left(x+1\right)}{2\left(x+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{5-15x}{2\left(x+1\right)}=\dfrac{5\left(3-x\right)}{2\left(x+1\right)}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-x\ge0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-x\le0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le3\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge3\\x< -1\end{matrix}\right.\end{matrix}\right.\)

Vậy suy ra tập nghiệm

b, (x+4)(5x+9)-x>4

\(\Leftrightarrow\)5x²+29x+36-x>4

\(\Leftrightarrow\)5x²+28x+36>4

\(\Leftrightarrow\)5x²+28x+32>0

\(\Leftrightarrow\)5(x²+\(\dfrac{28}{5}\)x+\(\dfrac{32}{5}\))>0

\(\Leftrightarrow\)x²+\(\dfrac{28}{5}\)x+\(\dfrac{32}{5}\)>0

\(\Leftrightarrow\)x²+2.\(\dfrac{14}{5}\)x+\(\dfrac{206}{25}\)+\(\dfrac{32}{5}\)-\(\dfrac{206}{25}\)>0

\(\Leftrightarrow\)(x+\(\dfrac{14}{5}\))²-\(\dfrac{46}{25}\)>0

\(\Leftrightarrow\)(x+\(\dfrac{14-\sqrt{46}}{5}\))(x+\(\dfrac{14+\sqrt{46}}{5}\))>0

\(\Leftrightarrow\)2 trường hợp

a: =>4x+12<=2x-1

=>2x<=-13

=>x<=-13/2

b: =>x^2-2x+1+4<0

=>(x-1)^2+4<0(loại)

c: =>(x-2+x+3)/(x+3)<0

=>(2x+1)/(x+3)<0

=>-3<x<-1/2

\(\text{a) }\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\\ \Leftrightarrow4\left(5x^2-3x\right)+5\left(3x+1\right)< 10x\left(2x+1\right)-15\\ \Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-15\\ \Leftrightarrow20x^2+3x-20x^2-10x< -15-5\\ \Leftrightarrow-7x< -20\\ \Leftrightarrow x>\dfrac{20}{7}\)

Vậy bất phương trình có nghiệm \(x>\dfrac{20}{7}\)

\(\text{b) }\dfrac{5x-20}{3}-\dfrac{2x^2+x}{2}\ge\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\\ \Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)\ge4x\left(1-3x\right)-15x\\ \Leftrightarrow20x-80-12x^2-6x\ge4x-12x^2-15x\\ \Leftrightarrow-12x^2+14x+12x^2+11x\ge80\\ \Leftrightarrow25x\ge80\\ \Leftrightarrow x\ge\dfrac{16}{5}\)

Vậy bất phương trình có nghiệm \(x\ge\dfrac{16}{5}\)

\(\text{c) }\left(x+3\right)^2\le x^2-7\\ \Leftrightarrow x^2+6x+9\le x^2-7\\ \Leftrightarrow x^2+6x-x^2\le-7-9\\ \Leftrightarrow6x\le-16\\ \Leftrightarrow x\le-\dfrac{8}{3}\)

Vậy bất phương trình có nghiệm \(x\le-\dfrac{8}{3}\)

a: \(\Leftrightarrow4\left(4x-2\right)+12\left(-x+3\right)< =3\left(1-5x\right)\)

=>16x-8-12x+36<=3-15x

=>4x+28<=3-15x

=>19x<=-25

hay x<=-25/19

b: \(\Leftrightarrow6\left(x+4\right)+30\left(-x-5\right)>=10\left(x+3\right)-15\left(x-2\right)\)

=>6x+24-30x-150<=10x+30-15x+30

=>-24x-126<=-5x+60

=>-19x<=186

hay x>=-186/19

\(a,\dfrac{4x-2}{3}-x+3\le\dfrac{1-5x}{4}\\ \Leftrightarrow\dfrac{4\left(4x-2\right)}{12}-\dfrac{12\left(x-3\right)}{12}\le\dfrac{3\left(1-5x\right)}{12}\\ \Leftrightarrow16x-8-12x+36\le3-15x\\ \Leftrightarrow4x+28\le3-15x\\ \Leftrightarrow19x+25\le0\\ \Leftrightarrow x\le-\dfrac{25}{19}\)

\(b,\dfrac{x+4}{5}-x-5\ge\dfrac{x+3}{3}-\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x+5\right)}{30}\ge\dfrac{10\left(x+3\right)}{30}-\dfrac{15\left(x-2\right)}{30}\\ \Leftrightarrow6x+24-30x-150\ge10x+30-15x+30\\ \Leftrightarrow-24x-126\ge-5x+60\\ \Leftrightarrow19x+186\le0\\ \Leftrightarrow x\le-\dfrac{186}{19}\)

Câu 1:

a) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)

\(\Leftrightarrow\dfrac{12x-2\left(5x+2\right)}{12}=\dfrac{3\left(7-3x\right)}{12}\)

\(\Leftrightarrow12x-10x-4=21-9x\)

\(\Leftrightarrow11x=25\)

\(\Leftrightarrow x=\dfrac{25}{11}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\Leftrightarrow x=\dfrac{1}{3}\\x-3=0\Leftrightarrow x=3\\7-2x=0\Leftrightarrow x=3,5\end{matrix}\right.\)

c) \(\left|3x\right|=4x+8\) (1)

Ta có: \(\left|3x\right|=3x\Leftrightarrow3x\ge0\Leftrightarrow x\ge0\)

\(\left|3x\right|=-3x\Leftrightarrow3x< 0\Leftrightarrow x< 0\)

Với \(x\ge0\), phương trình (1) có dạng:

\(3x=4x+8\Leftrightarrow-x=8\Leftrightarrow x=-8\)

(không thoả mãn điều kiện) \(\rightarrow\) loại

Với \(x< 0\), phương trình (1) có dạng:

\(-3x=4x+8\Leftrightarrow-7x=8\Leftrightarrow x=-\dfrac{8}{7}\)

(thoả mãn điều kiện) \(\rightarrow\) nhận

Vậy phương trình đã cho có 1 nghiệm \(x=-\dfrac{8}{7}\)

Câu 2:

\(2x\left(6x-1\right)\ge\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x\ge12x^2+9x-8x-6\)

\(\Leftrightarrow-3x\ge-6\)

\(\Leftrightarrow x\le2\)

Vậy bất phương trình đã cho có nghiệm \(x\le2\)

\(1.\)

\(a.\) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)

\(\Leftrightarrow\dfrac{24x}{24}-\dfrac{4\left(5x+2\right)}{24}=\dfrac{6\left(7-3x\right)}{24}\)

\(\Leftrightarrow24x-4\left(5x+2\right)=6\left(7-3x\right)\)

\(\Leftrightarrow24x-20x-8=42-18x\)

\(\Leftrightarrow24x-20x+18x=42+8\)

\(\Leftrightarrow22x=50\)

\(\Leftrightarrow x=\dfrac{50}{22}=\dfrac{25}{11}\)

Vậy : ...........

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

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

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

Vậy : ..............

\(c.\) \(\left|3x\right|=4x+8\) \(\left(1\right)\)

* Với \(3x< 0\Rightarrow x< 0\)

\(\left|3x\right|=-3x\)

Khi đó : \(\left(1\right)\Rightarrow-3x=4x+8\)

\(\Rightarrow-3x-4x=8\)

\(\Rightarrow-7x=8\)

\(\Rightarrow x=-\dfrac{8}{7}\) ( Thoả mãn điều kiện )

* Với \(3x\ge0\Rightarrow x\ge0\)

\(\left|3x\right|=3x\)

Khi đó : \(\left(1\right)\Rightarrow3x=4x+8\)

\(\Rightarrow3x-4x=8\)

\(\Rightarrow-x=8\)

\(\Rightarrow x=-8\) ( Không thoả mãn điều kiện )

Vậy : ..............

\(2.\)

\(2x\left(6x-1\right)\ge\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x\ge\left(12x^2+9x-8x-6\right)\)

\(\Leftrightarrow12x^2-2x-12x^2-9x+8x\ge-6\)

\(\Leftrightarrow-3x\ge-6\)

\(\Leftrightarrow x\le2\)

Vậy : ..............

Mk thấy mấy cái này dễ mà, toàn trong sách giáo khoa hết á. Bạn cố gắng đọc và lm đi. Sắp lên lớp 9 rồi đó

a)\(\dfrac{2x^2+10}{1-x}\le0\Rightarrow1-x< 0\Leftrightarrow x>1\)

b) \(\dfrac{3x-4}{x+2}\ge4\Leftrightarrow\dfrac{3x-4}{x+2}-\dfrac{4\left(x+2\right)}{x+2}\ge0\Leftrightarrow\dfrac{-x-12}{x+2}\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x-12\le0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-12\\x< -2\end{matrix}\right.\Leftrightarrow-12\le x< -2}}\\\left\{{}\begin{matrix}-x-12\ge0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-12\\x>-2\end{matrix}\right.\end{matrix}\right.\)\(S=\left\{x|-12\le x< -2\right\}\)

c) \(\dfrac{1}{x+4}\le\dfrac{1}{x-2}\Leftrightarrow\dfrac{6}{\left(x+4\right)\left(x-2\right)}\le0\Rightarrow\left(x+4\right)\left(x-2\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-4\\x< 2\end{matrix}\right.\Leftrightarrow-4< x< 2}}\\\left\{{}\begin{matrix}x+4< 0\\x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -4\\x>2\end{matrix}\right.\end{matrix}\right.\)

\(S=\left\{x|-4< x< 2\right\}\)

a.Ta có : \(\dfrac{x^2-4x+4}{x^3-2x^2-4x+8}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\)

Để \(\dfrac{1}{x+2}>0\) thì 1 và x+2 cùng dấu

mà 1>0

=>x + 2 > 0 <=> x > 2

\(\Rightarrow S=\left\{x|x>2\right\}\)

b, Ta có : \(x^2\ge0\Rightarrow x^2+1>0\)

Để \(\dfrac{7-8x}{x^2+1}>0\) thì 7 - 8x và \(x^2+1\) cùng dấu

mà \(x^2+1>0\Rightarrow7-8x>0\Leftrightarrow x< \dfrac{7}{8}\)

\(\Rightarrow S=\left\{x|x< \dfrac{7}{8}\right\}\)

c. Ta có bảng xét dấu:

Bổ xung câu c:

Vậy : \(-1< x\le\dfrac{-1}{2}\)

a) \(5x-3=7\)

\(\Leftrightarrow5x=7+3\)

\(\Leftrightarrow5x=10\)

\(\Leftrightarrow x=\dfrac{10}{5}\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

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

\(\Leftrightarrow x+3=0\) hoặc \(x-4=0\)

*) \(x+3=0\)

\(x=0-3\)

\(x=-3\)

*) \(x-4=0\)

\(x=0+4\)

\(x=4\)

Vậy \(S=\left\{-3;4\right\}\)

c) \(\left|x^2+2014\right|=1\)

\(\Leftrightarrow x^2+2014=1\) hoặc \(x^2+2014=-1\)

*) \(x^2+2014=1\)

\(\Leftrightarrow x^2=1-2014\)

\(\Leftrightarrow x^2=-2013\) (vô lý)

*) \(x^2+2014=-1\)

\(\Leftrightarrow x^2=-1-2014\)

\(\Leftrightarrow x^2=-2015\) (vô lý)

Vậy \(S=\varnothing\)

d) \(\dfrac{2}{x+1}-\dfrac{1}{x-3}=\dfrac{3x-11}{x^2-2x-3}\) (1)

ĐKXĐ: \(x\ne-1;x\ne3\)

\(\left(1\right)\Leftrightarrow2\left(x-3\right)-\left(x+1\right)=3x-11\)

\(\Leftrightarrow2x-6-x-1=3x-11\)

\(\Leftrightarrow-2x=-11+7\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\) (nhận)

Vậy \(S=\left\{2\right\}\)