Ôn tập: Bất phương trình bậc nhất một ẩn

a: \(\Leftrightarrow3x^2+x>3\left(x^2-4\right)\)

=>x>-12

b: \(\Leftrightarrow5x^2-x+20x-4>5x^2+16x+2\)

=>19x-4>16x+2

=>3x>6

hay x>2

a: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)

\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)

=>3x+5<10x-30

=>-7x<-35

hay x>5

b: \(\Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)

\(\Leftrightarrow20x-80-12x^2-6x>4x-12x^2-15x\)

=>14x-80>-11x

=>25x>80

hay x>16/5

a: \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\le\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

\(\Leftrightarrow2x-3+5x\left(x-2\right)\le5x^2-7\left(2x-3\right)\)

\(\Leftrightarrow2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

hay x<=4

b: \(\dfrac{6x+1}{18}+\dfrac{x+3}{12}>=\dfrac{5x+3}{6}+\dfrac{12-5x}{9}\)

=>2(6x+1)+3(x+3)>=6(5x+3)+4(12-5x)

=>12x+2+3x+9>=30x+18+48-20x

=>15x+11>=10x+66

=>5x>=55

hay x>=11

a,Vì \(x^2\ge0\Leftrightarrow-x^2\le0\left(\forall x\right)\)

Nên \(-x^2< 0\) với mọi x khác 0

b, Bấm máy tính fx 570 VN ra thôi

a)

\(\forall\) giá trị của x đều thỏa mãn

B)\(\left(x-2\right)\left(x-5\right)>0\)

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

=> PT vô nghiệm

a)\(\dfrac{x-2}{x+3}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\)

b) Vì 2>-5 =>x+2>x-5

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

câu b bạn có thể làm rõ hơn

a) ĐK: \(2x+2\ge0\Leftrightarrow x\ge-1\)

\(\left|2x+3\right|=2x+2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}0x=-1\left(vonghiem\right)\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}vonghiem\\x=\dfrac{-5}{4}\left(khongTMĐK\right)\end{matrix}\right.\)

vậy S=\(\varnothing\)

b)ĐK:\(5x-5\ge0\Leftrightarrow x\ge1\)

\(\left|5x-3\right|=5x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3=5x-5\\5x-3=5-5x\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}vonghiem\\x=0,8\left(KhongTMĐK\right)\end{matrix}\right.\)

Vậy S=\(\varnothing\)

a) \(\dfrac{2x-1}{x+3}>1\left(ĐKXĐ:x+3\ne0< =>x\ne-3\right)\)

\(< =>\dfrac{2x-1}{x+3}>\dfrac{x+3}{x+3}\\ < =>2x-1>x+3\\ < =>2x-x>3+1\\ < =>x>4\)

Vậy: Tập nghiệm của bất phương trình là S=\(\left\{x|x>4\right\}\)

b) \(\dfrac{2x-1}{x-2}< 3\left(ĐKXĐ:x-2\ne0< =>x\ne2\right)\)

\(< =>\dfrac{2x-1}{x-2}< \dfrac{3x-6}{x-2}\\ < =>2x-1< 3x-6\\ < =>2x-3x< -6+2\\ < =>-x< -4\\ < =>x>4\)

Vậy tập nghiệm của bất phương trình là S= \(\left\{x|x>4\right\}\)