Question

Câu 1: Giải các bất phương trình

1) \(4x^2-4x+1>9\)

2) \(\frac{x^2-3x+2}{3x^2+5x-8}< 0\)

Câu 2 : Tìm x để biểu thức sau có giá trị âm

\(A=\left(\frac{1-x}{x+3}-\frac{x+3}{x-1}\right):\left(\frac{x+3}{x-1}-\frac{x-1}{x+3}\right)\)

Answer 1

Em tranh câu 2 trước nha!:v Ko biết em có tính nhầm chỗ nào không đây:)

\(ĐK:x\ne-3;x\ne1;\frac{x+3}{x-1}\ne\frac{x-1}{x+3}\)

\(\Leftrightarrow x\ne-3;x\ne\pm1\)

Ta có \(A=\frac{\frac{1-x}{x+3}-\frac{x+3}{x-1}}{\frac{x+3}{x-1}-\frac{x-1}{x+3}}=\frac{\frac{1-x}{x+3}-\frac{x+3}{x-1}}{\frac{1-x}{x+3}+\frac{x+3}{x-1}}\)

\(=\frac{\left(1-x\right)\left(x-1\right)-\left(x+3\right)^2}{\left(1-x\right)\left(x-1\right)+\left(x+3\right)^2}\)

\(=\frac{-2\left(x^2+2x+5\right)}{8\left(x+1\right)}=-\frac{1}{4}\left(\frac{x^2+2x+5}{x+1}\right)\)

Để A nhận giá trị âm thì \(\frac{x^2+2x+5}{x+1}>0\Leftrightarrow x+1>0\left(\text{vì }x^2+2x+5>0\right)\Leftrightarrow x>-1\)

Vậy...

Answer 2

2) \(\frac{x^2-3x+2}{3x^2+5x-8}=\frac{x^2-x^2-2x+2}{3x^2-3x+8x-8}=\frac{\left(x-1\right)\left(x-2\right)}{\left(3x+8\right)\left(x-1\right)}=\frac{x-2}{3x+8}\)

(Với ĐK là \(x\ne-\frac{8}{3};x\ne1\))

Khi đó: \(\frac{x-2}{3x+8}< 0\) ⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\3x+8< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\3x+8>0\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x< -\frac{8}{3}\end{matrix}\right.\left(l\right)\\\left\{{}\begin{matrix}x< 2\\x>-\frac{8}{3}\end{matrix}\right.\end{matrix}\right.\)

Vậy \(-\frac{8}{3}< x< 2\) và \(x\ne1\)

Answer 3

@Vũ Huy Hoàng help me ple

Answer 4

@svtkvtm

Answer 5

Bài 1:

a) \(4x^2-4x+1>9\Leftrightarrow4x^2-4x-8>0\)

\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)>0\)

\(\Leftrightarrow x< -1\text{hoặc }x>2\)

b) TÍ làm sai đi, em bận rồi

Answer 6

\(4x^2-4x+1>9\Leftrightarrow\left(2x-1\right)^2>9\Leftrightarrow\left[{}\begin{matrix}2x-1>3\\2x-1< -3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>4\\2x< -2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)

Answer 7

\(\frac{x^2-3x+2}{3x^2+5x-8}< 0\Leftrightarrow\left\{{}\begin{matrix}x^2-3x+2>0\\3x^2+5x-8< 0\end{matrix}\right.hoac\left\{{}\begin{matrix}x^2-3x+2< 0\\3x^2+5x-8>0\end{matrix}\right.\)

\(+,\left\{{}\begin{matrix}x^2-3x+2>0\\3x^2+5x-8< 0\end{matrix}\right.;x^2-3x+2>0\Leftrightarrow\left(x+1,5\right)^2>0,25=\left(\pm0,5\right)^2\Leftrightarrow\left[{}\begin{matrix}x+1,5>0,5\\x+1,5< -0,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-1\\x< -2\end{matrix}\right.;3x^2+5x-8< 0\Leftrightarrow x^2+\frac{5}{3}x-\frac{8}{3}< 0\Leftrightarrow\left(x+\frac{5}{6}\right)^2-\frac{121}{36}< 0\Leftrightarrow\left(x+\frac{5}{6}\right)^2< \frac{121}{36}\Leftrightarrow\frac{-11}{6}< x+\frac{5}{6}< \frac{11}{6}\Leftrightarrow\frac{-8}{3}< x< 1;x< -2\Rightarrow\frac{-8}{3}< x< -2;x>-1\Rightarrow-1< x< 1\)

tưong tự TH con lại