giải bpt:\(\frac{4x+3}{2x+5}< 2\)
Những câu hỏi liên quan
Giải bpt:
a)x\(^2\)-4x+3≤0
b)9x\(^2\)-6x≥0
c)\(\frac{4x-3}{3}-\frac{2x+1}{4}< \frac{x-5}{-2}\)
a. \(x^2-4x+3\le0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(3x-3\right)\le0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\le0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\ge0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le1\\x\ge3\end{matrix}\right.\left(Vo.li\right)\\\left\{{}\begin{matrix}x\ge1\\x\le3\end{matrix}\right.\end{matrix}\right.\)
Vậy \(1\le x\le3\)
b. \(9x^2-6x\ge0\)
\(\Leftrightarrow3x\left(3x-2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\ge0\\3x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x\le0\\3x-2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge\frac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\le\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(0\le x\le\frac{2}{3}\)
c. Câu c cậu tự làm nha, tớ đang có việc. Quy đồng lên rồi tính bình thường thôi.
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giải pt và bpt sau
a, 2x(x-3)=x-3 b,\(\frac{x+2}{x-2}-\frac{5}{x}=\frac{8}{x^2-2x}\)
c,\(\frac{2x+1}{4}-\frac{x-5}{3}< \frac{4x-1}{12}+12\)
a,\(2x\left(x-3\right)=x-3.\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy .....
b, \(\frac{x+2}{x-2}-\frac{5}{x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{\left(x+2\right)\cdot x}{\left(x-2\right)\cdot x}-\frac{5\left(x-2\right)}{x\left(x-2\right)}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{x^2+2x-\left(5x-10\right)}{\left(x-2\right)x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{x^2+2x-5x+10}{x^2-2x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow x^2+2x-5x+10=8\)
\(\Leftrightarrow x^2-3x+10-8=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy ....
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\(\frac{2x+1}{4}-\frac{x-5}{3}< \frac{4x-1}{12}+12.\)
\(\Leftrightarrow\frac{\left(2x+1\right)\cdot3}{4\cdot3}-\frac{\left(x-5\right)\cdot4}{3\cdot4}< \frac{4x-1}{12}+12.\)
\(\Leftrightarrow\frac{6x+3}{12}-\frac{4x-20}{12}< \frac{4x-1}{12}+12\)
\(\Leftrightarrow\frac{6x+3-4x+20}{12}< \frac{4x-1}{12}+12\)
\(\Leftrightarrow\frac{2x+23}{12}< \frac{4x-1}{12}+12\)
\(\Leftrightarrow\frac{2x+23-4x+1}{12}< 12\)
\(\Leftrightarrow\frac{-2x+24}{12}< 12\)
\(\Leftrightarrow-2x+24< 144\)
\(\Leftrightarrow-2x< 120\)
\(\Leftrightarrow x< -60\)
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Giải các BPT sau
a) \(\dfrac{3-2x}{5}\)-\(\dfrac{4x+1}{3}\)<\(\dfrac{-2+x}{2}\)-\(\dfrac{1}{4}\)
b) (x+2)2-(5+x)2 < hoặc = -2(4x+5)
Giải BPT
\(\dfrac{x^5-2x^4+2x^3-4x^2-3x+6}{x^2+2x-8}\le0\)
\(\dfrac{x^5-2x^4+2x^3-4x^2-3x+6}{x^2+2x-8}\le0\)
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Xem thêm câu trả lời
Giải các bpt sau
a, \(\left(2x-3\right)\left(3x-4\right)\left(5x+2\right)>0\)
b, \(25-16x^2>8x^2-10x\)
c, \(\frac{4x\left(3x+2\right)}{2x+5}>0\)
d, \(\frac{2x-5}{3x+2}\le\frac{3x+2}{2x-5}\)
a/ \(\left(2x-3\right)\left(3x-4\right)\left(5x+2\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}-\frac{2}{3}< x< \frac{4}{3}\\x>\frac{3}{2}\end{matrix}\right.\)
b/ \(\Leftrightarrow24x^2-10x-25< 0\)
\(\Rightarrow-\frac{5}{6}< x< \frac{5}{4}\)
c/ \(\frac{4x\left(3x+2\right)}{2x+5}>0\Rightarrow\left[{}\begin{matrix}-\frac{5}{2}< x< -\frac{2}{3}\\x>0\end{matrix}\right.\)
d/ \(\Leftrightarrow\frac{3x+2}{2x-5}-\frac{2x-5}{3x+2}\ge0\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2-\left(2x-5\right)^2}{\left(2x-5\right)\left(3x+2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(5x-2\right)\left(x+7\right)}{\left(2x-5\right)\left(3x+2\right)}\ge0\Rightarrow\left[{}\begin{matrix}x\le-7\\-\frac{2}{3}< x\le\frac{2}{5}\\x>\frac{5}{2}\end{matrix}\right.\)
Giải BPT sau
-4(2x-1)2-5(72-8) < hoặc = (4x-2)
Giải bpt sau
a, \(( 2 x − 3 ) ( 3 x − 4 ) ( 5 x + 2 ) > 0 \)
b, \(25 − 16 x 2 > 8 x 2 − 10 x \)
c, \(\frac{4x\left(3x+2\right)}{2x+5}>0\)
d, \(\frac{2x-5}{3x+2}\le\frac{3x+2}{2x-5}\)
Dấu của nhị thức bậc nhất nhé
Giải bpt
\(\frac{\left|x^2-4x\right|+3}{x^2+\left|x-5\right|}\ge1\)
giải BPT sau
a,(4x-1)(x^2+12)(-x+4)>0
b,(2x-1)(5-2x)(1-x)<0
\(a,\left(4x-1\right)\left(x^2+12\right)\left(-x+4\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+12>0\left(LD\forall x\right)\\-x+4>0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x>1\\-x>-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{4}\\x< 4\end{matrix}\right.\)
Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)
\(b,\left(2x-1\right)\left(5-2x\right)\left(1-x\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1< 0\\5-2x< 0\\1-x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\)
Vậy \(S=\left\{x|1>x>\dfrac{5}{2}\right\}\)
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