Giai các bpt sau:
1) \(^{x^2\le|1-\frac{2}{x^2}|}\)
2) \(\frac{|x^2-4x|+3}{x^2+|x-5|}\ge1\)
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\(\frac{1}{x+2}-\frac{x+2}{3x-5}\ge0\)
\(\Leftrightarrow\frac{-x^2-x-9}{\left(x+2\right)\left(3x-5\right)}\ge0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-5\right)< 0\) (do \(-x^2-x-9< 0;\forall x\))
\(\Rightarrow-2< x< \frac{5}{3}\)
2/ ĐKXĐ: \(1\le x\le3\)
\(\Leftrightarrow-x^2+4x-3\le\left(x-1\right)^2\)
\(\Leftrightarrow2x^2-6x+4\ge0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le1\end{matrix}\right.\)
Kết hợp ĐKXĐ: \(\left[{}\begin{matrix}x=1\\2\le x\le3\end{matrix}\right.\)
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
\(a,\dfrac{4x-2}{3}-x+3\le\dfrac{1-5x}{4}\\ \Leftrightarrow\dfrac{4\left(4x-2\right)}{12}-\dfrac{12\left(x-3\right)}{12}\le\dfrac{3\left(1-5x\right)}{12}\\ \Leftrightarrow16x-8-12x+36\le3-15x\\ \Leftrightarrow4x+28\le3-15x\\ \Leftrightarrow19x+25\le0\\ \Leftrightarrow x\le-\dfrac{25}{19}\)
\(b,\dfrac{x+4}{5}-x-5\ge\dfrac{x+3}{3}-\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x+5\right)}{30}\ge\dfrac{10\left(x+3\right)}{30}-\dfrac{15\left(x-2\right)}{30}\\ \Leftrightarrow6x+24-30x-150\ge10x+30-15x+30\\ \Leftrightarrow-24x-126\ge-5x+60\\ \Leftrightarrow19x+186\le0\\ \Leftrightarrow x\le-\dfrac{186}{19}\)
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.
3: \(P=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(y+z\right)+\left(y+x\right)}+\dfrac{z}{\left(z+x\right)+\left(z+y\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{y+z}+\dfrac{y}{y+x}\right)+\dfrac{1}{4}\left(\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)=\dfrac{3}{2}\).
Đẳng thức xảy ra khi x = y = x = \(\dfrac{1}{3}\).
\(3x-\frac{x+2}{3}\le\frac{3\left(x-2\right)}{2}+5-x\)
\(\Leftrightarrow\frac{18x}{6}-\frac{2\left(x+2\right)}{6}\le\frac{9\left(x-2\right)}{6}+\frac{30}{6}-\frac{6x}{6}\)
\(\Rightarrow18x-2x-4\le9x-18+30-6x\)
\(\Leftrightarrow16x-4\le3x+12\)
\(\Leftrightarrow13x\le16\)
\(\Leftrightarrow x\le\frac{16}{13}\)
Vậy bất phương trình có tập nghiệm là: \(S=\left\{x|x\le\frac{16}{13}\right\}\)
1) ta có: \(x^2\le\left|1-\frac{2}{x^2}\right|\) ( *)
+ nếu \(x^2\ge2\)
từ (*) \(\Rightarrow x^2\le1-\frac{2}{x^2}\)
\(\Leftrightarrow x^2-1+\frac{2}{x^2}\le0\)
\(\Rightarrow x^4-x^2+2\le0\) (vì \(x^2\ge0\))
\(\Leftrightarrow\left(x^2-\frac{1}{4}\right)^2+\frac{7}{4}\le0\) ( vô lý )
+ nếu \(x^2\le2\)
tứ (*) \(\Rightarrow x^2\le\frac{2}{x^2}-1\)
\(\Leftrightarrow x^2-\frac{2}{x^2}+1\le0\)
\(\Rightarrow x^4-2+x^2\le0\) (vì \(x^2\ge0\))
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+2\right)\le0\)
\(\Leftrightarrow x^2-1\le0\) ( vì \(x^2+2\)> 0 )
\(\Leftrightarrow x^2\le1\)
\(\Leftrightarrow-1\le x\le1\)
Vậy: \(-1\le x\le1\)
Ta có : \(\frac{\left|x^2-4x\right|+3}{x^2+\left|x-5\right|}\ge1\)
\(\Leftrightarrow\left|x^2-4x\right|+3\ge x^2+\left|x-5\right|\)
\(\Leftrightarrow\left|x^2-4x\right|+3-x^2-\left|x-5\right|\ge0\) (1)
+ nếu x= 0. từ pt (1) => 3 \(\ge\)0 ( đúng )
+ nếu x < 4 và x \(\ne\)0.
từ pt (1) => 4x - x2 + 3 - x2 - ( 5 - x ) \(\ge\)0
\(\Leftrightarrow-2x^2+5x-2\ge0\)
\(\Leftrightarrow2x^2-5x+2\le0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\le0\)
\(\orbr{\begin{cases}\hept{\begin{cases}x-2\ge0\\2x-1\le0\end{cases}}\\\hept{\begin{cases}x-2\le0\\2x-1\ge0\end{cases}}\end{cases}}\) TH 1:
\(\hept{\begin{cases}x-2\ge0\\2x-1\le0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge2\\x\le\frac{1}{2}\end{cases}}\)( vô lý )
TH 2:
\(\hept{\begin{cases}x-2\le0\\2x-1\ge0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\le2\\x\ge\frac{1}{2}\end{cases}}\)\(\Leftrightarrow\)\(\frac{1}{2}\le x\le2\) ( thỏa mãn x< 4 )
+ nếu \(4\le x< 5\)
từ pt (1) => x2 - 4x + 3 - x2 - ( 5 - x ) \(\ge0\)
\(\Leftrightarrow-3x-2\ge0\)
\(\Leftrightarrow3x+2\le0\)
\(\Leftrightarrow x\le-\frac{2}{3}\)( không thỏa man \(4\le x< 5\))
+ nếu \(x\ge5\)
từ pt (1) => x2 - 4x + 3 - x2 - ( x -5 ) \(\ge\)0
\(\Leftrightarrow-5x+8\ge0\)
\(\Leftrightarrow5x\le8\)
\(\Leftrightarrow x\le\frac{8}{5}\) ( không thỏa mãn \(x\ge5\))
vậy: bpt có nghiệm là \(\frac{1}{2}\le x\le2\)