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a/ \(x\ge-\frac{5}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+7=2x+5\\4x+7=-2x-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
b/ \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+12=0\\x^2=22\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=6\\x=\pm\sqrt{22}\end{matrix}\right.\)
c/ \(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\2x^2-7x+5=0\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{2}\)
d/ \(\left|x-1\right|+\left|2x+1\right|\ge\left|x-1+2x+1\right|=\left|3x\right|\)
Dấu "=" xảy ra khi và chỉ khi: \(\left(x-1\right)\left(2x+1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge1\end{matrix}\right.\)
Vậy nghiệm của pt là \(\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge1\end{matrix}\right.\)
\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)
Mà \(-3< 0.\)
\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)
\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)
Kết hợp ĐKXĐ.
\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)
\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)
Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)
Ta có bảng sau:
| \(x\) | \(-\infty\) \(-\dfrac{3}{2}\) \(-\dfrac{1}{2}\) \(1\) \(2\) \(+\infty\) |
| \(1+2x\) | - | - 0 + | + | + |
| \(x-2\) | - | - | - | - 0 + |
| \(2x+3\) | - 0 + | + | + | + |
| \(1-x\) | + | + | + 0 - | - |
| \(f\left(x\right)\) | - || + 0 - || + 0 - |
Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].
a/ ĐKXĐ:...
\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-1\right)^2}+2\left|\frac{x-2}{x-1}\right|-3=0\)
\(\Leftrightarrow\left|\frac{x-2}{x-1}\right|^2+2\left|\frac{x-2}{x-1}\right|-3=0\)
Đặt \(\left|\frac{x-2}{x-1}\right|=t\left(t\ge0\right)\)
\(\Rightarrow pt\Leftrightarrow t^2+2t-3=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
Ok tự giải nốt
b/ viết lại đề bài đi cậu
Casio:
a/ \(\Leftrightarrow\left(x^2-5x-2\right)\left(x^2-2x-2\right)=0\)
b/ \(\Leftrightarrow2\left(2x^2+3x+3\right)^2+6\left(x+\frac{2}{3}\right)^2+\frac{28}{3}=0\)
Vế trái luôn dương nên pt vô nghiệm
c/ Câu này đề sai, pt này ko thể tách ra được nên chắc chắn là ko giải được
d/ Câu này chắc đề cũng ko đúng: đặt \(2x-4=a\Rightarrow2x=a+4\)
\(\Rightarrow\left(a+5\right)\left(a+1\right)\left(a+2\right)\left(a+10\right)=100\)
\(\Leftrightarrow a\left(a^3+18a^2+97a+180\right)=0\)
Dù pt có nghiệm \(a=0\) nhưng pt bậc 3 đằng sau lại ko thể giải
e/ Câu này giống câu trên
\(\Leftrightarrow x\left(16x^3+16x^2-93x+12\right)=0\)
Pt bậc 3 phía sau ko giải được
a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x+2=3\\x+2=-3\end{matrix}\right.\)
b/
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)
\(\Leftrightarrow\left(\left|x+2\right|+1\right)\left(\left|x+2\right|-4\right)=0\)
\(\Leftrightarrow\left|x+2\right|-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)
c/
\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)
\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)
d/ ĐKXĐ: ...
\(\Leftrightarrow\frac{\left|x-2\right|^2}{\left(x-1\right)^2}+\frac{2\left|x-4\right|}{x-1}=3\)
Đặt \(\frac{\left|x-2\right|}{x-1}=a\)
\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)
e/ ĐKXĐ: ...
Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)
\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)
a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)
b) \(\dfrac{x+3}{x-2}\le0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow-3\le x< 2\)
d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)
\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)
\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)
\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)
a: \(x^2-2x+\left|x-1\right|-1=0\)
\(\Leftrightarrow x^2-2x+1+\left|x-1\right|-2=0\)
\(\Leftrightarrow\left(\left|x-1\right|\right)^2+\left|x-1\right|-2=0\)
\(\Leftrightarrow\left(\left|x-1\right|+2\right)\left(\left|x-1\right|-1\right)=0\)
=>|x-1|=1
=>x-1=1 hoặc x-1=-1
=>x=2 hoặc x=0
b: \(4x^2-4x-\left|2x-1\right|-1=0\)
\(\Leftrightarrow4x^2-4x+1-\left|2x-1\right|-2=0\)
\(\Leftrightarrow\left(\left|2x-1\right|\right)^2-\left|2x-1\right|-2=0\)
\(\Leftrightarrow\left(\left|2x-1\right|-2\right)\left(\left|2x-1\right|+1\right)=0\)
=>|2x-1|=2
=>2x-1=2 hoặc 2x-1=-2
=>x=3/2 hoặc x=-1/2
c: \(\left|2x-5\right|+\left|2x^2-7x+5\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\\left(2x-5\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{5}{2}\)
d: \(x^2-2x-5\left|x-1\right|-5=0\)
\(\Leftrightarrow x^2-2x+1-5\left|x-1\right|-6=0\)
\(\Leftrightarrow\left(\left|x-1\right|\right)^2-5\left|x-1\right|-6=0\)
\(\Leftrightarrow\left(\left|x-1\right|-6\right)\left(\left|x-1\right|+1\right)=0\)
=>|x-1|=6
=>x-1=6 hoặc x-1=-6
=>x=7 hoặc x=-5