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a: =>|x+3|=|2x-1|
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+3\\2x-1=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\3x=-2\end{matrix}\right.\Leftrightarrow x\in\left\{4;-\dfrac{2}{3}\right\}\)
b: \(\left|x^2-2x\right|=\left|2x^2-x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-x-2=x^2-2x\\2x^2-x-2=-x^2+2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\3x^2+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-1\right)=0\\\left(x+1\right)\left(3x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1;-1;\dfrac{2}{3}\right\}\)
c: \(\left|3x^2-2x\right|=\left|6-x^2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\end{matrix}\right.\)
\(\Leftrightarrow2x^2-x-3=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x+1\right)=0\)
=>x=3/2 hoặc x=-1
d: \(\left|2x^2-3x-5\right|=\left|x^2-4x-5\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=x^2-4x-5\\2x^2-3x-5=4x+5-x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\\3x^2-7x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=0\\3x^2-10x+3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=0\\\left(3x-10\right)\left(x+1\right)=0\end{matrix}\right.\)
hay \(x\in\left\{\dfrac{10}{3};-1\right\}\)
e: |5x+1|=|2x-3|
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=2x-3\\5x+1=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\7x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{7}\end{matrix}\right.\)
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
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/
\(\Leftrightarrow4x^2-12x+9=\left(3x-2\right)^2\)
\(\Leftrightarrow5x^2-5=0\Rightarrow x=\pm1\)
b/
\(\Leftrightarrow25x^2-10x+1=\left(x+6\right)^2\)
\(\Leftrightarrow24x^2-22x-35=0\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=-\frac{5}{6}\end{matrix}\right.\)
c/
\(\Leftrightarrow16x^2-8x+1=\left(x-3\right)^2\)
\(\Leftrightarrow15x^2-2x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-\frac{2}{3}\end{matrix}\right.\)
d/ \(x\ge\frac{3}{2}\)
\(\Leftrightarrow\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Leftrightarrow21x^2+22x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{2}{7}\\x=-\frac{4}{3}\end{matrix}\right.\)
e/
\(\Leftrightarrow\left[{}\begin{matrix}3x-4=x-2\\3x-4=2-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\4x=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)
f/
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)
g/
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=2x^2-x-2\\x^2-2x=-2x^2+x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\3x^2-3x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\frac{3\pm\sqrt{33}}{6}\\\end{matrix}\right.\)
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
\(\left(2x^2-x\right)^2\left(2x+1\right)\left(x-1\right)=280\)
\(\Leftrightarrow\left(2x^2-x\right)^2\left(2x^2-x-1\right)=280\)
Đặt \(2x^2-x=a\)
\(pt\Leftrightarrow a^2\left(a-1\right)=280\)
\(\Leftrightarrow a^3-a^2-280=0\)
Bạn xem lại đề chứ nghiệm xấu quá.