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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 \(\left|\frac{2x-1}{x+2}\right|=a>0\)
\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)
Câu 1:
a/ \(x\ge-11\)
Đặt \(\sqrt{x+11}=a\ge0\Rightarrow11=a^2-x\), pt đã cho trở thành:
\(x^2+a=a^2-x\Leftrightarrow x^2-a^2+x+a=0\Leftrightarrow\left(x+a\right)\left(x-a+1\right)=0\)
TH1: \(x+a=0\Leftrightarrow x+\sqrt{x+11}=0\Leftrightarrow-x=\sqrt{x+11}\)
\(\Leftrightarrow\left[{}\begin{matrix}-x\ge0\\x^2=x+11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^2-x-11=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1-3\sqrt{5}}{2}\)
TH2: \(x-a+1=0\Leftrightarrow x+1=\sqrt{x+11}\) \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\\left(x+1\right)^2=x+11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-10=0\end{matrix}\right.\) \(\Rightarrow x=\frac{-1+\sqrt{41}}{2}\)
b/ \(\sqrt{9+x}=x-9\Leftrightarrow\left\{{}\begin{matrix}x-9\ge0\\9+x=\left(x-9\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge9\\x^2-19x+72=0\end{matrix}\right.\) \(\Rightarrow x=\frac{19+\sqrt{73}}{2}\)
Câu 2:
a/
\(f\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-1\right)\left(x-4\right)}=\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-4\right)}\)
Lập bảng xét dấu ta được:
\(f\left(x\right)>0\) khi \(\left[{}\begin{matrix}x< -1\\x>4\\1< x< 3\end{matrix}\right.\)
\(f\left(x\right)< 0\) khi \(\left[{}\begin{matrix}-1< x< 1\\3< x< 4\end{matrix}\right.\)
\(f\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
\(f\left(x\right)\) ko xác định tại \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b/ \(h\left(x\right)=\frac{-x^2+3x-1}{\left(x^2-2x+3\right)\left(x+2\right)}\)
Lập bảng xét dấu ta được:
\(f\left(x\right)>0\) khi \(\left[{}\begin{matrix}x< -2\\\frac{3-\sqrt{5}}{2}< x< \frac{3+\sqrt{5}}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\) khi \(\left[{}\begin{matrix}-2< x< \frac{3-\sqrt{5}}{2}\\x>\frac{3+\sqrt{5}}{2}\end{matrix}\right.\)
\(f\left(x\right)=0\) tại \(x=\frac{3\pm\sqrt{5}}{2}\)
\(f\left(x\right)\) ko xác định tại \(x=-2\)
Giải PT:
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
PT cho tđuong với: (x^2 +9). (x^2 + 9x) = 22 (x-1)^2
Đặt t = [x^2 + 9 + x^2 + 9x]/2 hay t= x^2 + (9x + 9)/2.
Khi đó: x^2 + 9 = t - 9(x-1)/2
x^2 + 9x = t + 9(x-1)/2
PT cho trở thành: [t - 9(x-1)/2]. [t + 9(x-1)/2] = 22(x-1)^2
<=> t^2 -(81/4)(x-1)^2 = 22(x-1)^2
<=> t^2 = (169/4)(x-1)^2
<=> t = 13/2. (x-1) hoặc t= -13/2. (x-1)
<=> 2t =13x -13 hoặc 2t =-13x + 13
hay 2x^2 + 9x+ 9 =13x -13 hoặc 2x^2 + 9x +9 = -13x +13
hay 2x^2 - 4x +22 =0 hoặc 2x^2 + 22x - 4 =0
PT bậc hai thứ nhất vô nghiệm, PT bậc hai thứ hai cho ta hai nghiệm là:
x= (-11 +căn(129))/2 , x= (-11 - căn(129))/2.
cách 2:đặt x-1=k
pt trở thành (k+1)(k2+2k+10)(k+10)=22k2
<=>(k2+2k+10)(k2+11k+10)=22k2
tự làm tiếp