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a,|2x-3|=x-5
th1:2x-3=x-5
➜ x=-2
th2:2x-3=5-x
➜ 3x=8
➜x 8/3
bạn giải giúp mình mấy câu còn lại với , mình sẽ tick cho
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:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
a/ ĐKXĐ: \(0\le x\le4\)
\(\left(x^2-4x\right)\sqrt{-x^2+4x}+x^2-4x+2=0\)
Đặt \(\sqrt{-x^2+4x}=a\ge0\)
\(-a^2.a-a^2+2=0\)
\(\Leftrightarrow a^3+a^2-2=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+2a+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a^2+2a+2=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-x^2+4x}=1\Leftrightarrow x^2-4x+1=0\Rightarrow...\)
b/ \(x^4+2x^2+x\sqrt{2x^2+4}-4=0\)
Đặt \(x\sqrt{2x^2+4}=a\Rightarrow x^2\left(2x^2+4\right)=a^2\Rightarrow x^4+2x^2=\frac{a^2}{2}\)
\(\frac{a^2}{2}+a-4=0\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=2\left(x>0\right)\\x\sqrt{2x^2+4}=-4\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^4+4x^2=4\\2x^4+4x^2=16\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=\sqrt{3}-1\\x^2=-\sqrt{3}-1\left(l\right)\\x^2=2\\x^2=-4\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)
c/ Đặt \(\sqrt[3]{2x^2+3x-10}=a\Rightarrow2x^2+3x=a^3+10\)
\(a^3+10-14=2a\)
\(\Leftrightarrow a^3-2a-4=0\)
\(\Leftrightarrow\left(a-2\right)\left(a^2+2a+2\right)=0\Rightarrow a=2\)
\(\Rightarrow\sqrt[3]{2x^2+3x-10}=2\Rightarrow2x^2+3x-18=0\Rightarrow...\)
d/ \(\Leftrightarrow2\left(3x^2+x+4\right)+\sqrt[3]{3x^2+x+4}-18=0\)
Đặt \(\sqrt[3]{3x^2+x+4}=a\)
\(2a^3+a-18=0\)
\(\Leftrightarrow\left(a-2\right)\left(2a^2+4a+9\right)=0\Rightarrow a=2\)
\(\Rightarrow\sqrt[3]{3x^2+x+4}=2\Rightarrow3x^2+x-4=0\Rightarrow...\)
e/ \(\Leftrightarrow x^2+5x+2-3\sqrt{x^2+5x+2}-2=0\)
Đặt \(\sqrt{x^2+5x+2}=a\ge0\)
\(a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=\frac{3+\sqrt{17}}{2}\\a=\frac{3-\sqrt{17}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+5x+2}=\frac{3+\sqrt{17}}{2}\Rightarrow x^2+5x-\frac{9+3\sqrt{17}}{2}=0\)
Bài cuối xấu quá, chắc nhầm số liệu
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/
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=2x^2-x-2\\x^2-2x=-2x^2+x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\3x^2-3x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\frac{3\pm\sqrt{33}}{6}\\\end{matrix}\right.\)
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/ \(x\ge-\frac{5}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+7=2x+5\\4x+7=-2x-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
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 \(\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge1\end{matrix}\right.\)
|3x+4)/(x-2)| <=3
<=>|3 +10/(x-2) | <=3
10/(x-2) =t
<=> |3+t| <=3
9 +6t +t^2 <=9 <=> -6<=t <=0
10/(x-2) <=0 => x<2
10/(x-2) >=-6 <=>5/(x-2)>=-3
<=>5 <=-3(x-2) <=>3x <=10-5 =5 => x <=5/3
kết luận x<= 5/3
a) \(\left|\frac{3x+4}{x-2}\right|< =3̸\) đk: x\(\ne\) 2
BPT \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\frac{3x+4}{x-2}\ge-3\\\frac{3x+4}{x-2}\le3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{3x+4}{x-2}+3\ge0\\\frac{3x+4}{x-2}-3\le0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}\frac{6x-2}{x-2}\ge0\\\frac{10}{x-2}\le0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le\frac{1}{3}\\x>2\end{matrix}\right.\\x< 2\end{matrix}\right.\Rightarrow}x\le\frac{1}{3}}\)
b) \(\left|\frac{2x-1}{x-3}\right|\ge1\) đk: x\(\ne\) 3
BPT \(\Leftrightarrow\left[{}\begin{matrix}\frac{2x-3}{x-3}\le-1\\\frac{2x-3}{x-3}\ge1\end{matrix}\right.\)
ta có:
+) \(\frac{2x-3}{x-3}\le-1\Leftrightarrow\frac{2x-3}{x-3}+1\le0\Leftrightarrow\frac{3x-6}{x-3}\le0\Leftrightarrow2\le x< 3\)
+) \(\frac{2x-3}{x-3}\ge1\Leftrightarrow\frac{2x-3}{x-3}-1\ge0\Leftrightarrow\frac{x}{x-3}\ge0\Leftrightarrow\left[{}\begin{matrix}x\le0\\x>3\end{matrix}\right.\)
vậy tập nghiệm là: \((-\infty;0]\cup[2;3)\cup(3;+\infty)\)
a/ \(x\ge-3\)
\(\Leftrightarrow\left(2x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow3x^2-10x-8=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)
b/ \(x\ge-\frac{5}{2}\)
\(\Leftrightarrow\left(4x+7\right)^2=\left(2x+5\right)^2\)
\(\Leftrightarrow x^2+3x+2=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
c/ \(x\ge1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=5x-5\\2x^2-3x-5=5-5x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-8x=0\\2x^2+2x-10=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=4\\x=\frac{-1+\sqrt{21}}{2}\\x=\frac{-1-\sqrt{21}}{2}\left(l\right)\end{matrix}\right.\)
d/ \(x\ge\frac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+12=0\\x^2=22\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\left(l\right)\\x=\sqrt{22}\\x=-\sqrt{22}\left(l\right)\end{matrix}\right.\)
e/ \(\left[{}\begin{matrix}x\ge1\\x\le-\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-x-2=x-2\\3x^2-x-2=2-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=0\\3x^2=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=\frac{2}{3}\left(l\right)\\x=\frac{2\sqrt{3}}{3}\\x=\frac{-2\sqrt{3}}{3}\end{matrix}\right.\)