giải phương trình:
a)\(\frac{2-x}{x}+2=\frac{3}{x}\)
b)\(\frac{x}{x-2}-\frac{2}{x+2}=\frac{2x+4}{x^2-4}\)
giải hộ e vs ạ
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NH
Nguyễn Huy Tú
2 tháng 7 2021
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NH
Nguyễn Huy Tú
2 tháng 7 2021
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Những câu hỏi liên quan
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{5\pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)
b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k4\pi \\x = - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
d) \(2\cos 3x + 5 = 3\)
\(\begin{array}{l} \Leftrightarrow \cos 3x = - 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi + k2\pi \\3x = - \pi + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
QA
21 tháng 7 2021
Trả lời:
a, \(A=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)
b, \(B=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)
c, \(C=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2\left(x+2\right)}=\frac{x+2}{2}\)
d, \(D=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x}{x+2}\)
e, \(E=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
JS
24 tháng 2 2020
d, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\) (Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\) ≠ 0)
\(\Leftrightarrow x=-10\)
Vậy x = -10 là nghiệm của phương trình.
24 tháng 7 2021
\(ĐKXĐ:x\ne2;x\ne-2;x\ne0\)
\(a,P=\left(\frac{-1}{2-x}-\frac{2x}{4-x^2}+\frac{1}{2+x}\right)\left(\frac{2}{x}-1\right)\)
\(P=\left(\frac{-2-x+2-x-2x}{\left(2-x\right)\left(2+x\right)}\right)\left(\frac{2-x}{x}\right)\)
\(P=\frac{-4x}{\left(2-x\right)\left(2+x\right)}\frac{2-x}{x}\)
\(P=\frac{-4}{2+x}\)
\(b,P=\frac{-4}{2+x}=\frac{1}{2}\)
\(2+x=-8\)
\(x=-10\)
\(c,P=-\frac{4}{2+x}\)
\(< =>-4⋮x+2\)
lập bảng ra thì bạn ra đc \(x=\left\{2;-1;-3;-6\right\}\)
AD
24 tháng 7 2021
a)\(P=\left(\frac{1}{x-2}-\frac{2x}{4-x^2}+\frac{1}{2+x}\right)\left(\frac{2}{x}-1\right)\)
\(P=\left(\frac{1}{x-2}+\frac{2x}{\left(x+2\right)\left(x-2\right)}+\frac{1}{2+x}\right).\frac{2-x}{x}\)
\(P=\frac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)
\(P=\frac{4x}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)
\(P=\frac{-4}{x+2}\)
b) Để P=1/2
\(\Rightarrow-\frac{4}{x+2}=\frac{1}{2}\)
\(\Leftrightarrow-8=x+2\)
\(\Leftrightarrow x=-10\)
c) Để P nhận GT nguyên
\(\Rightarrow\left(x+2\right)\inƯ_{\left(-4\right)}\)
\(\Rightarrow\left(x+2\right)\in\left\{-1;1;-2;2;-4;4\right\}\)
\(\Rightarrow x=\left\{-3;-1;-4;0;-6;2\right\}\)
#H
13 tháng 2 2019
Làm đc 2 bài đầu chưa, t làm câu cuối cho, hai câu đầu dễ í mà
BL
10 tháng 3 2020
1. \(1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\)
\(\Rightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\)
\(\Rightarrow x^2-x-2x+2+2x^2-2x-5x+5-3x^2+6x+5x-10=0\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
2. \(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}-\frac{16}{5}=0\)
\(\Rightarrow x^2-4x-3x+12-x^2+4x-4-16=0\)
\(\Rightarrow-3x-8=0\Rightarrow x=\frac{-8}{3}\)
3. \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
\(\Rightarrow\frac{\left(x-2\right)^2}{x^2-4}-\frac{3\left(x+2\right)}{x^2-4}-\frac{2\left(x-11\right)}{x^2-4}=0\)
\(\Rightarrow x^2-4x+4-3x-6-2x+22=0\)
\(\Rightarrow x^2-9x+20=0\)
\(\Rightarrow x^2-4x-5x+20=0\)
\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
KN
28 tháng 1 2020
\(ĐKXĐ:x\ne\pm5\)
\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
\(\Rightarrow\frac{3\left(x+5\right)}{4\left(x-5\right)\left(x+5\right)}+\frac{30}{4\left(25-x^2\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x+15}{4\left(x-5\right)\left(x+5\right)}+\frac{-30}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x+15-30}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x-15}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3\left(x-5\right)}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3}{4\left(x+5\right)}=\frac{-7}{6\left(x+5\right)}\)
\(\Rightarrow18\left(x+5\right)=-28\left(x+5\right)\)
\(\Rightarrow18\left(x+5\right)+28\left(x+5\right)=0\)
\(\Rightarrow46\left(x+5\right)=0\Leftrightarrow x+5=0\Leftrightarrow x=-5\)(ktm)
Vậy pt vô nghiệm
MQ
11 tháng 5 2019
a.(x+2)2-x(x+2)=0
\(\Leftrightarrow\)(x+2)(x-2-x)=0
\(\Leftrightarrow\)(x+2)*2=0
\(\Leftrightarrow\)x+2=0
\(\Leftrightarrow\)x=-2
vay s={-2}
b.\(\frac{2x+7}{3}\)-\(\frac{x-2}{4}\)=2
\(\Leftrightarrow\)\(\frac{4\left(2x+7\right)}{12}\)+\(\frac{-3\left(x-2\right)}{12}\)=\(\frac{24}{12}\)
\(\Leftrightarrow\)8x+28-3x+6=24
\(\Leftrightarrow\)5x=-10
\(\Leftrightarrow\)x=-2
vay s={-2}
c.|x+5|=3x+1
neu x+5\(\ge\)0 thi |x+5|=x+5
\(\Leftrightarrow\)x\(\ge\)-5
ta co phuong trinh
x+5=3x+1
\(\Leftrightarrow\)-2x=-4
\(\Leftrightarrow\)x=2( thoa man dieu kien x\(\ge\)-5)
neu x+5<0 thi |x+5|=5-x
\(\Leftrightarrow\)x<-5
ta co phuong trinh
5-x=3x+1
\(\Leftrightarrow\)-4x=-4
\(\Leftrightarrow\)x=1 (k thoa man dieu kien x<5)
vay s={2}
chuc bn hoc tot