bài 1: xét dấu các biểu thức sau:
a) k(x)=(1-2x)(x+1)(x-1)
b) l(x)=\(\frac{x-1}{x+1}\)
c) m(x)=\(\frac{4-2x}{3+x}\)
K
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TN
Thiện Nguyễn
23 tháng 3 2020
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AH
Akai Haruma
Giáo viên
28 tháng 6 2019
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
AH
Akai Haruma
Giáo viên
28 tháng 6 2019
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
H
12 tháng 3 2019
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{8x}{x^2-1}\right):\left(\frac{2x-2x^2-6}{x^2-1}-\frac{2}{x-1}\right)\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{8x}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{2x-2x^2-6}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1-8x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x-2x^2-6-2x-2}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{4x-8x}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{-2x^2-8}\)
..........
NN
12 tháng 3 2019
\(\frac{x+32}{2008}+\frac{x+31}{2009}+\frac{x+29}{2011}+\frac{x+28}{2012}+\frac{x+2056}{4}=0\) \(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+1+\frac{x+31}{2009}+1+\frac{x+29}{2011}+1\)\(+\frac{x+28}{2012}+1+\frac{x+2056}{4}-4\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+\frac{2008}{2008}+\frac{x+31}{2009}+\frac{2009}{2009}+\)\(\frac{x+29}{2011}+\frac{2011}{2011}+\frac{x+28}{2012}+\frac{2012}{2012}+\)\(\frac{x+2056}{4}-\frac{16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32+2008}{2008}+\frac{x+31+2009}{2009}\)\(+\frac{x+29+2011}{2011}+\frac{x+28+2012}{2012}\)\(+\frac{x+2056-16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+2040}{2008}+\frac{x+2040}{2009}+\frac{x+2040}{2011}\)\(+\frac{x+2040}{2012}+\frac{x+2040}{4}=0\)
\(\Leftrightarrow\)\(\left(x+2040\right).\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+2040=0\\\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}=0\end{cases}}\)(vô lí)
\(\Leftrightarrow\)\(x=-2040\)
Vậy phương trình có nghiệm là : x = -2040
AV
17 tháng 2 2020
a)\(F\left(x\right)>0\) khi x thuộc \(\left(\frac{-9}{8};\frac{-1}{3}\right)\cup\left(2;-\infty\right)\)
b) ta có công thức ax2+bx+c=0 thì có a(x-x1)(x-x2)
với x là nghiệm của phương trình trên
vây f(x)>0 khi x thuộc\(\left(-\infty;\frac{-1}{2}\right)\cup\left(\frac{1}{2};+\infty\right)\)
c)f(x)>0 khi x thuộc \(\left(-2;\frac{-1}{2}\right)\cup\left(1:+\infty\right)\)
MR
1 tháng 4 2020
a) f (x) = \(\frac{-4}{3x+1}-\frac{3}{2-x}\)
= \(\frac{-4\left(2-x\right)-3\left(3x+1\right)}{\left(3x+1\right)\left(2-x\right)}=\frac{-8+4-9x-3}{\left(3x+1\right)\left(2-x\right)}\) = \(\frac{-5x-11}{\left(3x+1\right)\left(2-x\right)}\)
BXD : x \(\frac{-11}{5}\) \(\frac{-1}{3}\) 2
f(x) - 0 + \(||\) - \(||\) +
Vậy f(x) < 0 <=> x ∈ ( -∞ ; \(\frac{-11}{5}\) ) U (\(\frac{-1}{3}\) ; 2)
f(x) > 0 <=> x ∈ ( \(\frac{-11}{5}\); \(\frac{-1}{3}\) ) U (2 ; +∞)
b) f(x) = 4x2 -1
f(x) = (2x-1)(2x+1)
2x-1 =0 <=> x = \(\frac{1}{2}\)
2x +1 =0 <=> x= \(\frac{-1}{2}\)
BXD : x \(\frac{-1}{2}\) \(\frac{1}{2}\)
f(x) + 0 - 0 +
f(x) >0 khi x ∈ ( -∞ ; \(\frac{-1}{2}\)) U ( \(\frac{1}{2}\); +∞)
f(x) <0 khi x ∈ ( \(\frac{-1}{2}\); \(\frac{1}{2}\))
c) f(x) = \(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\)
2x +1 = 0 <=> x= \(\frac{-1}{2}\)
x-1 =0 <=> x = 1
x+2 =0 <=> x = -2
BXD : x -2 \(\frac{-1}{2}\) 1
f(x) + \(||\) - 0 + \(||\) -
Vậy f(x) >0 khi x ∈ ( -∞ ;-2) U ( \(\frac{-1}{2}\) ; 1)
f(x)<0 khi x ∈ ( -2 ; \(\frac{-1}{2}\)) U ( 1; +∞)