Phương trình \(0,{1^{2{\rm{x}} - 1}} = 100\) có nghiệm là:
A. \( - \frac{1}{2}\).
B. \(\frac{1}{3}\).
C. \(1\frac{1}{2}\).
D. \(2\frac{1}{3}\).
K
Khách
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HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
a)
\(\begin{array}{l}\frac{2}{{3{\rm{x}}}} + \frac{x}{{x - 1}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{2}{{3{\rm{x}}}} - \frac{x}{{1 - x}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4\left( {1 - x} \right) - 6{{\rm{x}}^2} + 3\left( {6{{\rm{x}}^2} - 4} \right)}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4 - 4{\rm{x}} - 6{{\rm{x}}^2} + 18{{\rm{x}}^2} - 12}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{12{{\rm{x}}^2} - 4{\rm{x}} - 8}}{{6{\rm{x}}\left( {1 - x} \right)}}\end{array}\)
b)
\(\begin{array}{l}\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1}}{{{x^3} - 1}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1 + x\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{ - {x^3} - 1 + {x^3} + {x^2} + x - {x^2} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{x}{{{x^3} - 1}}\end{array}\)
c)
\(\begin{array}{l}\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2\left( {1 - x} \right) - 2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2 - 2{\rm{x}} - 2{\rm{x}} - 4}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{ - 4{\rm{x - 2}}}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{\left( { - 4{\rm{x}} - 2} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 8{\rm{x}} - 2{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 6{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {4{{\rm{x}}^2} - 1} \right)}}\end{array}\)
d)
\(\begin{array}{l}1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}.\frac{{1 + x - 1}}{{1 - {x^2}}}\\ = 1 + \frac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\frac{x}{{1 - {x^2}}}\\ = 1 + \frac{{ - {x^2}\left( {{x^2} - 1} \right)}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}\\ = 1 + \frac{{ - {x^2}}}{{{x^2} + 1}}\\ = \frac{{{x^2} + 1 - {x^2}}}{{{x^2} + 1}}\\ = \frac{1}{{{x^2} + 1}}\end{array}\)
HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
\(\begin{array}{l}a)\frac{1}{x} + \frac{2}{{x + 1}} + \frac{3}{{x + 2}} - \frac{1}{x} - \frac{2}{{x - 1}} - \frac{3}{{x + 2}}\\ = \left( {\frac{1}{x} - \frac{1}{x}} \right) + \left( {\frac{2}{{x + 1}} - \frac{2}{{x - 1}}} \right) + \left( {\frac{3}{{x + 2}} - \frac{3}{{x + 2}}} \right)\\ = 0 + \frac{2}{{x + 1}} - \frac{2}{{x - 1}} + 0\\ = \frac{{2\left( {x - 1} \right) - 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{2{\rm{x}} - 2 - 2{\rm{x}} - 2}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{ - 4}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\end{array}\)
\(\begin{array}{l}b)\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{3}{{{x^2} - 9}} + \frac{{1 - 2{\rm{x}}}}{x} + \frac{{x - 1}}{{2{\rm{x}} + 1}} - \frac{3}{{x + 3}}\\ = \left( {\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - 2{\rm{x}}}}{x}} \right) + \left( {\frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{{x - 1}}{{2{\rm{x}} + 1}}} \right) + \left( {\frac{3}{{{x^2} - 9}} - \frac{3}{{x + 3}}} \right)\\ = 0 + 0 + \frac{3}{{\left( {x + 3} \right)\left( {x - 3} \right)}} - \frac{3}{{x + 3}}\\ = \frac{{3 - 3\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{{12 - 3{\rm{x}}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\end{array}\)
13 tháng 2 2019
d) x+1/2019 + x+3/2017 = x+5/2015 + x+7/2013
<=> x+1/2019 + x+3/2017 - x+5/2015 - x+7/2013 =0
<=> ( x+1/2019 + 1) + ( x+3/2017 + 1) - ( x+5/2015 + 1) - ( x+7/2013 +1) = 0
<=> ( x+1+2019/2019) +(x+3+2017/2017) - ( x+5+2015/2015) - ( x+7+2013/2013) =0
<=> x+2020/2019 + x+2020/2017 - x+2020/2015 - x+2020/2013 =0
<=> (x+2020)× ( 1/2019 + 1/2017 - 1/2015 - 1/2013) =0
Mà 1/2019 + 1/2017 - 1/2015 - 1/2013 khác 0
=> x+2020 =0
=> x = -2020
13 tháng 2 2019
\(\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow\left(x-1\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
HOẶC\(x-1=0\Leftrightarrow x=1\)(NHẬN)
HOẶC\(x-3=0\Leftrightarrow x=3\)(NHẬN)
VẬY: tập ngiệm của pt là S={1;3}
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(y' = 2.3{{\rm{x}}^2} - \frac{1}{2}.2{\rm{x}} + 4.1 - 0 = 6{{\rm{x}}^2} - x + 4\).
b) \(y' = \frac{{{{\left( { - 2{\rm{x}} + 3} \right)}^\prime }.\left( {{\rm{x}} - 4} \right) - \left( { - 2{\rm{x}} + 3} \right).{{\left( {{\rm{x}} - 4} \right)}^\prime }}}{{{{\left( {{\rm{x}} - 4} \right)}^2}}}\)
\( = \frac{{ - 2\left( {{\rm{x}} - 4} \right) - \left( { - 2{\rm{x}} + 3} \right).1}}{{{{\left( {{\rm{x}} - 4} \right)}^2}}}\)
\( = \frac{{ - 2{\rm{x}} + 8 + 2{\rm{x}} - 3}}{{{{\left( {{\rm{x}} - 4} \right)}^2}}} = \frac{5}{{{{\left( {{\rm{x}} - 4} \right)}^2}}}\)
c) \(y' = \frac{{{{\left( {{x^2} - 2{\rm{x}} + 3} \right)}^\prime }\left( {{\rm{x}} - 1} \right) - \left( {{x^2} - 2{\rm{x}} + 3} \right){{\left( {{\rm{x}} - 1} \right)}^\prime }}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}\)
\( = \frac{{\left( {2{\rm{x}} - 2} \right)\left( {{\rm{x}} - 1} \right) - \left( {{x^2} - 2{\rm{x}} + 3} \right).1}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}\) \( = \frac{{2{{\rm{x}}^2} - 2{\rm{x}} - 2{\rm{x}} + 2 - {x^2} + 2{\rm{x}} - 3}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}\)
\( = \frac{{{x^2} - 2{\rm{x}} - 1}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}\)
d) \(y' = {\left( {\sqrt 5 .\sqrt x } \right)^\prime } = \sqrt 5 .\frac{1}{{2\sqrt x }} = \frac{{\sqrt 5 }}{{2\sqrt x }} = \frac{5}{{2\sqrt {5x} }}\).
H9
HT.Phong (9A5)
CTVHS
26 tháng 8 2023
\(0,5^{3x-1}>0,25\)
\(\Leftrightarrow0,5^{3x-1}>0,5^2\)
\(\Leftrightarrow3x-1< 2\)
\(\Leftrightarrow3x< 3\)
\(\Leftrightarrow x< \dfrac{3}{3}\)
\(\Leftrightarrow x< 1\)
Vậy: \(\left(-\infty;1\right)\)
Chọn A
VN
18 tháng 1 2017
Nhìn sơ qua thì thấy bài 3, b thay -2 vào x rồi giải bình thường tìm m
VN
18 tháng 1 2017
Bài 2:
a) \(x+x^2=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=0-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(0x-3=0\)
\(\Leftrightarrow0x=3\)
\(\Rightarrow vonghiem\)
c) \(3y=0\)
\(\Leftrightarrow y=0\)
\(0,1^{2x-1}=100\)
\(\Leftrightarrow0,1^{2x-1}=0,1^{log_{0,1}100}\)
\(\Leftrightarrow2x-1=log_{0,1}100\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Chọn A