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G
8 tháng 9 2017
Giải:
a) \(\left(x+2\right)^5=\left(x+2\right)^2\)
Vì \(x+2=x+2\)
Mà \(5\ne2\)
\(\Leftrightarrow\left(x+2\right)=\left\{0;1\right\}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)
Vậy \(x=-2\) hoặc \(x=-1\)
b) \(2^{3x+4}=4^{x+8}\)
\(\Leftrightarrow2^{3x+4}=\left(2^2\right)^{x+8}\)
\(\Leftrightarrow2^{3x+4}=2^{2x+16}\)
Vì \(2=2\)
Nên \(3x+4=2x+16\)
\(\Leftrightarrow3x-2x=16-4\)
\(\Leftrightarrow x=12\)
Vậy \(x=12\).
c) \(3^{x+2}=9^{x-1}\)
\(\Leftrightarrow3^{x+2}=\left(3^2\right)^{x-1}\)
\(\Leftrightarrow3^{x+2}=3^{2x-2}\)
Vì \(3=3\)
Nên \(x+2=2x-2\)
\(\Leftrightarrow2x-2=2+2\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\).
d) \(9^{x-2}=27^{x-4}\)
\(\Leftrightarrow\left(3^2\right)^{x-2}=\left(3^3\right)^{x-4}\)
\(\Leftrightarrow3^{2x-4}=3^{3x-12}\)
Vì \(3=3\)
Nên \(2x-4=3x-12\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\).
Chúc bạn học tốt!!!
9 tháng 9 2017
a)\(\left(x+2\right)^5=\left(x+2\right)^2\)
=>\(\left(x+2\right)^5-\left(x+2\right)^2=0\)
=>\(\left(x+2\right)^2\left(\left(x+2\right)^3-1\right)=0\)
=>\(\left(x+2\right)^2=0\) hoặc \(\left(x+2\right)^3-1=0\)
=>\(x+2=0\) hoặc \(\left(x+2\right)^3=1\)
=>\(x=-2\) hoặc \(x+2=1\)
=>\(x=-2hoặcx=-1\)
Vậy...
Các câu sau tương tự
8 tháng 9 2020
a) \(32< 2^x< 128\)
=> \(2^5< 2^x< 2^7\)
=> x = 6
b) \(2^{x-1}+4\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^{2+x}=9\cdot2^5\)
=> 9.2x-1 = 9.25
=> 2x-1 = \(\frac{9\cdot2^5}{9}=2^5\)
=> x - 1 = 5 => x = 6
c) \(9\cdot27\le3^x\le243\)
=> \(243\le3^x\le243\)
=> x = 5
d) Giống câu b)
e) \(3^{x-1}+5\cdot3^{x-2}=216\)
=> 8.3x-2 = 216
=> 3x-2 = 27
=> 3x-2 = 33
=> x - 2 = 3 => x = 5
f) 27x-3 = 9x+3
=> 27x-3 = 9x+3
=> (33)x-3 = (32)x+3
=> 33x-9 = 32x + 6
=> không thỏa mãn x vì x là phân số mà theo đề bài là số nguyên
g) x2019 = x => x2019 - x = 0 => x(x2018 - 1) = 0 => x = 0 hoặc x = 1
8 tháng 9 2020
a)
\(2^5< 2^x< 2^7\)
\(5< x< 7\)
\(x=6\)
b)
\(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
\(2^{x-1}+2^{2+x}=9\cdot2^5\)
\(2^{x-1}\left(1+2^3\right)=9\cdot2^5\)
\(2^{x-1}\cdot9=9\cdot2^5\)
\(2^{x-1}=2^5\)
\(x-1=5\)
\(x=6\)
TT
1
3 tháng 11 2018
\(\frac{x^2}{9}=\frac{9^2}{18}\)
\(\frac{x^2}{9}=\frac{\left(3^2\right)^2}{18}\)
\(\frac{x^2}{9}=\frac{3^{2.2}}{18}\)
\(\frac{x^2}{9}=\frac{3^4}{18}\)
\(\frac{x^2}{9}=\frac{3^4}{2.3^2}\)
\(\frac{x^2}{9}=\frac{3^4}{2.3^2}\)
\(\frac{x^2}{9}=\frac{3^{4-2}}{2}\)
\(\frac{x^2}{9}=\frac{3^2}{2}\)
\(\frac{x^2}{3^2}=\frac{3^2}{2}\)
\(2\left(x^2\right)=9\left(3^2\right)\)
\(\frac{2\left(x^2\right)}{2}=\frac{9\left(3^2\right)}{2}\)
\(x^2=\frac{3^2.3^2}{2}\)
\(x^2=\frac{3^{2+2}}{2}\)
\(x^2=\frac{3^4}{2}\)
\(\left(x^2\right)^{\frac{1}{2}}=\left(\frac{3^4}{2}\right)^{\frac{1}{2}}\)
\(x^{2.\frac{1}{2}}=\frac{\left(3^4\right)^{\frac{1}{2}}}{2^{\frac{1}{2}}}\)
\(x^{\frac{2}{2}}=\frac{3^{4.\frac{1}{2}}}{2^{\frac{1}{2}}}\)
\(x=\frac{3^{\frac{4}{2}}}{2^{\frac{1}{2}}}\)
\(x=\frac{3^{\frac{2^2}{2}}}{2^{\frac{1}{2}}}\)
\(x=\frac{3^{2^{2-1}}}{2^{\frac{1}{2}}}\)
\(x=\frac{3^2}{2^{\frac{1}{2}}}\)
\(x=\frac{9}{2^{\frac{1}{2}}}\)
\(x=\frac{9}{\frac{5}{2}}\)
\(x=9:\left(5:2\right)\)
\(x=9:2,5\)
\(x=3,6\)
\(3,6^2+y^2=100\)
\(12,96+y^2=100\)
\(y^2=100-12,96\)
\(y^2=87,04\)
\(y=\sqrt{87,04}\)
26 tháng 8 2018
3. Từ \(\dfrac{x-2}{27}=\dfrac{3}{x-2}\Rightarrow\left(x-2\right)^2=81\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=-9\\x-2=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
Vậy x = -7 hoặc x = 11
4. Từ \(\dfrac{2x+5}{9-2x}=\dfrac{2}{5}\)
\(\Rightarrow5\left(2x+5\right)=2\left(9-2x\right)\\ \Leftrightarrow10x+25=18-4x\\ \Leftrightarrow14x=-7\\ \Rightarrow x=-\dfrac{1}{2}\)
5. Từ \(\dfrac{x-7}{x+8}=\dfrac{x-8}{x+9}\)
\(\Rightarrow\left(x-7\right)\left(x+9\right)=\left(x-8\right)\left(x+8\right)\\ \Leftrightarrow x^2+2x-63=x^2-64\\ \Leftrightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
12 tháng 9 2018
a) \(\sqrt{x^2-4x+4}=\sqrt{\left(x-2\right)^2}=3\Leftrightarrow x-2=3\Leftrightarrow x=5\)
b) \(\sqrt{x^2-12}=2\) \(\Leftrightarrow x^2-12=4\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)
c) \(\sqrt{x+3}=x+3\Leftrightarrow x+3-\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+3}-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+3=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
mấy câu còn lại bn làm tương tự
MX
1
G
3 tháng 3 2018
Câu 2 đây:
\(|x^2+|x-1||=x^2+2\)
\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)
\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
KT
3 tháng 3 2018
a) \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)
\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)
\(=0\)
27 tháng 5 2022
a: \(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\)
=>|x-1|=1
=>x-1=1 hoặc x-1=-1
=>x=2 hoặc x=0
b: \(\dfrac{x-2}{-4}=\dfrac{-9}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=36\)
=>x-2=6 hoặc x-2=-6
=>x=8 hoặc x=-4
c: \(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\)
\(\Leftrightarrow\dfrac{x-5}{3}=\dfrac{12}{x-5}\)
\(\Leftrightarrow\left(x-5\right)^2=36\)
=>x-5=6 hoặc x-5=-6
=>x=11 hoặc x=-1
(\(x+2)\)\(^2\) - 9
= (\(x+2\))(\(x+2\)) -9
= \(x\left(x+2\right)\) + 2(\(x\) + 2) - 9
= \(x^2\) + 2\(x\) + 2\(x\) + 4 - 9
= \(x^2\) + (2\(x+2x\)) - (9 - 4)
= \(x^2\) + 4\(x\) - 5
(x + 2)\(^2\) - 9
= (x + 2) . (x + 2) - 9
= x (x + 2) + 2 (x + 2) - 9
= (\(x^2\) + 2x) + (2x + 4) - 9
= \(x^2\) + 2x + 2x + 4 - 9
= \(x^2\) + 4x + 4 - 9
= \(x^2\) + 4x - 5