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HH
1
DH
24 tháng 8 2017
a, \(4.3^{2x}-2.9^x-54=0\)
\(\Rightarrow3^{2x}\left(4-2\right)=54\)
\(\Rightarrow3^{2x}=27=3^3\)
Vì \(3\ne\pm1;3\ne0\) nên \(2x=3\Rightarrow x=\dfrac{3}{2}\)
b, \(\dfrac{1}{2}.2^x+4.2^x-288=0\)
\(\Rightarrow2^x\left(\dfrac{1}{2}+4\right)=288\)
\(\Rightarrow2^x=64=2^6\)
Vì \(2\ne\pm1;2\ne0\) nên \(x=6\)
VN
2
13 tháng 9 2017
\(\frac{25}{5^x}=\frac{1}{125}\Rightarrow25.125=5^x.1\)
\(3125=5^x\)
\(5^5=5^x\)
\(\Rightarrow x=5\)
MH
24 tháng 9 2015
a.\(\left(\frac{1}{3}-\frac{1}{2}\right)^{x-1}=\frac{1}{36}\)
\(\Rightarrow\left(-\frac{1}{6}\right)^{x-1}=\frac{1}{36}\)
\(\Rightarrow \left(-\frac{1}{6}\right)^{x-1}=\left(-\frac{1}{6}\right)^2\)
=> x-1=2
=> x=2+1
Vậy x=3.
b.\(81^{-2x}.27^x=9^5\)
\(\Rightarrow\left(3^4\right)^{-2x}.\left(3^3\right)^x=\left(3^2\right)^5\)
\(\Rightarrow3^{4.\left(-2x\right)}.3^{3x}=3^{10}\)
\(\Rightarrow3^{-8x}.3^{3x}=3^{10}\)
\(\Rightarrow3^{-5x}=3^{10}\)
=> -5x=10
=> x=10:(-5)
Vậy x=-2.
c.\(2^x+2^{x+3}=288\)
\(\Rightarrow2^x.\left(1+2^3\right)=288\)
\(\Rightarrow2^x.9=288\)
\(\Rightarrow2^x=288:9\)
\(\Rightarrow2^x=32\)
=> 2x=25
Vậy x=5.
29 tháng 10 2017
\(4^{x+1}.2=32\)
\(4^{x+1}=32:2\)
\(4^{x+1}=16\)
\(4^{x+1}=4^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
vậy \(x=1\)
\(\left(x-\frac{2}{3}\right)^2=\frac{25}{81}\)
\(\left(x-\frac{2}{3}\right)^2=\left(\frac{5}{9}\right)^2\)
\(\Rightarrow x-\frac{2}{3}=\frac{5}{9}\)
\(\Rightarrow x=\frac{11}{9}\)
vậy \(x=\frac{11}{9}\)
\(500^{300}=\left(500^3\right)^{100}=125000000^{100}\)
\(300^{500}=\left(300^5\right)^{100}\)
vì \(\left(500^3\right)^{100}< \left(300^3\right)^{100}\)nên\(500^{300}< 300^{500}\)
\(4^{45}=\left(4^9\right)^5=262144^5\)
\(3^{60}=\left(3^{12}\right)^5=531441^5\)
vì \(262144^5< 531441^5\) nên \(4^{45}< 3^{60}\)
VM
17 tháng 10 2019
a) \(x^2-2=0\)
\(\Rightarrow x^2-\left(\sqrt{2}\right)^2=0\)
\(\Rightarrow\left(x-\sqrt{2}\right).\left(x+\sqrt{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\x+\sqrt{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+\sqrt{2}\\x=0-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}.\)
b) \(x^2+\frac{7}{4}=\frac{23}{4}\)
\(\Rightarrow x^2=\frac{23}{4}-\frac{7}{4}\)
\(\Rightarrow x^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}.\)
c) \(\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2=0^2\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=0+1\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
g) \(\sqrt{x}=0\)
\(\Rightarrow x=0\)
Vậy \(x=0.\)
h) \(\sqrt{x}=4\)
\(\Rightarrow\sqrt{x}=\left(\sqrt{4}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{16}\)
\(\Rightarrow x=16\)
Vậy \(x=16.\)
i) \(\sqrt{x}-\frac{1}{7}=0\)
\(\Rightarrow\sqrt{x}=0+\frac{1}{7}\)
\(\Rightarrow\sqrt{x}=\frac{1}{7}\)
\(\Rightarrow\sqrt{x}=\left(\sqrt{\frac{1}{7}}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\frac{1}{49}}\)
\(\Rightarrow x=\frac{1}{49}\)
Vậy \(x=\frac{1}{49}.\)
Chúc bạn học tốt!
28 tháng 8 2017
mấy cái này đơn dãng vô cùng nhưng có đều bn ra đề dài quá nha
a) \(3x+4\ge7\Leftrightarrow3x\ge7-4\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\) vậy \(x\ge1\)
b) \(-5x+1< 11\Leftrightarrow-5x< 11-1\Leftrightarrow-5x< 10\Leftrightarrow x>\dfrac{10}{-5}\)
\(\Leftrightarrow x>-2\) vậy \(x>-2\)
c) \(\dfrac{5}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\) vậy \(x< 3\)
d) \(\dfrac{-7}{2-x}\ge0\Leftrightarrow2-x\le0\Leftrightarrow x\ge2\) vậy \(x\ge2\)
e) \(x^2+4x>0\Leftrightarrow x\left(x+4\right)>0\) \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x>-4\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< -4\end{matrix}\right.\) vậy \(x>0\) hoặc \(x< -4\)
f) \(\dfrac{x-2}{x-6}< 0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-2>0\\x-6>0\end{matrix}\right.\\\left[{}\begin{matrix}x-2< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x>6\end{matrix}\right.\\\left[{}\begin{matrix}x< 2\\x< 6\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>6\\x< 2\end{matrix}\right.\)
vậy \(x>6\) hoặc \(x< 2\)
g) \(\left(x-1\right)\left(x+2\right)\left(3-x\right)< 0\Leftrightarrow-\left[\left(x-1\right)\left(x+2\right)\left(x-3\right)\right]< 0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)>0\)
th1: 3 số hạng đều dương : \(\Leftrightarrow\left[{}\begin{matrix}x-1>0\\x+2>0\\x-3>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>1\\x>-2\\x>3\end{matrix}\right.\) \(\Rightarrow x>3\)
th2: 2 âm 1 dương : (vì trong 3 số hạng ta có : \(\left(x+2\right)\) lớn nhất \(\Rightarrow\left(x+2\right)\) dương)
\(\Leftrightarrow\left[{}\begin{matrix}x-1< 0\\x+2>0\\x-3< 0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x>-2\\x< 3\end{matrix}\right.\) \(\Rightarrow-2< x< 1\)
vậy \(x>3\) hoặc \(-2< x< 1\)
h) \(\dfrac{x^2-1}{x}>0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2-1>0\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x^2-1< 0\\x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2>1\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x^2< 1\\x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}-1< x< 1\\x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1\\-1< x< 0\end{matrix}\right.\) vậy \(x>1\) hoặc \(-1< x< 0\)
i) \(x^2+x-2< 0\Leftrightarrow x^2+x+\dfrac{1}{4}-\dfrac{9}{4}< 0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{9}{4}< 0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2< \dfrac{9}{4}\Leftrightarrow\dfrac{-3}{2}< \left(x+\dfrac{1}{2}\right)< \dfrac{3}{2}\Leftrightarrow-2< x< 1\)
vậy \(-2< x< 1\)
27 tháng 8 2017
Mysterious Person, Đoàn Đức Hiếu, Nguyễn Đình Dũng , ... giúp mình!
NB
25 tháng 2 2020
a)\(x^2=0\\ \Leftrightarrow x=0\)
vậy...
b)\(x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
vậy...
c)\(x^2=2\\ \Rightarrow x^2=\left(\pm\sqrt{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
vậy...
d)\(x^2=6\left(x>0\right)\\ \Rightarrow x^2=\left(\pm\sqrt{6}\right)^2\\ màx>0\\ \Rightarrow x=\sqrt{6}\)
vậy...
e)\(x^2=7\left(x< 0\right)\)
\(wtf\) ????? thông minh đấy \(x^2\ge0\) mà điều kiện lại là x < 0 ??? :D
rỗng r
f) \(\left(x+1\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
vậy....
g)\(\left(x-2\right)^2=2\\ \Rightarrow\left(x-2\right)^2=\left(\pm\sqrt{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=\sqrt{2}\\x-2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}+2\\x=-\sqrt{2}+2\end{matrix}\right.\)
tự tính :D
vậy..
h)\(\left(x+\sqrt{3}\right)^2=5\\ \Leftrightarrow\left(x+\sqrt{3}\right)^2=\left(\pm\sqrt{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\sqrt{3}=\sqrt{5}\\x+\sqrt{3}=-\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\\x=\end{matrix}\right.\)
tự tính lười lắm
29 tháng 10 2019
X:(\(\frac{2}{9}-\frac{1}{5}\))=\(\frac{8}{16}\)
x:\(\frac{1}{45}\) =\(\frac{8}{16}\)
x: =\(\frac{8}{16}.\frac{1}{45}\)
x: =\(\frac{1}{90}\)