d) (x - 5)^4 = (x-5)^6
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\(a,\dfrac{3}{4}:\dfrac{6}{x}=\dfrac{5}{4}\\ \dfrac{6}{x}=\dfrac{3}{4}:\dfrac{5}{4}\\ \dfrac{6}{x}=\dfrac{3}{5}\\ x=6:\dfrac{3}{5}\\ x=10\\ b,\left(x+\dfrac{7}{4}\right)\times\dfrac{2}{3}=6\\ x+\dfrac{7}{4}=6:\dfrac{2}{3}\\x+\dfrac{7}{4}=9\\ x=9-\dfrac{7}{4}\\ x=\dfrac{29}{4}\)
\(x\times\dfrac{3}{5}=\dfrac{2}{3}+\dfrac{1}{6}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}:\dfrac{3}{5}=\dfrac{25}{18}\)
a, \(\dfrac{4x}{7}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{3+10}{15}=\dfrac{13}{15}\Rightarrow60x=91\Leftrightarrow x=\dfrac{91}{60}\)
b, \(\dfrac{5}{7}:x=\dfrac{1}{6}+\dfrac{4}{5}=\dfrac{5+24}{30}=\dfrac{29}{30}\Leftrightarrow x=\dfrac{5}{7}:\dfrac{29}{30}=\dfrac{150}{203}\)
\(\left(y-\dfrac{2}{3}\right):\dfrac{4}{5}=\dfrac{5}{6}\)
\(y-\dfrac{2}{3}=\dfrac{5}{6}\times\dfrac{4}{5}\)
\(y-\dfrac{2}{3}=\dfrac{2}{3}\)
\(y=\dfrac{2}{3}+\dfrac{2}{3}\)
\(y=\dfrac{4}{3}\)
a) 6/-x=x/-24
=> -x.x=6.-24
=>-x.x=-144
=>x=12 hay x=-12
b)9/x=-35/105
=>9/x=-1/3
=>x=9.3/-1=-27
=>x=-27
c)x-1/8=5/8
=>x=5/8+1/8
=>x=3/4
d)x-1/2-(3/2+x)=-2
=>-x+4/2=-2
=>-x/2=0
=>x=0
e)x+1/3=-12/5.10/6
x+1/3=-4
x=-4-1/3
x= -13/3
#YM
\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{29}\cdot9^{10}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{29}\cdot3^{20}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3^2-7\right)}\)
\(=\dfrac{10-9}{5\cdot9-7}=\dfrac{1}{38}\)
15/8−4/9×6/8+1/6=15/8−24/72+1/6=15/8−1/3+1/6
=45/24−8/24+4/24=(45−8+4)/24=41/24
Ta có thể chia hai vế cho \(\left(\right. x - 5 \left.\right)^{4}\) (với điều kiện \(x \neq 5\)):
\(\frac{\left(\right. x - 5 \left.\right)^{4}}{\left(\right. x - 5 \left.\right)^{4}} = \frac{\left(\right. x - 5 \left.\right)^{6}}{\left(\right. x - 5 \left.\right)^{4}} \Rightarrow 1 = \left(\right. x - 5 \left.\right)^{2}\)\(\mid x-5\mid=1\Rightarrow x-5=\pm1\Rightarrow x=6hoặcx=4\)Ngoài ra, với x = 5 thì cả hai vế bằng 0 nên phương trình cũng đúng.
vậy
x = 4, x = 5, x = 6
nhé bạn
(\(x-5\))\(^4\) = (\(x-5)^6\)
(\(x-5\))\(^4\) - (\(x-5)^6\) = 0
(\(x-5\))\(^4\)[1 - (\(x-5\))\(^2\)] = 0
\(\left[\begin{array}{l}x-5=0\\ x-5=-1\\ x-5=1\end{array}\right.\)
\(\left[\begin{array}{l}x=5\\ x=4\\ x=6\end{array}\right.\)
Vậy \(x\in\) {4; 5; 6}