x mũ 10 = x
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a ) \(\left(3x+\frac{1}{4}\right)^3=-27\)
\(\left(3x+\frac{1}{4}\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+\frac{1}{4}=-3\)
\(\Rightarrow3x=-3-\frac{1}{4}=-\frac{13}{4}\)
\(\Rightarrow x=-\frac{13}{4}:3=-\frac{13}{12}\)
Vậy x = \(-\frac{13}{12}\)
cho em hoi câu này xin các anh chị:
10mux x+4y = 2013
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
a: =30-22=8
b: =10*(23+17)=10*40=400
c: =21*3-7*(-14)
=63+98=161
d: =-20-[10*10*5^2+16]
=-20-100*25-16
=-36-2500
=-2536
a: x^9=x^29
=>x^29-x^9=0
=>x^9*(x^20-1)=0
=>x^9=0 hoặc x^20-1=0
=>x=0; x=1;x=-1
b: x^10=x^7
=>x^7(x^3-1)=0
=>x=0 hoặc x=1
\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{29}\cdot9^{10}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{29}\cdot3^{20}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3^2-7\right)}\)
\(=\dfrac{10-9}{5\cdot9-7}=\dfrac{1}{38}\)
=>x(x^9-1)=0
=>+)x=0
+)x^9-1=0 =>x^9=1 =>x=1
x\(^{10}\)= x
x\(^9\). x = x
Nếu x\(^9\). x = 0 thì \(\Rightarrow\)x = 0
Nếu x\(^9\). x = 1 thì \(\Rightarrow\)x = 1
Chúc bạn học tốt!