Tìm x biết: x - 234 = 12 ?
A. 246
B. 264
C. 462
D. 426
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a)Ta có:
\(3^x-3^{x-3}=-234\)
\(\Rightarrow3^x-3^x\cdot3^3=-234\)
\(\Rightarrow3^x\cdot\left(1-3^3\right)=-234\)
\(\Rightarrow3^x\cdot\left(-26\right)=-234\)
\(\Rightarrow3^x=9\)
\(\Rightarrow x=2\)
Vậy x=2
\(\Rightarrow3^x=3^2\)
b) Ta có:
\(2^{x+1}\cdot3^x-6^x=216\)
\(\Rightarrow2^x\cdot2\cdot3^x-2^x\cdot3^x=216\)
\(\Rightarrow\left(2^x\cdot3^x\right)\cdot\left(2-1\right)=216\)
\(\Rightarrow6^x\cdot1=216\)
\(\Rightarrow6^x=6^3\)
\(\Rightarrow x=3\)
Vậy x=3
a) 4 1 5 + 3 2 5 = 21 5 + 17 5 = 38 5
b) 1 2 − 1 8 − 1 16 = 8 16 − 2 16 − 1 16 = 5 16
c) 426 × 305 = 129930
d) 72306 : 351 = 206
a) \(\left(x+1\right)\left(x+2\right)=272\)
\(\Rightarrow x^2+3x+2=272\)
\(\Rightarrow x^2+3x-270=0\)
\(\Rightarrow x^2+18x-15x-270=0\)
\(\Rightarrow x\left(x+18\right)-15\left(x+18\right)=0\)
\(\Rightarrow\left(x+18\right)\left(x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+18=0\\x-15=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-18\\x=15\end{matrix}\right.\)
d) \(\left(x+4\right)\left(x+5\right)=552\)
\(\Rightarrow x^2+9x+20=552\)
\(\Rightarrow x^2+9x-532=0\)
\(\Rightarrow x^2+28x-19x-532=0\)
\(\Rightarrow x\left(x+28\right)-19\left(x+28\right)=0\)
\(\Rightarrow\left(x+28\right)\left(x-19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+28=0\\x-19=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-28\\x=19\end{matrix}\right.\)
1,
a,= (25.4).(-2.5).(-37)= 100.(-10).(-37)= 37000
b,= 34(156-56)= 34.100=3400
c,= 47(-1-234+234)= 47.(-1)= -47
d,= (25.4).(-5.(-2)).13= 100.10.13= 13000
e,= 45(123-23)= 45.100= 4500
f,= 65(-1+246-246)= 65.(-1)= -65
2,
a, -> x= -21+15= -6
b, -> x= \(\dfrac{-28+12}{4}\)= -4
c, -> x= -9-5-6= -20
d, -> x= -31+22= -9
e, -> x= -7-8-4= -19
f, -> x= \(\dfrac{31-25}{-3}\)= -2
\(126=2\cdot3^2\cdot7\\ 462=2\cdot3\cdot7\cdot11\\ 264=2^3\cdot3\cdot11\\ ƯCLN\left(126,264,462\right)=2\cdot3=6\)
Đáp án A