tính nhanh:
18.73+24.56+81.27+75.44
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a) Đề:........
= 18. (73 + 15 + 12)
= 18. 100
= 18000
b) Đề nên sửa 23 thành 25
= 100 : { 250 : [ 450 - (4. 125 - 4.25)]}
= 100 : { 250 : [ 450 - (500 - 100)]}
= 100 : { 250 : [ 450 - 400 ]}
= 100 : { 250 : 50 }
= 100 : 5
= 20
a) 18.73+15.18+12.18
=18.(73+15+12)
=18.100
=1800
b) (37+63):{250:[450-(4.53-22.23)]}
=(37+63):{250:[450-(22.53-22.23)]}
=100:{250:[450-22.(53-23)]}
=100:{250:[450-22.102]}
=100:{250:[450-408]}
=100:{250:42}
=100:\(\frac{125}{21}\)
=\(\frac{84}{5}\)
a,714+382+286+318
\(=\left(714+286\right)+\left(382+318\right)\)
\(=1000+700\)
= \(1700\)
b,18.73+15.18+12.18
\(=18.\left(73+15+12\right)\)
\(=18.100\)
\(=1800\)
c,(37+63) : {250:[450-(4.5^3-2^2.25)]}
\(=100:\left\{250:\left[450-\left(2^2.125-2^2.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(2^2.\left(125-25\right)\right)\right]\right\}\)
\(=100:\left\{250:\left[450-400\right]\right\}\)
\(=100:\left\{250:50\right\}\)
\(=100:5\\ =20\)
CHÚC BANJ HỌC TỐT!!
a) 714+382+286+318
=(714+286)+(382+318)
=1000+700
=1700
b) 18.73+15.18+12.18
=18.(73+15+12)
=18.100
=1800
c) (37+63):{250:[450-(4.5^3-2^2.25)]}
=100:{250:[450-(4.125-4.25)]}
=100:{250:[450-4.(125-25)]}
=100:{250:[450-4.100]}
=100:{250:[450-400]}
=100:{250:50}
=100:5
=20
-Chúc bạn học tốt-
a/
\(9.3^2.\frac{1}{81}.27=\frac{9.3^2.27}{81}=\frac{3^2.3^2.3^3}{3^4}=\frac{3^7}{3^4}=3^3\)
b/
\(4.32:\left(2^3.\frac{1}{16}\right)=4.32:\left(\frac{2^3}{16}\right)=4.32:\left(\frac{2^3}{2^4}\right)=4.32:\frac{1}{2}=4.32.2=4.64=4.4^3=4^4\)
c/
\(3^4.3^5:\frac{1}{27}=3^4.3^5.27=3^4.3^5.3^3=3^{12}\)
d/(ý bạn là (-2)^2 hay -2^2 , mình làm theo cách (-2)^2 nhé!)
\(2^2.4.\frac{32}{\left(-2\right)^2}.2^5=2^2.2^2.\frac{2^5}{2^2}.2^5=2^2.2^2.2^3.2^5=2^{12}\)
\(a,\left(\dfrac{1}{4}\right)^{x-2}=\sqrt{8}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^{2x-4}=\left(\dfrac{1}{2}\right)^{-\dfrac{3}{2}}\\ \Leftrightarrow2x-4=-\dfrac{3}{2}\\ \Leftrightarrow2x=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{4}\)
\(b,9^{2x-1}=81\cdot27^x\\ \Leftrightarrow3^{4x-2}=3^{4+3x}\\ \Leftrightarrow4x-2=4+3x\\ \Leftrightarrow x=6\)
c, ĐK: \(x-2>0\Rightarrow x>2\)
\(2log_5\left(x-2\right)=log_59\\
\Leftrightarrow log_5\left(x-2\right)^2=log_59\\
\Leftrightarrow\left(x-2\right)^2=3^2\\
\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\\
\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = 5.
d, ĐK: \(x-1>0\Leftrightarrow x>1\)
\(log_2\left(3x+1\right)=2-log_2\left(x-1\right)\\ \Leftrightarrow log_2\left(3x+1\right)\left(x-1\right)=2\\ \Leftrightarrow3x^2-2x-1=4\\ \Leftrightarrow3x^2-2x-5=0\\ \Leftrightarrow\left(3x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm \(x=\dfrac{5}{3}\)
18,73+24,56+81,27+75,44
=[18,73 +81,27]+[24,56+75,44]
= 10 ,00 + 10,00
= 10+10
= 20
CHO MÌNH NHÉ