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a: \(\frac{x-100}{24}+\frac{x-98}{26}+\frac{x-96}{28}=3\)
=>\(\left(\frac{x-100}{24}-1\right)+\left(\frac{x-98}{26}-1\right)+\left(\frac{x-96}{28}-1\right)=0\)
=>\(\frac{x-124}{24}+\frac{x-124}{26}+\frac{x-124}{28}=0\)
=>\(\left(x-124\right)\left(\frac{1}{24}+\frac{1}{26}+\frac{1}{28}\right)=0\)
=>x-124=0
=>x=124
b: \(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)
=>\(\left(\frac{x-1}{65}-1\right)+\left(\frac{x-3}{63}-1\right)=\left(\frac{x-5}{61}-1\right)+\left(\frac{x-7}{59}-1\right)\)
=>\(\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)
=>\(\left(x-66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
=>x-66=0
=>x=66
c: \(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}=3\)
=>\(\left(\frac{x-28-124}{2011}-1\right)+\left(\frac{x-124-2011}{28}-1\right)+\left(\frac{x-28-2011}{124}-1\right)=0\)
=>x-28-124-2011=0
=>x=2011+124+28
=>x=2163
Ta có: \(\hat{bFE}+\hat{bFc}=180^0\) (hai góc kề bù)
=>\(\hat{bFE}=180^0-120^0=60^0\)
Ta có: \(\hat{bFE}=\hat{xEF}\left(=60^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên ax//by






Bài 2:
1: \(2^4\cdot5-\left\lbrack131-\left(13-4\right)^2\right\rbrack\)
\(=16\cdot5-\left\lbrack131-9^2\right\rbrack\)
=80-(131-81)
=80-50
=30
2: \(2^3+3\cdot\left(\frac12\right)^0-1+\left\lbrack\left(-2\right)^2:\frac12\right\rbrack-8\)
\(=8+3\cdot1-1+4\cdot2-8\)
=8+3-1+8-8
=8+3-1
=8+2
=10
3: \(\left(\frac14\right)^2+25\cdot\left\lbrack\left(\frac34\right)^3:\left(\frac54\right)^3\right\rbrack:\left(\frac32\right)^3\)
\(=\frac{1}{16}+25\cdot\left\lbrack\left(\frac34:\frac54\right)^3\right\rbrack:\left(\frac32\right)^3\)
\(=\frac{1}{16}+25\cdot\left(\frac35\right)^3\cdot\left(\frac23\right)^3=\frac{1}{16}+25\cdot\left(\frac35\cdot\frac23\right)^3\)
\(=\frac{1}{16}+25\cdot\left(\frac25\right)^3=\frac{1}{16}+25\cdot\frac{8}{125}=\frac{1}{16}+\frac85\)
\(=\frac{5}{80}+\frac{126}{80}=\frac{131}{80}\)
4: \(5-\left(-\frac{5}{11}\right)^0+\left(\frac13\right)^2:3\)
\(=5-1+\frac19:3\)
\(=4+\frac{1}{27}=\frac{109}{27}\)
5: \(\left(-8\right)^2:\left\lbrace25-18:\left\lbrack\left(5^2+2^3\right):11-2018^0\right\rbrack\right\rbrace\)
\(=64:\left\lbrace25-18:\left\lbrack\frac{\left(25+8\right)}{11}-1\right\rbrack\right\rbrace\)
\(=64:\left\lbrace25-18:\left(\frac{33}{11}-1\right)\right\rbrace=64:\left\lbrace25-18:2\right\rbrace=\frac{64}{25-9}=\frac{64}{16}=4\)
6: \(2^4+8\cdot\left\lbrack\left(-2\right)^2:\frac12\right\rbrack^0-2^{-2}\cdot4+\left(-2\right)^2\)
\(=16+8-\frac14\cdot4+4\)
=24+4-1
=24+3
=27
Bài 1:
1: \(\left(-\frac12+\frac13\right)^3=\left(-\frac36+\frac26\right)^3=\left(-\frac16\right)^3=-\frac{1}{216}\)
2: \(\left(-\frac12-1\right)^2=\left(-\frac32\right)^2=\frac{\left(-3\right)^2}{2^2}=\frac94\)
3: \(\left(-\frac13\right)^6:\left(-\frac13\right)^4:\left(-\frac13\right)=\left(-\frac13\right)^{6-4-1}=\left(-\frac13\right)^1=-\frac13\)
4: \(\left(1-\frac13\right)^4=\left(\frac23\right)^4=\frac{2^4}{3^4}=\frac{16}{81}\)
5: \(\left(1-\frac47\right)^2=\left(\frac37\right)^2=\frac{3^2}{7^2}=\frac{9}{49}\)
6: \(\left(-1\frac12\right)^3:\left(\frac12\right)^3=\left(-\frac32\right)^3:\left(\frac12\right)^3=\left(-\frac32:\frac12\right)^3=\left(-3\right)^3=-27\)
7: \(\left(-\frac56\right)^7:\left(-\frac56\right)^5=\left(-\frac56\right)^{7-5}=\left(-\frac56\right)^2=\frac{25}{36}\)
8: \(\left(\frac27\right)^8:\left(\frac27\right)^7\cdot\frac27=\left(\frac27\right)^{8-7+1}=\left(\frac27\right)^2=\frac{4}{49}\)
9: \(\left(-1,5\right)^4:\left(-1,5\right)^4\cdot\left(-1.5\right)^2=\left(-1,5\right)^2=2,25\)
10: \(\left(-2\right)^3\cdot2^5:\left(2^2\right)^3=-2^3\cdot2^5:2^6=-2^{3+5-6}=-2^2=-4\)
11: \(\left(\frac25\right)^8:\left(\frac{4}{25}\right)^3=\left(\frac25\right)^8:\left(\frac25\right)^6=\left(\frac25\right)^{8-6}=\left(\frac25\right)^2=\frac{4}{25}\)
12: \(\left(\frac13\right)^5:\frac{1}{81}=\left(\frac13\right)^5:\left(\frac13\right)^4=\left(\frac13\right)^{5-4}=\left(\frac13\right)^1=\frac13\)