Bài 1: a)

\(\left\{\left[\left(2\div x+14\right)\div2^2-3\right]\div2\right\}-1=0\)

\(\Rightarrow\left[\left(2\div x+14\right)\div2^2-3\right]\div2=1\)

\(\Rightarrow\left(2\div x+14\right)\div2^2-3=2\)

\(\Rightarrow\left(2\div x+14\right)\div2^2=5\)

\(\Rightarrow2\div x+14=5\times2^2=20\)

\(\Rightarrow2\div x=6\Rightarrow x=\dfrac{1}{3}\)

Vậy \(x=\dfrac{1}{3}\)

Câu b tương tự nhan bạn!!!

Trên hình vẽ ta có :

\(\widehat{xAy}\) kề bù với góc \(\widehat{x'Ay}\) , nên

\(\widehat{xAy}+\widehat{x'Ay}=180^o\)

\(\Rightarrow\widehat{x'Ay}=180^o-\widehat{xAy}=180^o-30^o=150^o\)

Vậy góc x'Ay là \(150^o\)

Dựa vào " hai góc đối đỉnh" thì, ta có:

\(\widehat{xAy}\) đối đỉnh \(\widehat{x'Ay'}\) ( = \(30^o\) )

và \(\widehat{x'Ay}\) đối đỉnh \(\widehat{xAy'}\) ( = \(150^o\) )

~ Học tốt ~

2) Ta có :

\(a=2016\times2016=\left(2014+2\right)\times2016=2016\times2014+4032\)

\(b=2014\times2018=2014\times\left(2016+2\right)=2014\times2016+4028\)

Vì \(2016\times2014+4032>2014\times2016+4028\)

Nên \(2016\times2016>2014\times2018\)

Vậy \(2016\times2016>2014\times2018\)

~ Học tốt ~

1) \(1+2+3+4+....+x=55\)

\(\Rightarrow\left(x+1\right)x\div2=55\)

\(\Rightarrow\left(x+1\right)x=55\times2=110\)

\(\Rightarrow\left(x+1\right)x=11\times10\)

\(\Rightarrow x=10\)

Vậy x = 10

2/ Ta có : \(A=\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{99\times100}\)

\(=\left(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}\right)+\left(\dfrac{1}{5\times6}+....+\dfrac{1}{99\times100}\right)\)

\(=\dfrac{7}{12}+\left(\dfrac{1}{5\times6}+....+\dfrac{1}{99\times100}\right)>\dfrac{7}{12}\) (1)

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+.....+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{100}\right)\times2\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)\)

\(=\dfrac{1}{51}+\dfrac{1}{52}+....+\dfrac{1}{100}\)

Dãy số trên có : \(100-51+1=50\) số hạng.

mà 50 chia hết cho 10 nên ta nhóm 10 số vào một nhóm.

\(A=\left(\dfrac{1}{51}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+...+\dfrac{1}{70}\right)+\left(\dfrac{1}{71}+...+\dfrac{1}{80}\right)+\left(\dfrac{1}{81}+...+\dfrac{1}{90}\right)+\left(\dfrac{1}{91}+...+\dfrac{1}{100}\right)\)

\(< \dfrac{1}{50.10}+\dfrac{1}{60.10}+\dfrac{1}{70.10}+\dfrac{1}{80.10}+\dfrac{1}{90.10}=\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7.3}=\dfrac{167}{210}< \dfrac{175}{210}=\dfrac{5}{6}\)

\(\Rightarrow A< \dfrac{5}{6}\) (2)

Từ (1) và (2) \(\Rightarrow\) đpcm

1/ Ta có :

\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{49\times50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{50}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\times2\)

\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)

\(\Rightarrow\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}\)

Hay \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...+\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)

~ Học tốt nha ~

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy \(x=-2004\)

1) Ta có : \(5^{27}=\left(5^3\right)^9=125^9\)

\(2^{63}=\left(2^7\right)^9=128^9\)

Vì \(125^9< 128^9\Rightarrow5^{27}< 2^{63}\)

Vậy \(5^{27}< 2^{63}\)

2) Ta có : \(2^{63}=\left(2^9\right)^7=512^7\)

\(5^{28}=\left(5^4\right)^7=625^7\)

Vì \(512^7< 625^7\Rightarrow2^{63}< 5^{28}\)

Vậy \(2^{63}< 5^{28}\)

Chúc bạn học tốt !!

Giải : Phân số chỉ bạn thứ ba góp được là :

\(1-\left(\dfrac{1}{5}+60\%\right)=\dfrac{1}{5}\)

Cả ba bạn góp được :

\(16000:\dfrac{1}{5}=80000\) (đồng)

Vậy.......

tick cho minh roi minh lam cho