(1 + 2/3) x (1 + 2/4) x (1 + 2/5) x ... x (1 + 2/97) x (1 + 2/98)
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a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)
\(\Leftrightarrow x+100=0\)
hay x=-100
câu 1
=> x+1/2+x+1/3+x+1/4-x-1/5-x-1/6=0
=> (x+x+x-x-x)+(1/2+1/3+1/4-1/5-1/6)=0
=> x+43/60=0
=> x = -43/60
câu dưới làm tương tự bạn nhé!
Kết quả : ( 1 + 2/3 ) x ( 1 + 2/4 ) x ( 1 + 2/5 ) x ... x ( 1 + 2/97 ) x ( 1 + 2/98 )
= ( 3/3 + 2/3 ) x ( 4/4 + 2/4 ) x ( 5/5 + 2/5 ) x .... x ( 97/97 + 3/97 ) x ( 98/98 + 2/98 )
= 5/3 x 6/4 x 7/5 x ... x 99/97 x 100/98
= ( 5 x 6 x 7 x ...x 99 x 100 ) / ( 3 x 4 x 5 x ... x 97 x 98 )
= 99 x 100 / 3 x 4
= 33 x 25 / 1 x 1
= 825
Tham khảo nha !!!
Ai nhanh và đúng nhất mình k cho.
Mình đang cần gấp!!!!!
Bài 1:
a: Tổng là:
(-19+19)+(-18+18)+...+20=20
b: Tổng là:
-18+(-17+17)+...+0=-18
\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\)
<=> x=-100
ko chép đề nhé
\(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95} \)
=> \((x+100)(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95})=0\)
vì \((\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}) khác 0\)
=>\(x+100=0\)
=>x=-100
Ta có \(1\frac{1}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{6}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{28}{15}x=-\frac{56}{125}\)
<=> \(x=-\frac{2}{15}\)
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
<=> \(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
<=> x + 100 = 0
<=> x = -100
( 1 + 2/3 ) x ( 1 + 2/4 ) x ( 1 + 2/5 ) x ... x ( 1 + 2/97 ) x ( 1 + 2/98 )
= ( 3/3 + 2/3 ) x ( 4/4 + 2/4 ) x ( 5/5 + 2/5 ) x .... x ( 97/97 + 3/97 ) x ( 98/98 + 2/98 )
= 5/3 x 6/4 x 7/5 x ... x 99/97 x 100/98
= ( 5 x 6 x 7 x ...x 99 x 100 ) / ( 3 x 4 x 5 x ... x 97 x 98 )
= 99 x 100 / 3 x 4
= 33 x 25 / 1 x 1
= 825