x=1-2+3-4+...+97-98
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a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
Chúc bạn học tốt
có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai
`(x+1)/99+(x+2)/98+(x+3)/97+(x+4)/96=-4`
`=>(x+1)/99+1+(x+2)/98+1+(x+3)/97+1+(x+4)/96+1=-4+4`
`=>(x+100)/99+(x+100)/98+(x+100)/97+(x+100)/96=0`
`=>(x+100)(1/99+1/98+1/97+1/96)=0`
`=>x+100=0` (Vì `1/99+1/98+1/97+1/96\ne0`)
`=>x=-100`
Vậy ...
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\\ \dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\\ \left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\\ \dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\\ \left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)
mà `1/99+1/98+1/97+1/96 \ne 0`
nên `x+100=0`
`x=-100`
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
(x+1)/99 + (x+2)/98 + (x+3)/97 + (x+4)/96 = -4`
x+199+x+298+x+397+x+496+4=0𝑥+199+𝑥+298+𝑥+397+𝑥+496+4=0
x+199+1+x+298+1+x+397+1 +x+496+1=0𝑥+199+1+𝑥+298+1+𝑥+397+1 +𝑥+496+1=0
x+10099+x+10098+x+10097+x+10096=0𝑥+10099+𝑥+10098+𝑥+10097+𝑥+10096=0
(x+100)(199+198+197+196)=0(𝑥+100)(199+198+197+196)=0
Vì 199+198+197+196≠0199+198+197+196≠0
⇒ x+100=0⇒ 𝑥+100=0
⇒x=0−100⇒𝑥=0-100
⇒x=−100⇒𝑥=-100
Vậy 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
ko phai menh de tap hop dau nha to bam lon day . moi nguoi giup to bai nay voi
Ta có:
x= (1+3+...+97)-(2+4+...+98)
Đặt A=1+3+...+97
B= 2+4+...+98
Tổng A có số số hạng là: (97-1):2+1=49 (số hạng)
A= (97+1).49:2=2401.
Tổng B có số số hạng là: (98-2):2+1=49 (số hạng)
B= (98+2).49:2=2450.
Vậy x= A-B=2401-2450=-49.
(Bạn có thể làm bằng cách nhóm 2 số hạng có cùng hiệu)
Chúc bạn học tốt!