giúp mình với
1,Tìm x
a,|x| = 1/2 = 3/4 b,|x – 1/2| – 3/4 = 1/4
c,1/20 – |3/2- x| = 1/40 d, ( x + 1/3 ) ² + 4/9 = 8/9
2.Tính hợp lí
a,A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8
b,B = 2/1.3 + 2/3.5 + …… + 2/11.13
Làm nhanh nha mik đang cần rất rất gấp
Đáp án:
1.
a) `x in { -5/4 ; 5/4 }`
b) `x in { -1/2 ; 3/2 }`
c) `x in { 59/40 ; 61/40 }`
d) `x in { -1 ; 1/3 }`
2.
a) `A = 7/8`
b) `B = 12/13`
Giải thích các bước giải:
1.
a) `|x| – 1/2 = 3/4`
`to |x| = 3/4 + 1/2`
`to |x| = 5/4`
`to x = 5/4` or `x = -5/4`
Vậy `x in { -5/4 ; 5/4 }`
b) `|x-1/2| – 3/4 = 1/4`
`to |x-1/2| = 1/4 + 3/4`
`to |x-1/2| = 1`
`to` \(\left[ \begin{array}{l}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=\dfrac{3}{2} \\x= -\dfrac{1}{2}\end{array} \right.\)
Vậy `x in { -1/2 ; 3/2 }`
c) `1/20 – | 3/2 – x | = 1/40`
`to | 3/2 – x | = 1/2 – 1/40`
`to | 3/2 – x | = 1/40`
`to ` \(\left[ \begin{array}{l}\dfrac{3}{2}-x=\dfrac{1}{40}\\\dfrac{3}{2}-x=-\dfrac{1}{40}\end{array} \right.\) `to `\(\left[ \begin{array}{l}x = \dfrac{59}{40} \\x=\dfrac{61}{40}\end{array} \right.\)
Vậy `x in { 59/40 ; 61/40 }`
d) `(x+1/3)^2 + 4/9 = 8/9`
`to (x+1/3)^2 = 8/9 – 4/9`
`to (x+1/3)^2 = 4/9`
`to ` \(\left[ \begin{array}{l}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=-\dfrac{2}{3}\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-1\end{array} \right.\)
Vậy `x in { -1 ; 1/3 }`
$\\$
2.
a) `A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8`
`= 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + 1/5 – 1/6 + 1/6 – 1/7 + 1/7 – 1/8`
`= 1 – 1/8 = 7/8`
Vậy `A = 7/8`
b) `B = 2/1.3 + 2/3.5 + … + 2/11.13`
`= 1 – 1/3 + 1/3 – 1/5 + ….. + 1/11 – 1/13`
`= 1 – 1/13 = 12/13`
Vậy `B = 12/13`
$1$.
$a$) `|x| – 1/2 = 3/4`
`⇔ |x| = 3/4 + 2/4`
`⇔ |x| = 5/4`
`⇒` ` x = ± 5/4`
$b$) `|x-1/2| – 3/4 = 1/4`
`⇔ |x-1/2| = 1/4 + 3/4`
`⇔ |x-1/2| = 1`
`⇒` \(\left[ \begin{array}{l}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=\dfrac{3}{2} \\x= -\dfrac{1}{2}\end{array} \right.\)
$c$) `1/{20} – | 3/2 – x| = 1/{40}`
`⇔ |3/2 – x| = 1/{20} – 1/{40}`
`⇔ |3/2 – x| = 1/{40}`
`⇒` \(\left[ \begin{array}{l}\dfrac{3}{2}-x=\dfrac{1}{40}\\\dfrac{3}{2}-x=\dfrac{-1}{40}\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x = \dfrac{59}{40} \\x=\dfrac{61}{40}\end{array} \right.\)
$d$) `(x+1/3)^2 + 4/9 = 8/9`
`⇔ (x+1/3)^2 = 8/9 – 4/9`
`⇔ (x+1/3)^2 = 4/9`
`⇔ (x+1/3)^2 = (± 2/3)^2`
`⇒` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-1\end{array} \right.\)
$2$.
$a$) `A = 1/{1.2} + 1/{2.3} + 1/{3.4} + 1/{4.5} + 1/{5.6} + 1/{6.7} + 1/{7.8}`
`⇔ A = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + 1/5 – 1/6 + 1/6 – 1/7 + 1/7 – 1/8`
`⇔ A = 1 – 1/8`
`⇔ A = 7/8`.
$b$) `B = 2/{1.3} + 2/{3.5} + ….. + 2/{11.13}`
`⇔ B = 1 – 1/3 + 1/3 – 1/5 + ….. + 1/11 – 1/13`
`⇔ B = 1 – 1/{13}`
`⇔ B = 12/13`.