2 x 5 x 8
15 x 10 x 16
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đáp án:
Giải thích các bước giải:
Ta có:
Vậy .
a ) − 13 30 ≤ x < 1 = > x = 0. b ) 5 9 < x ≤ 3 = > x ∈ {1;2;3}
1 6 − 2 5 < x 30 ≤ 1 2 − 8 15 ⇒ 5 30 − 12 30 < x 30 ≤ 15 30 − 16 30 ⇒ − 7 30 < x 30 ≤ − 1 30 ⇒ − 7 < x ≤ − 1 ⇒ x ∈ − 6 ; − 5 ; − 4 ; − 3 ; − 2 ; − 1
a) − 2 5 + 1 6 + − 1 5 ≤ x < − 3 4 + 9 7 + − 1 4 + 5 7 ⇔ − 13 30 ≤ x ≤ 1 ⇔ x ∈ 0 ; 1
b) 5 17 + − 4 9 + 12 17 < x ≤ − 3 7 + 7 15 + 4 − 7 + 8 15 + 9 3 ⇔ 5 9 < x ≤ 3 ⇔ x ∈ 1 ; 2 ; 3
a) x-32:16=48
x-2=48
x=48+2
x=50
Vậy x \(\varepsilon\) {50}
b)(x-32):16=48
x-32=48.16
x-32=768
x=768+32
x=800
Vậy x \(\varepsilon\) {800}
c)Thiếu đầu mối.
d)815-(x-306)=713
x-306=815-713
x-306=102
x=102+306
x=408
Vậy x \(\varepsilon\) {408}.
Tích đúng nếu bạn thấy nó đúng chuẩn nhé.
1)
`x^2 -144=0`
`<=> x^2 =144`
\(< =>\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\)
2)`
`2x^2 -72=0`
`<=>2x^2 =72`
`<=>x^2=36`
\(< =>\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
3)
`5x^2 -125=0`
`<=> 5x^2 =125`
`<=>x^2 =25`
\(< =>\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
4)
`-x^2 +81=0`
`<=> x^2 -81=0`
`<=> x^2 =81`
\(< =>\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
5)
`x(2x-18)=0`
\(< =>\left[{}\begin{matrix}x=0\\2x-18=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
a)\(\frac{2x}{x+5}+\frac{10}{x+5}=\frac{2x+10}{x+5}=\frac{2\left(x+5\right)}{x+5}=2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{8\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{x-2}\)
a) \(\frac{2x}{x+5}+\frac{10}{x+5}\)=\(\frac{2x+10}{x+5}\)=\(\frac{2\left(x+5\right)}{x+5}\)=\(2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}\)=\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{\left(x+2-x+2\right)\left(x+2+x-2\right)+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4\times2x+16}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8}{x-2}\)
\(\frac{2\times5\times8}{15\times10\times16}\)\(=\)\(\frac{2\times5\times8}{3\times5\times2\times5\times2\times8}\)
\(=\)\(\frac{1}{3\times5\times2}\)
\(=\)\(\frac{1}{30}\)
2x5x8/15x10x16=2x5x8/15x2x5x8x2=1/15x2=1/30 Mình không viết dấu gạch ngang được nên thay bằng dấu gạch chéo nhé.