5/x-2=10/16
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NT
1
TD
3
27 tháng 12 2017
a)\(\frac{2x}{x+5}+\frac{10}{x+5}=\frac{2x+10}{x+5}=\frac{2\left(x+5\right)}{x+5}=2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{8\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{x-2}\)
PD
27 tháng 12 2017
a) \(\frac{2x}{x+5}+\frac{10}{x+5}\)=\(\frac{2x+10}{x+5}\)=\(\frac{2\left(x+5\right)}{x+5}\)=\(2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}\)=\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{\left(x+2-x+2\right)\left(x+2+x-2\right)+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4\times2x+16}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8}{x-2}\)
NG
1
28 tháng 4 2019
\(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}+\frac{x}{13\cdot16}=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[1-\frac{1}{16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\cdot\frac{15}{16}=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{2}:\frac{15}{16}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{2}\cdot\frac{16}{15}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{1}\cdot\frac{8}{3}\)
\(\Rightarrow\frac{x}{3}=\frac{8}{3}\Leftrightarrow x=8\)
NS
1
3 tháng 3 2022
a: =>x/16=3/8+7/2=3/8+28/8=31/8=62/16
=>x=62
b: =>3/5:x=17/10-2/5=17/10-4/10=13/10
=>x=3/5:13/10=3/5x10/13=30/65=6/13
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 9 2024
(-15)\(x\) = - 45
\(x\) = - 45 : (-15)
\(x\) = 3
Vậy \(x=3\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 9 2024
(-15)\(x\) = 10.(-40).(-5)
-15\(x\) = -400.(-5)
-15\(x\) = 2000
\(x\) = 2000 : (-15)
\(x\) = - \(\dfrac{400}{3}\)
Vậy \(x=-\dfrac{400}{3}\)
27 tháng 4 2020
gọi A là VT
Ta có : \(A=\left[\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)-x^4y^4\right]+\left[\frac{1}{4}\left(x^{16}+y^{16}\right)-2x^2y^2\right]-1\)
Áp dụng BĐT Cô-si,ta có :
\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)\ge\frac{1}{2}2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}=x^4y^4\Rightarrow\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)-x^4y^4\ge0\)
\(\frac{x^{16}+y^{16}}{4}\ge\frac{x^8y^8}{2}=\left(\frac{x^8y^8}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)-\frac{3}{2}\ge4\sqrt[4]{\frac{x^8y^8}{16}}-\frac{3}{2}==2x^2y^2-\frac{3}{2}\)
\(\Rightarrow\frac{1}{4}\left(x^{16}+y^{16}\right)-2x^2y^2\ge\frac{-3}{2}\)
Từ đó ta có : \(A\ge0-\frac{3}{2}-1=\frac{-5}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=y\\x^2y^2=1\end{cases}\Leftrightarrow x=y=\pm1}\)
`5/[x-2]=10/16`
`10(x-2)=16.5=80`
`x-2=80:10=8`
`x=8+2=10`
\(\dfrac{5}{x-2}=\dfrac{10}{16}\)
\(\Rightarrow5.16=\left(x-2\right).10\)
\(\Rightarrow\left(x-2\right).10=80\)
\(\Rightarrow x-2=80:10\)
\(\Rightarrow x-2=8\)
\(\Rightarrow x=8+2\)
\(\Rightarrow x=10\)