K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
AD
2
T
9 tháng 7 2019
#)Giải :
\(\left(x+2\right)^{n+1}=\left(x+2\right)^{n+11}\)
\(\Rightarrow\left(x+2\right)^{n+1}-\left(x+2\right)^{n+11}=0\)
\(\Rightarrow\left(x+2\right)^{n+1}.\left[1-\left(x+2\right)^{10}\right]=0\)
\(\Rightarrow\left(x+2\right)^{n+1}=0\)hoặc \(1-\left(x+2\right)^{10}=0\)
Với \(1-\left(x+2\right)^{10}=0\Rightarrow x+2=0\Rightarrow x=-2\)
Với \(1-\left(x+2\right)^{n+1}=0\Rightarrow\left(x+2\right)^{10}=1\Rightarrow\orbr{\begin{cases}x+2=1\\x+2=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}}\)
H
1
N
9 tháng 8 2019
\(2x-\frac{5}{9}=\frac{1}{3}+\text{ }[\frac{11}{3}-4+\frac{2}{3}]\)
\(2x-\frac{5}{9}=\frac{1}{3}+\text{ }[-\frac{1}{3}+\frac{2}{3}]\)
\(2x-\frac{5}{9}=\frac{1}{3}+\frac{1}{3}\)
\(2x-\frac{5}{9}=\frac{2}{3}\)
\(2x=\frac{11}{9}\)
\(x=\frac{11}{18}\)
Vậy \(x=\frac{11}{18}\)
A
18 tháng 8 2021
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.\)
\(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)(cộng 2 vế cho 3)
\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}+\frac{x+3}{2007}+\frac{2007}{2007}=\frac{x+10}{2000}+\frac{2000}{2000}+\frac{x+11}{1999}+\frac{1999}{1999}+\frac{x+12}{1998}+\frac{1998}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
x+2010=0
x=-2010
18 tháng 8 2021
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)
\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)
\(=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
16 tháng 11 2019
a) ta có: \(VT=\left|x-2\right|+\left|x+7,5\right|=\left|2-x\right|+\left|x+7,5\right|\le\left|2-x+x+7,5\right|=9=VP.\)
Muốn \(\left|x-2\right|+\left|x+7,5\right|=9\)thì \(x=0\)
b) (Cái này mình không biết đúng hay không, nếu không thì các bạn ý kiến nha!)
+) Giả sử x = 0: \(PT\Rightarrow2\left|0+3\right|+\left|2\cdot0\right|+5=6+5=11\)(đúng)
+) Giả sử x > 0:
\(PT\Leftrightarrow2\left(x+3\right)+2x+5=11\)
\(\Leftrightarrow2x+6+2x+5=11\)
\(\Leftrightarrow4x+11=11\)
\(\Leftrightarrow4x=0\Rightarrow x=0\)
+) Giả sử x < 0:
\(PT\Leftrightarrow-2\left(x+3\right)-2x-5=11\)
\(\Leftrightarrow-2x-6-2x-5=11\)
\(\Leftrightarrow-4x-11=11\)
\(\Leftrightarrow-4x=22\Rightarrow x=-\frac{11}{2}\)
Thử lại: \(2\left|-\frac{11}{2}+3\right|+\left|-\frac{2.11}{2}+5\right|=\frac{2.5}{2}+6=5+6=11\)(đúng)
Vậy x = 0 hoặc \(x=-\frac{11}{2}\)
30 tháng 8 2015
câu b nha
B= 1/100 - (1/2.1 + 1/3.2 + ... + 1/98.97 + 1/99.98 + 1/100.99)
B=1/100 - (1 - 1/2 + 1/2 - 1/3 + 1/3 - ... - 1/99 + 1/99 - 1/100)
B=1/100-(1-1/100)
B=1/100-99/100
B= - 98/100
B= - 49/50
đ ú g nha
\(\dfrac{11}{5}-\left(0,35+x\right)=1\dfrac{1}{2}\\ \dfrac{11}{5}-\left(\dfrac{7}{20}+x\right)=\dfrac{3}{2}\\ \dfrac{11}{5}-\dfrac{7}{20}-x=\dfrac{3}{2}\\ \dfrac{44}{20}-\dfrac{7}{20}-x=\dfrac{3}{2}\\ \dfrac{37}{20}-x=\dfrac{3}{2}\\ x=\dfrac{37}{20}-\dfrac{3}{2}\\ x=\dfrac{7}{20}\)