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Ta có:A=\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(\frac{1}{2}A\)=\(\frac{1}{2}\)\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{4}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(\frac{1}{2}A\)=\(\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^5+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\frac{1}{2}A-A\)=\(\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^5+...+\left(\frac{1}{2}\right)^{100}\right]\)-\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(-\frac{1}{2}A\)=\(\left(\frac{1}{2}^{100}\right)-\frac{1}{2}\)
\(-\frac{1}{2}A\)=\(-\frac{1}{2}\)
A=\(-\frac{1}{2}:\left(-\frac{1}{2}\right)\)
A=1
Chúc bạn học tốt!
\(S=\left(\frac{1}{7}\right)^2+\left(\frac{2}{7}\right)^2+\left(\frac{3}{7}\right)^2+...+\left(\frac{10}{7}\right)^2\)
\(=\frac{1^2}{7^2}+\frac{2^2}{7^2}+\frac{3^2}{7^2}+...+\frac{10^2}{7^2}\)
\(=\frac{1^2+2^2+3^2+...+10^2}{7^2}\)
\(=\frac{385}{49}=\frac{55}{7}\)
Vậy S = \(\frac{55}{7}\)
Ta có : 49S= \(1^2+2^2+...+10^2\)
49S= 385
S = \(\frac{385}{49}=\frac{55}{7}.\)
Đề sai bạn nhé. Đưa dữ kiện 3 ẩn bắt tính biểu thức chứa 2 ẩn làm sao làm được ?
Bạn kiểm tra lại nha
\(\left(x-3\right)^2+\left|y^2-9\right|=0\)
Vì \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left|y^2-9\right|\ge0\forall y\end{matrix}\right.\)
để bt = 0 \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left|y^2-9\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y^2-9=0\Rightarrow y^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\\left[{}\begin{matrix}y=3\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Vậy.....
\(\left(x-3\right)^2+\left|y^2-9\right|=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left|y^2-9\right|=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\y^2-9=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\y^2=9\left[{}\begin{matrix}y=3\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=3\\y=3hoặcy=-3\end{matrix}\right.\)
2.Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow a+b+c-a-b+c=0\)
\(\Rightarrow2c=0\)
\(\Rightarrow c=0\)
Vậy c=0
BT5: Ta có: f(1)=1.a+b=1 =>a+b=1 (1)
f(2)=2a+b=4 (2)
Trừ (1) cho (2) ta có: 2a+b-a-b=4-1 => a=3
Với a=3 thay vào (1) ta có: 3+b=1 => b=-2
Vậy a=3, b=-2
Ta có : \(\left\{\begin{matrix}Q=-\left(x-7\right)^2-6\\-\left(x-7\right)^2\le0\\-6=-6\end{matrix}\right.\)
\(\Rightarrow Q=-\left(x-7\right)^2-6\le0-6=-6\)
Vậy GTLN của \(Q=-\left(x-7\right)^2-6\) là \(-6\)
\(\sqrt{x^2}.\left|x+2\right|=x\)
\(\Rightarrow x.\left|x+2\right|=x\)
\(\Rightarrow\left|x+2\right|=1\)
\(\Rightarrow\left[\begin{matrix}x+2=1\\x+2=-1\end{matrix}\right.\) \(\Rightarrow\)\(\left[\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
6-/2-x/=4
\(\Rightarrow\)/2-x/=6-4
\(\Rightarrow\)/2-x/=2
\(\Rightarrow\left\{\begin{matrix}2-x=2\\2-x=-2\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=2-2\\x=2-\left(-2\right)\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Rightarrow x=\left\{0;4\right\}\)
0
x=2