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Ta có: \(\left(x-y+z\right)^2=x^2-y^2+z^2\)
<=> \(x^2+y^2+z^2-2xy-2yz+2zx=x^2-y^2+z^2\)
<=> \(2y^2-2xy-2yz+2zx=0\)
<=> \(\left(2y^2-2yz\right)-\left(2xy-2xz\right)=0\)
<=>\(2y\left(y-z\right)-2x\left(y-z\right)=0\)
<=>\(2\left(y-x\right)\left(y-z\right)=0\)
<=> \(\left[\begin{array}{nghiempt}y-x=0\\y-z=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}y=x\\y=z\end{array}\right.\)
Với y=x thì mọi giá trị của z đều thỏa mãn.
Với y=z ta có: \(\left(x-2y\right)^2=x^2\)
<=> \(\left[\begin{array}{nghiempt}x-2y=-x\\x-2y=x\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=y\\x=-y\end{array}\right.\)
=> x=y=z hoặc -x=y=z.
Đặt \(A=1+2+2^2+...+2^{32}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{33}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{33}\right)-\left(1+2+2^2+...+2^{32}\right)\)
\(\Rightarrow A=2^{33}-1\)
\(\Rightarrow n=33\)
Vậy n = 33
_Chúc bạn học tốt_
1+2+2^2+2^3+2^4+...+2^32=2^n-1 (1)
=>2+2^2+2^3+...+2^33=2^(n+1)-2 (2)
=>trừ (2) cho (1) ta có : 2^33-1=(2-1)*(2^n-1)
=>2^33-1=2^n-1
=>n=33
vậy n=33
k cho mình nha
\(3.3^{n-1}\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)
\(\Leftrightarrow18.3^{2n+1}+3.3^n-2.3^{2n+3}+2.3^n=405\)
\(\Leftrightarrow54.3^{2n}+5.3^n-2.3^3.3^{2n}=405\)
\(\Leftrightarrow3^n=81\)
\(\Leftrightarrow n=4\)
\(3.3^{n-1}\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)
\(\Leftrightarrow18.3^{2n+1}+3.3^n-2.3^{2n+3}+2.3^n=405\)
\(\Leftrightarrow54.3^{2n}+5.3^n-2.3^3.3^{2n}=405\)
\(\Leftrightarrow3^n=81\)
\(\Leftrightarrow n=4\)
\(d=\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)....\left(1+\dfrac{1}{n^2+2n}\right)\)
\(d=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...........\dfrac{n^2+2n+1}{n^2+2n}\)
\(d=\dfrac{2^2}{3}.\dfrac{3^2}{8}.\dfrac{4^2}{15}......\dfrac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(d=\dfrac{2^2.3^2.4^2......\left(n+1\right)^2}{3.8.15.....n\left(n+2\right)}\)
\(d=\dfrac{2.2.3.3.4.4......\left(n+1\right)\left(n+1\right)}{1.3.2.4.3.5......n\left(n+2\right)}\)
\(d=\dfrac{2.3.4......\left(n+1\right)}{1.2.3......n}.\dfrac{2.3.4.....\left(n+1\right)}{3.4.5.....\left(n+2\right)}\)
\(d=\left(n+1\right)\dfrac{2}{n+2}\)
\(d=\dfrac{2n+2}{n+2}\)
Rút gọn vế trái ta được 2^33-1
=> n = 33
Đặt \(A=1+2+2^2+2^3+......+2^{31}+2^{32}\)
Ta có:
\(A=1+2+2^2+2^3+......+2^{31}+2^{32}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{32}+2^{33}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{32}+2^{33}\right)-\left(1+2+2^2+2^3+......+2^{31}+2^{32}\right)\)
\(\Leftrightarrow A=2^{33}-1\)
Mặt khác \(A=2^n-1\)
\(\Rightarrow2^{33}-1=2^n-1\)
\(\Rightarrow2^{33}-1-2^n+1=0\)
\(\Rightarrow2^{33}-2^n=0\)
\(\Rightarrow2^{33}=2^n\)
\(\Rightarrow n=33\)
Vậy n=33