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Bài 1:
\(3^{-1}.3^n+4.3^n=13.3^5\)
\(\Rightarrow3^{n-1}+4.3.3^{n-1}=13.3^5\)
\(\Rightarrow3^{n-1}\left(1+4.3\right)=13.3^5\)
\(\Rightarrow3^{n-1}.13=13.3^5\)
\(\Rightarrow3^{n-1}=3^5\)
\(\Rightarrow n-1=5\)
\(\Rightarrow n=6\)
Vậy n = 6
Bài 2a: Câu hỏi của Nguyễn Trọng Phúc - Toán lớp 7 | Học trực tuyến
Ta có:\(\dfrac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\dfrac{x^{2010}}{a^2}=\dfrac{y^{2010}}{b^2}=\dfrac{z^{2010}}{c^2}=\dfrac{t^{2010}}{d^2}\)
\(\Rightarrow\dfrac{x^{2010}}{a^2}+\dfrac{y^{2010}}{b^2}+\dfrac{z^{2010}}{c^2}+\dfrac{t^{2010}}{d^2}=\dfrac{x^{2010}}{a^2}\)
\(\Rightarrow\dfrac{y^{2010}}{b^2}+\dfrac{z^{2010}}{c^2}+\dfrac{t^{2010}}{d^2}=0\)
\(\Leftrightarrow3\cdot\dfrac{y^{2010}}{b^2}=0\)
\(\Leftrightarrow y^{2010}=0\)
\(\Leftrightarrow y=0\)
CMTT\(\Rightarrow x=z=t=0\)
\(\Rightarrow T=0\)
Ta có \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2(a+b+c)}{a+b+c}=2 \)
=> a+b=c
b+c=a
c+a=b
M=\(\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}=\frac{(a+b)(b+c)(c+a)}{abc}=2.2.2=8 \)
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{-10-6}{-8}=\dfrac{-16}{-8}=2\)\(\Rightarrow\left\{{}\begin{matrix}x=2.2+1=5\\y=2.3+2=8\\z=2.4+3=11\end{matrix}\right.\)
Theo đề bài ta có:
\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=k\)
ta có:
\(\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=k^3=\dfrac{a}{d}\)
Và \(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
Ta có đpcm
Câu 1:
Ta có: \(\left[\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{65.68}\right]x-\dfrac{7}{34}=\dfrac{19}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{65.68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\dfrac{11}{68}x=\dfrac{33}{68}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)