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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
Sửa: CMR \(\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\left(\dfrac{a+c-m}{b+d-n}\right)^3\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{m}{n}=k\Rightarrow a=kb;c=kd;m=kn\)
\(\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\dfrac{k^3b^3+k^3d^3+k^3n^3}{b^3+d^3+n^3}=\dfrac{k^3\left(b^3+d^3+n^3\right)}{b^3+d^3+n^3}=k^3\)
\(\left(\dfrac{a+c-m}{b+d-m}\right)^3=\left(\dfrac{kb+kd-kn}{b+d-n}\right)^3=\left(\dfrac{k\left(b+d-n\right)}{b+d-n}\right)^3=k^3\)
\(\Rightarrow\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\left(\dfrac{a+c-m}{b+d-n}\right)^3\left(=k^3\right)\)
theo bài ra ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\) \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\) \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\) vì a; b;c ; d khác 0
=> a = b = c = d
=> \(M=1+1+1+1=4\)
vậy M = 4
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}=\dfrac{2a+b+c+d-a-2b-c-d}{a-b}=1\)
\(\Rightarrow\left\{{}\begin{matrix}-a=b+c+d\\-b=a+c+d\\-c=b+c+d\\-d=a+b+c\end{matrix}\right.\)\(\Rightarrow a=b=c=d\)
\(\Rightarrow M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow M=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}\)
\(\Rightarrow M=1+1+1+1\)
\(\Rightarrow M=4\)
Vậy \(M=4\)
Ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
⇔ \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1\)
\(=\dfrac{a+b+c+2d}{d}-1\)
⇔ \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
Nếu a+b+c+d=0
⇒a+b=−(c+d);c+b=−(a+d);c+d=−(a+b);a+d=−(c+b)
Thay vào M, ta có:
\(M=\dfrac{a+b}{-\left(a+b\right)}=\dfrac{b+c}{-\left(b+c\right)}=\dfrac{c+d}{-\left(c+d\right)}=\dfrac{a+d}{-\left(a+d\right)}=-1\)
Nếu a+b+c+d ≠0
⇒ \(a=b=c=d\)
Thay vào M, ta có
\(M=\dfrac{a+b}{a+b}=\dfrac{b+c}{b+c}=\dfrac{c+d}{c+d}=\dfrac{d+a}{d+a}=1\)
\(\dfrac{a}{b+c+d}=\dfrac{b}{c+a+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)
+)Xét a+b+c+d=0 thì a+d=-c-d
b+c=-d-a
c+d=-b-a
d+a=-b-c
Do đó:
\(P=\dfrac{-c-d}{c+d}+\dfrac{-a-b}{a+b}+\dfrac{-b-c}{b+c}+\dfrac{-d-a}{a+d}\\ =-1+-1+-1+-1=-4\)
+)Xét a+b+c+d khác 0
áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+c+d}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
=>\(a=\dfrac{1}{3}\left(d+b+c\right)\)
\(b=\dfrac{1}{3}\left(a+c+d\right)\)
\(c=\dfrac{1}{3}\left(a+b+d\right)\)
\(d=\dfrac{1}{3}\left(a+b+c\right)\)
Bạn thay vào r tính
Ta có : \(\dfrac{a}{b+c+d}=\dfrac{b}{c+d+a}=\dfrac{c}{d+a+b}=\dfrac{d}{a+b+c}\)
\(\Rightarrow\)\(\dfrac{a}{b+c+d}+1=\dfrac{b}{c+d+a}+1=\dfrac{c}{d+a+b}+1=\dfrac{d}{a+b+c}+1\)
\(\Rightarrow\)\(\dfrac{a+b+c+d}{b+c+d}=\dfrac{b+c+d+a}{c+d+a}=\dfrac{c+d+a+b}{d+a+b}=\dfrac{d+a+b+c}{a+b+c}\)
TH1 : \(a+b+c+d\ne0\)\(\Rightarrow\) \(a=b=c=d\)
\(\Rightarrow\) P= \(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{b+a}+\dfrac{d+a}{b+c}=1+1+1+1=4\)
TH2 : \(a+b+c+d=0\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{b+a}+\dfrac{d+a}{b+c}=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
TH1:
\(\dfrac{a}{b+c+d}=\dfrac{b}{c+d+a}=\dfrac{c}{d+a+b}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}3a=b+c+d\\3b=a+c+d\\3c=a+b+d\\3d=a+b+c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3\left(a-b\right)=b-a\\3\left(b-c\right)=c-b\\3\left(c-d\right)=d-c\\3\left(d-a\right)=a-d\end{matrix}\right.\) \(\Rightarrow a=b=c=d\)
\(\Rightarrow P=\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}=1+1+1+1=4\)
TH2: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a-b}{a-a}=-1\)
\(\Rightarrow-a=b+c+d\Rightarrow a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\a+c=-\left(b+d\right)\\a+d=-\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{b+c}{-\left(b+c\right)}+\dfrac{c+d}{-\left(c+d\right)}+\dfrac{-\left(b+c\right)}{b+c}=-1+-1+-1+-1=-4\)
Vậy \(\left[{}\begin{matrix}P=4\\P=-4\end{matrix}\right.\)
Lời giải:
Ta có:
\(M=\frac{a}{a+b+c}+\frac{b}{a+b+d}+\frac{c}{b+c+d}+\frac{d}{a+d+c}\)
\(> \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}\)
\(\Leftrightarrow M>\frac{a+b+c+d}{a+b+c+d}=1(1)\)
Mặt khác:
\(M=1-\frac{b+c}{a+b+c}+1-\frac{a+d}{a+b+d}+1-\frac{b+d}{b+c+d}+1-\frac{a+c}{a+d+c}\)
\(\Leftrightarrow M=4-\underbrace{\left(\frac{b+c}{a+b+c}+\frac{a+d}{a+b+d}+\frac{b+d}{b+c+d}+\frac{a+c}{a+d+c}\right)}_{N}\)
Có: \(N>\frac{b+c}{a+b+c+d}+\frac{a+d}{a+b+c+d}+\frac{b+d}{a+b+c+d}+\frac{a+c}{a+b+c+d}\)
\(\Leftrightarrow N>\frac{2(a+b+c+d)}{a+b+c+d}=2\)
\(\Rightarrow M=4-N< 4-2\Leftrightarrow M< 2(2)\)
Từ \((1);(2)\Rightarrow 1< M< 2\Rightarrow M\not\in \mathbb{N}\)