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Bài 2 :
Ta có :
\(\dfrac{2a+b+c}{a}=\dfrac{a+2b+c}{b}=\dfrac{a+b+2c}{c}\)
\(\Rightarrow\dfrac{2a+b+c}{a}-1=\dfrac{a+2b+c}{b}-1=\dfrac{a+b+2c}{c}-1\)\(\Rightarrow\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{c}\)
* Nếu \(a+b+c=0\), Ta suy ra các đẳng thức sau :
\(\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
Thay các đẳng thức vừa tìm được vào N, ta có :
\(N=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)
\(\Leftrightarrow N=\dfrac{-c}{c}+\dfrac{-a}{a}+\dfrac{-b}{b}\)
\(\Leftrightarrow N=-1+\left(-1\right)+\left(-1\right)=-3\)
* Nếu \(a+b+c\ne0\)
Để \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{c}\)
\(\Rightarrow a=b=c\)
\(\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
Thay các đẳng thức vào N ta có :
\(N=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)
\(\Leftrightarrow N=\dfrac{2c}{c}+\dfrac{2a}{a}+\dfrac{2b}{b}=2+2+2=6\)
Vậy.....
tik mik nha !!!
Ta có \(\frac{2a+b+c}{b+c}=\frac{2b+c+a}{c+a}=\frac{2c+a+b}{a+b}\Rightarrow\frac{2a}{b+c}+1=\frac{2b}{a+c}+1=\frac{2c}{a+b}+1\)
=> \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}\)
^_^
Bài 1: Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\)
\(\Rightarrow\hept{\begin{cases}a=2016k\\b=2017k\\c=2018k\end{cases}}\).Thay vào M,ta có:
\(M=4\left(2016k-2017k\right)\left(2017k-2018k\right)-\left(2018k-2016k\right)^2\)
\(=4.\left(-1k\right)\left(-1k\right)-\left(2k\right)^2\)
\(=4k^2-4k^2=0\)
\(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}=\frac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\)
\(=\frac{2a+2b+2c}{a+b+c}=2\)
+ Từ \(\frac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow3a-2b=c\) và \(3a-c=2b\)
+ Tương tự ta cũng có \(3b-2c=a\) và \(3b-a=2c\)
Và \(3c-2a=b\); \(3c-b=2a\)
Thay vào P
\(P=\frac{c.a.b}{2.b.2.c.2.a}=\frac{1}{8}\)
a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)
=> \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{k.b^2}{k.d^2}=\frac{b^2}{d^2}\) (1)
Từ (1) và (2) => \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
Ta có: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Mà: \(k^3=\frac{a}{d}\) => \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
a)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\left(đpcm\right)\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=> a = b = c = d
=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)
D = 1 + 1 + 1 + 1 = 4
Bài 2: Mình nghĩ câu a là a+2b-3c=-20
a) Ta có: a/2 = b/3 = c/4 = 2b/6 = 3c/12 = a + 2b - 3c/ 2 + 6 - 12 = -20/-4 = 5
a/2 = 5 => a = 2 . 5 = 10
b/3 = 5 => b = 5 . 3 = 15
c/4 = 5 => c = 5 . 4 = 20
Vậy a = 10; b = 15; c = 20
b) Ta có: a/2 = b/3 => a/10 = b/15
b/5 = c/4 => b/15 = c/12
=> a/10 = b/15 = c/12 = a - b + c / 10 - 15 + 12 = -49/7 = -7
a/10 = -7 => a = -7 . 10 = -70
b/15 = -7 => b = -7 . 15 = -105
c/12 = -7 => c = -7 . 12 = -84
Vậy a = -70; b = -105; c = -84.
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Leftrightarrow\)\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Leftrightarrow\)\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+) Xét \(a+b+c+d=0\)
Suy ra :
\(a+b=-\left(c+d\right)\)
\(b+c=-\left(d+a\right)\)
\(c+a=-\left(b+d\right)\)
\(d+a=-\left(b+c\right)\)
Do đó : \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{c+b}\)
\(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}\)
\(M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(M=-4\)
+) Xét \(a+b+c+d\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=4\)
Do đó :
\(\frac{a+b+c+d}{a}=4\)\(\Leftrightarrow\)\(a+b+c+d=4a\) \(\left(1\right)\)
\(\frac{a+b+c+d}{b}=4\)\(\Leftrightarrow\)\(a+b+c+d=4b\) \(\left(2\right)\)
\(\frac{a+b+c+d}{c}=4\)\(\Leftrightarrow\)\(a+b+c+d=4c\) \(\left(3\right)\)
\(\frac{a+b+c+d}{d}=4\)\(\Leftrightarrow\)\(a+b+c+d=4d\) \(\left(4\right)\)
Từ (1), (2), (3) và (4) suy ra \(4a=4b=4c=4d\) \(\left(=a+b+c+d\right)\)
\(\Leftrightarrow\)\(a=b=c=d\)
\(\Rightarrow\)\(M=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)
\(\Rightarrow\)\(M=1+1+1+1=4\)
Vậy \(M=-4\) hoặc \(M=4\)
Chúc bạn học tốt ~
Ta có :
\(2a+2b+2c=by+cz+ax+cz+ax+by\)
\(\Leftrightarrow\)\(2\left(a+b+c\right)=2\left(ax+by+cz\right)\)
\(\Leftrightarrow\)\(a+b+c=ax+by+cz\)
+) \(a+b+c=ax+\left(by+cz\right)=ax+2a=a\left(x+2\right)\)
\(\Rightarrow\)\(\frac{1}{x+2}=\frac{a}{a+b+c}\) \(\left(1\right)\)
+) \(a+b+c=by+\left(ax+cz\right)=by+2b=b\left(y+2\right)\)
\(\Rightarrow\)\(\frac{1}{y+2}=\frac{b}{a+b+c}\) \(\left(2\right)\)
+) \(a+b+c=cz+\left(ax+by\right)=cz+2c=c\left(z+2\right)\)
\(\Rightarrow\)\(\frac{1}{z+2}=\frac{c}{a+b+c}\) \(\left(3\right)\)
Từ (1), (2) và (3) suy ra \(M=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)
\(M=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\)
\(M=\frac{a+b+c}{a+b+c}=1\)
Vậy \(M=1\)
Chúc bạn học tốt ~
BÀI 2: Áp dụng tc của dãy tỉ số bằng nhau, ta có:
\(\frac{2a+b+c}{a}=\frac{a+2b+c}{b}=\frac{a+b+2c}{c}=\frac{4a+4b+4c}{a+b+c}=4\)
\(\Rightarrow2+\frac{b+c}{a}=2+\frac{a+c}{b}=2+\frac{a+b}{c}=4\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=2\)
Vậy N = 6
BÀI 1: Theo đề bài, ta có:
\(ac+c^2=b^2+bd\Rightarrow c\left(a+c\right)=b\left(b+d\right)\Rightarrow c\left(a+c\right)+bc=b\left(b+d\right)+bc\)\(\Rightarrow c\left(a+b+c\right)=b\left(b+c+d\right)\)\(\Rightarrow\frac{a+b+c}{b+c+d}=\frac{b}{c}\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\left(\frac{b}{c}\right)^3=\frac{b^2b}{c^2c}=\frac{acb}{bdc}=\frac{a}{d}\).