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Theo bài ra \(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\)
\(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)\(\Rightarrow\)\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)+)Nếu a+b+c+d=0
\(\Rightarrow\)a=b=c=d
\(\Rightarrow\)M=1+1+1+1=4
+)Nếu a+b+c+d=0
\(\Rightarrow\)\(\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow\)M=(-1)+(-1)+(-1)+(-1)=-4
Vậy M=4 hoặc M=-4
Câu 1:
\(\frac{a^{2016}+b^{2016}}{c^{2016}+d^{2016}}=\frac{a^{2016}-b^{2016}}{c^{2016}-d^{2016}}\)
\(\Rightarrow (a^{2016}+b^{2016})(c^{2016}-d^{2016})=(a^{2016}-b^{2016})(c^{2016}+d^{2016})\)
\(\Leftrightarrow 2(bc)^{2016}=2(ad)^{2016}\Rightarrow (bc)^{2016}=(ad)^{2016}\)
\(\Rightarrow (\frac{a}{b})^{2016}=(\frac{c}{d})^{2016}\)
\(\Rightarrow \frac{a}{b}=\pm \frac{c}{d}\) (đpcm)
Câu 2:
Nếu $a+b+c+d=0$ thì: \(\left\{\begin{matrix} a+b=-(c+d)\\ b+c=-(d+a)\\ c+d=-(a+b)\\ d+a=-(b+c)\end{matrix}\right.\)
\(\Rightarrow M=(-1)+(-1)+(-1)+(-1)=-4\)
Nếu $a+b+c+d\neq 0$
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{5(a+b+c+d)}{a+b+c+d}=5\)
\(\Rightarrow \left\{\begin{matrix} 2a+b+c+d=5a\\ a+2b+c+d=5b\\ a+b+2c+d=5c\\ a+b+c+2d=5d\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} b+c+d=3a(1)\\ a+c+d=3b(2)\\ a+b+d=3c(3)\\ a+b+c=3d(4)\end{matrix}\right.\)
Từ \((1);(2)\Rightarrow b+a+2(c+d)=3(a+b)\Rightarrow c+d=a+b\)
\(\Rightarrow \frac{a+b}{c+d}=1\)
Tương tự: \(\frac{b+c}{d+a}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\Rightarrow M=1+1+1+1=4\)
Câu 2:A= 75.(42004+42003+.....+42+4+1)+25=75.|(42005-1):3+25=25.(42005-1+1)=25.42005chia hết 100
Suy ra A chia hết cho 100
CHÚC BẠN HỌC TỐT NHÉ !!!!!!!!!
Ta có:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{2a}{2c}=\frac{b}{d}\Rightarrow\frac{a}{c}=\frac{2b}{2d}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{2a}{2c}=\frac{b}{d}=\frac{2a-b}{2c-d}\)\(\left(1\right)\)và \(\frac{a}{c}=\frac{2b}{2d}=\frac{a+2b}{c+2d}\)\(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(\frac{2a-b}{2c-d}=\frac{a+2b}{c+2d}\)
\(\Rightarrow\frac{2a-b}{a+2b}=\frac{2c-d}{c+2d}\)\(\left(đpcm\right)\)
Lập luận không chắc !
\(\text{Ta có: }\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{2b}{2d}\)
\(\text{Áp dụng t/c dãy tỉ số bằng nhau, ta có:}\)
\(\cdot\frac{2a}{2c}=\frac{b}{d}=\frac{2a-b}{2c-d}\)
\(\cdot\frac{a}{c}=\frac{2b}{2d}=\frac{a+2b}{c+2d}\)
\(\text{Mà }\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{2b}{2d}\)
\(\Rightarrow\frac{2a-b}{2c-d}=\frac{a+2b}{c+2d}\)
\(\text{Vậy: }\frac{2a-b}{a+2b}=\frac{2c-d}{c+2d}\left(\text{ĐPCM}\right)\)
a/ Nhân cả 2 vế với a+b+c+d
\(\Rightarrow\frac{a+b+c+d}{a+b+c}+\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{c+d+a}+\frac{a+b+c+d}{d+a+b}=\frac{a+b+c+d}{40}.\)
\(\Rightarrow1+\frac{d}{a+b+c}+1+\frac{a}{b+c+d}+1+\frac{b}{c+d+a}+1+\frac{c}{d+a+b}=\frac{2000}{40}=50\)
\(\Rightarrow S=46\)
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Rightarrow\frac{q^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4\)
=> \(\frac{a^2}{4}=4\Rightarrow a^2=4.4=16\Rightarrow a=+-4\)
=>\(\frac{b^2}{9}=4\Rightarrow b^2=4.9=36\Rightarrow b=+-6\)
=>\(\frac{2c^2}{32}=4\Rightarrow c^2=4.32:2=64\Rightarrow c=+-8\)
Câu 2 :
Ta có : \(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
a) Giải:
Ta có: \(a,b,c>0\Rightarrow a+b+c>0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{2b+c}=\frac{b}{2c+a}=\frac{c}{2a+b}=\frac{a+b+c}{2b+c+2c+a+2a+b}=\frac{a+b+c}{3a+3b+3c}=\frac{a+b+c}{3\left(a+b+c\right)}=\frac{1}{3}\)
Vậy \(\frac{a}{2b+c}=\frac{b}{2c+a}=\frac{c}{2a+b}=\frac{1}{3}\)
Mỗi tỉ số đã cho đều bớt đi 1 ta được :
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
=> \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+ Nếu a + b + c + d \(\ne\)0 thì a = b = c = d , lúc đó Q = 1 + 1 + 1 + 1 = 4
+ Nếu a + b + c + d = 0 thì
a + b = -\((c+d)\); b + c = -\((d+a)\); c + d = -\((a+b)\); d + a = -\((b+c)\)
Lúc đó Q = \((-1)+(-1)+(-1)+(-1)=-4\)