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\(\Rightarrow3+\frac{y+z-2x}{x}=3+\frac{x+z-2y}{y}=3+\frac{x+y-2z}{z}\)
\(\Rightarrow\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\)
\(TH1:x+y+z=0\)
\(\Rightarrow x=-\left(y+z\right),y=-\left(x+z\right),z=-\left(x+y\right)\)
\(A=\left(1+\frac{-y-z}{y}\right).\left(1+\frac{-x-z}{z}\right).\left(1+\frac{-x-y}{x}\right)\)
\(A=-\left(\frac{z}{y}\cdot\frac{x}{z}\cdot\frac{y}{x}\right)=-1\)
\(TH2:x+y+z\ne0\)
\(\Rightarrow x=y=z\Rightarrow A=2^3=8\)
sai đề ròi: tớ làm 2 trường hợp luôn vì trường hợp x+y+z khác 0 thì A mới t/m thuộc N
mà đề là x+y+z khác 0 -.-
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
Theo giả thiết suy ra \(\frac{a\left(y+z\right)}{abc}=\frac{b\left(z+x\right)}{abc}=\frac{c\left(x+y\right)}{abc}\)\(\Rightarrow\)\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}=\frac{z+x-\left(y+z\right)}{ac-bc}=\frac{x-y}{c\left(a-b\right)}\) (1)
\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}=\frac{y+z-\left(x+y\right)}{bc-ab}=\frac{z-x}{b\left(c-a\right)}\) (2)
\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}=\frac{x+y-\left(z+x\right)}{ab-ac}=\frac{y-z}{a\left(b-c\right)}\) (3)
Từ (1), (2), (3) suy ra \(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\) (đpcm).
Thay \(3,7=3\frac{7}{10}\)vào biểu thức:
A = \(\left[3+\frac{7}{10}\right]+\left[3+\frac{9}{10}\right]+\left[3+\frac{11}{10}\right]+\left[3+\frac{13}{10}\right]+\left[3+\frac{15}{10}\right]\)
A = 3 + 3 + 4 +4 + 4 = 18
B = \(\left[5x\right]=\left[5.3,7\right]=\left[18,5\right]=18\)
Vậy A = B
1) c)
\(\left[\frac{1000}{3}\right]+\left[\frac{1000}{3^2}\right]+\left[\frac{1000}{3^3}\right]+\left[\frac{1000}{3^4}\right]=33+11+3+1=48\)
ồ a khác b
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