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a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
1) Ta có : Đặt M = 3x + 1 + 3x + 2 + ... + 3x + 100
= 3x(3 + 32 + ... + 3100)
= 3x[(3 + 32 + 33 + 34) + (35 + 36 + 37 + 38) + ... + (397 398 + 399 + 3100)]
= 3x[(3 + 32 + 33 + 34) + 34.(3 + 32 + 33 + 34) + ... + 396.(3 + 32 + 33 + 34)]
= 3x(120 + 34.120 + .... + 396.120)
= 3x.120.(1 + 34 + .... + 396)
=> \(M⋮120\)(ĐPCM)
2) Ta có \(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)
\(\Rightarrow\frac{3a+b+c}{a}-2=\frac{a+3b+c}{b}-2=\frac{a+b+3c}{c}-2\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Nếu a + b + c = 0
=> a + b = - c
b + c = -a
c + a = -b
Khi đó P = \(\frac{-c}{c}+\frac{-a}{a}+\frac{-b}{b}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
Nếu a + b + c \(\ne\)0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Khi đó P = \(\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
Vậy nếu a + b + c = 0 thì P = -3
nếu a + b + c \(\ne\)0 thì P = 6
Ta có :
\(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)
\(=\left(3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}\right)+...\)\(+\left(3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100}\right)\)
\(=3^x\left(3+3^2+3^3+3^4\right)+...+3^{x+96}\left(3+3^2+3^3+3^4\right)\)
\(=3^x.120+3^{x+4}.120+...+3^{x+96}.120\)
\(=120.\left(3^x+3^{x+4}+...+3^{x+96}\right)\)
Vì \(120⋮120\)
\(\Rightarrow120.\left(3^x+3^{x+4}+...+3^{x+96}\right)⋮120\)
\(\Rightarrow3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}⋮120\left(\forall x\inℕ\right)\left(đpcm\right)\)
Áp dụng ta đc:
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{5a+5b+5c}{a+b+c}=5\left(\text{vì: a,b,c khác 0}\right)\)
\(\Rightarrow\hept{\begin{cases}b+c=2a\\c+a=2b\\a+b=2c\end{cases}}\Rightarrow a=b=c\)
\(\Rightarrow P=6\)
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)
\(\Rightarrow\frac{3a+b+c}{a}-2=\frac{a+3b+c}{b}-2=\frac{a+b+3c}{c}-2\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Xét \(a+b+c\ne0\)
\(\Rightarrow a=b=c\)
Thay vào P ta được P=6
Xét \(a+b+c=0\)
\(\Rightarrow a+b=-c;b+c=-a;a+c=-b\)
Thay vào P ta được P= -3
Vậy P có 2 gtri là ...........
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)
suy ra: x/5 = 45 => x = 225
y/7 = 45 => y = 315
z/9 = 45 => z = 405