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Ta có: \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{b+a+d}=\frac{d}{c+b+a}\)
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{b+a+d}+1=\frac{d}{c+b+a}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{b+a+d}=\frac{a+b+c+d}{c+b+a}\)
Mà a+b+c+d khác 0
=> b+c+d = a+c+d = b+a+d = c+b+a
=> b = a = c = d
Ta có:
\(P=\frac{2a+5b}{3c+4d}-\frac{2b+5c}{3d+4a}-\frac{2c+5d}{3a+4b}-\frac{2d+5a}{3c+4b}\)
\(P=\frac{2a+5a}{3a+4a}-\frac{2b+5b}{3b+4b}-\frac{2c+5d}{3c+4c}-\frac{2d+5d}{3d+4d}\)
\(P=\frac{7a}{7a}-\frac{7b}{7b}-\frac{7c}{7c}-\frac{7d}{7d}\)
\(P=1-1-1-1=-2\)
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: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
Ta có: \(\frac{3a-b}{c}=\frac{3b-c}{a}=\frac{3c-a}{b}\)
Áp dụng tính chất tỉ dãy số bằng nhau. Ta có:
\(\frac{3a-b}{c}=\frac{3b-c}{a}=\frac{3c-a}{b}\Leftrightarrow\frac{3a-b+3b-c+3c-a}{a+b+c}=\frac{3\left(a+b+c\right)-\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow\frac{\left(a+b+c\right)\left(3-1\right)}{a+b+c}=\frac{\left(a+b+c\right)2}{a+b+c}=2\).Do:
\(\frac{3a-b}{c}=\frac{3b-c}{a}=\frac{3c-a}{b}=2\) nên:
\(\Rightarrow3a-b=2c\) (1)
\(\Rightarrow3b-c=2a\) (2)
\(\Rightarrow3c-a=2b\)(3)
Thế (1) ; (2) ; (3) vào A. Ta có:
\(\frac{a}{2b-3c}+\frac{b}{2c-3a}+\frac{c}{2a-3b}\)
\(\Leftrightarrow A=\frac{a}{3c-a-3c}+\frac{b}{3a-b-3a}+\frac{c}{3b-c-3b}\)
\(\Leftrightarrow A=\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}\). Do: \(\frac{3a-b}{c}=\frac{3b-c}{a}=\frac{3c-a}{b}\Rightarrow\frac{a}{-a}=\frac{b}{-b}=\frac{c}{-c}=\left(-1\right)\)
\(\Leftrightarrow A=\left(-1\right)+\left(-1\right)+\left(-1\right)=\left(-3\right)\)
P/s: Mình không chắc nên nếu sai thì bạn thông cảm nha
Mình làm thử các bạn xem có đúng ko nhé
Ta có :
\(\frac{3a-b}{c}=\frac{3b-c}{a}=\frac{3c-a}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{3a-b}{c}=\frac{3b-c}{a}=\frac{3c-a}{b}=\frac{3a-b+3b-c+3c-a}{a+b+c}=\frac{3a+3b+3c-a-b-c}{a+b+c}\)
\(=\frac{3\left(a+b+c\right)-\left(a+b+c\right)}{a+b+c}=\frac{\left(a+b+c\right)\left(3-1\right)}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=\frac{2}{1}=2\)
Do đó :
\(\frac{3a-b}{c}=2\)\(\Rightarrow\)\(3a-b=2c\)\(\left(1\right)\)
\(\frac{3b-c}{a}=2\)\(\Rightarrow\)\(3b-c=2a\)\(\left(2\right)\)
\(\frac{3c-a}{b}=2\)\(\Rightarrow\)\(3c-a=2b\)\(\left(3\right)\)
Thay (1), (2) và (3) vào A ta có :
\(A=\frac{a}{2b-3c}+\frac{b}{2c-3a}+\frac{c}{2a-3b}\)
\(A=\frac{a}{3c-a-3c}+\frac{b}{3a-b-3a}+\frac{c}{3b-c-3b}\)
\(A=\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}\)
\(A=\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(A=-3\)
Vậy \(A=-3\)
Nếu đúng thì thui, sai thì đừng có k sai cho mình nha :)
Ta có :
\(\frac{3a-b}{c}=\frac{3b-c}{a}=\frac{3c-a}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{3a-b}{c}=\frac{3b-c}{a}=\frac{3c-a}{b}=\frac{3a-b+3b-c+3c-a}{a+b+c}=\frac{3\left(a+b+c\right)-\left(a+b+c\right)}{a+b+c}\)
\(=\frac{\left(a+b+c\right)\left(3-1\right)}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=\frac{2}{1}=2\)
Do đó :
\(\frac{3a-b}{c}=2\)\(\Rightarrow\)\(3a-b=2c\)\(\left(1\right)\)
\(\frac{3b-c}{a}=2\)\(\Rightarrow\)\(3b-c=2a\)\(\left(2\right)\)
\(\frac{3c-a}{b}=2\)\(\Rightarrow\)\(3c-a=2b\)\(\left(3\right)\)
Thay (1), (2) và (3) vào A ta có :
\(A=\frac{a}{2b-3c}+\frac{b}{2c-3a}+\frac{c}{2a-3b}\)
\(A=\frac{a}{3c-a-3c}+\frac{b}{3a-b-3a}+\frac{c}{3b-c-3b}\)
\(A=\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}\)
\(A=\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(A=-3\)
Vậy \(A=-3\)
Chúc bạn học tốt
\(\frac{a}{b}=\frac{-2}{3}\Rightarrow3a=-2b\)
thay vào C ta đc:
\(C=\frac{-2b-2b}{-2b+b}+\frac{-2b+2b}{2\left(-2b\right)-b}\)
=> \(C=\frac{-4b}{-b}+0=4\)
a/ Ta có \(a\left(2a-5c\right)=2a^2-5ac=2bc-5ac=c\left(2b-5a\right)\Rightarrow\frac{c}{2a-5c}=\frac{a}{2b-5a}\)
Các câu khác làm tương tự
Áp dụng tính chất dãy tỉ số bằng nhau ta có : \(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}=\frac{\left(2b+c-a\right)+\left(2c-b+a\right)+\left(2a+b-c\right)}{a+b+c}\)\(=\frac{2a+2c+2a}{a+b+c}=2\)
vậy : \(\frac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b+c-3a=0\Rightarrow3a-2c=c\Rightarrow3a-c=2b\)
\(\frac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c+a-3b=0\Rightarrow3b-2c=a\Rightarrow3b-a=2c\)
\(\frac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a+b-3c=0\Rightarrow3c-2a=b\Rightarrow3c-b=2a\)
Vậy \(P=\frac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}=\frac{c.a.b}{2b.2c.2a}=\frac{1}{8}\)
đặt \(\frac{a}{2}=\frac{b}{-3}=\frac{c}{-4,5}=k\)
\(\Rightarrow a=2k,b=-3k,c=-4,5k\)
thay vào biểu thức P ta có:
\(P=\frac{3.2k-2.\left(-3k\right)}{8.2k-\left(-3k\right)+3.\left(-4,5k\right)}=\frac{6k+6k}{7,5k}=\frac{12}{7,5}=\frac{8}{5}\)