cho a,b,c khac 0 va a+b-c/c =b+c-a/a=c+a-b/b tinh p=(1+b/a)(1+c/b)(1+a/c)
cho a b c khac 0 va a-b-c=0 tinh gia tri bieu thuc A=(1-c/a) (1-a/b) (1+b/c)
1, cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\) va a+b+c khac 0 tinh b,c
Theo dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\) (vì \(a+b+c\ne0\))
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c=\pm1\)
cho a,b,c khac 0 va a+b+c=0 . tinh Q=\(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\)
a + b + c = 0 => c = -a - b ; b= -a - c ; a = - b - c
Thay vào Q ta có :
\(Q=\frac{1}{a^2+b^2-\left(a+b\right)^2}+\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{a^2+c^2-\left(a+c\right)^2}\)
\(Q=\frac{1}{a^2+b^2-a^2-b^2-2ab}+\frac{1}{b^2+c^2-b^2-c^2-2bc}+\frac{1}{c^2+a^2-c^2-a^2-2ac}\)
\(Q=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{c+a+b}{-2abc}=0\)
cho a , b ,c khac 0 thoa man a + b + c = 0 tinh A = (1+a/b)(1+b/c)(1+c/a)
Cho a,b,c khac 0 va a+b+c=0. Tính (1+a/b)(1+b/c)(1+c/a)
cho a, b>0 va c khac 0. cmr neu 1/a+1/b+1/c=0 thi can(a+b)=can(b+c)+can(c+a)
biet a/a"=b/b"=c/c''=4 va a"+b"+c"khac 0 tinh a+b+c/a"+b"+c"va a-3b+2c /a"-3b"+2c"
cho a/x=b/y=c/z=1/5 va x+y+z khac 0 tinh A=x+y+z/a+b+c
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}=\frac{1}{5}\)
\(\Rightarrow A=\frac{x+y+z}{a+b+c}=\frac{5}{1}=5\)
Vậy A = 5
Cho 3 so a,b,c khac 0 va doi mot khac nhau thoa man a^2.(b+c)=b^2.(a+c)=2015 Tinh c^2.(a+b)