cho 3 số a,b,c thỏa mãn
\(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}=a+b+c\)
tính giá trị biểu thức A=\(\dfrac{a^2+b^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{b^2+c^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{c^2+a^2}{\left(c+b\right)\left(a+b\right)}\)
CMR:\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
Cho a + b + c = 4m. Chứng minh rằng:
\(\left(\dfrac{a+b-c}{2}\right)^2+\left(\dfrac{a-b+c}{2}\right)^2+\left(\dfrac{-a+b+c}{2}\right)^2=a^2+b^2+c^2-4m^2\)
1) Cho \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
CMR: \(a=b=c=1\)
2) CMR: nếu \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\) thì \(\dfrac{a}{x}=\dfrac{b}{y}\)
3) Cho \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
CMR: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Cho \(a,b,c>0.\)CMR:
\(\left(a^3+b^3+c^3\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right)^2\)
1,Cho \(a^2+b^2+c^2+3=2\left(a+b+c\right)\) .Cmr: \(a=b=c=1\)
2,Cho \(\left(a+b+c\right)^2=3\left(ab+ac+bc\right)\) .Cmr: \(a=b=c\)
3,Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2\) .Cmr: \(a=b=c\)
4,Cho a,b,c,d là các số khác 0 và:
\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\) .Cmr: \(\dfrac{a}{c}=\dfrac{b}{d}\)
5,Cho \(x^2-y^2-z^2=0\) .Cmr: \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
HELP ME!mik cần gấp lắm rồi!Thank trước nhé!
Cho biểu thức A=\(\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)
a) Rút gọn A
b) Tính các cặp gia trị nguyên (x.y)để A=-3
Cho biểu thức A=\(\dfrac{x^2}{\left(x+y\right)\left(1-y\right)}-\dfrac{y^2}{\left(x+y\right)\left(1+x\right)}-\dfrac{x^2y^2}{\left(1+x\right)\left(1-y\right)}\)
a) Rút gọn A
b) Tính các cặp gia trị nguyên (x.y)để A=-3
Chứng minh:
a) \(x\ne0,y\ne0\) và \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)\) thì \(\dfrac{a}{x}=\dfrac{b}{y}\)
b) \(x\ne0,y\ne0,z\ne0\) và \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\) thì \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)