Chứng minh rằng: Nếu \(\dfrac{x}{y}\)=\(\dfrac{z}{t}\) thì \(\left(\dfrac{x-y}{z-t}\right)^{1996}=\dfrac{x^{1996}+y^{1996}}{z^{1996}+x^{1996}}\) với các điều kiện các mẫu đều khác 0
Giúp mk vs ạ!
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= có x +y+z=a=>x2+y2+z2+2(xy+yz+xz)=a2
Thay vào a2=b+3992=>xy+zy+xz=1996
thay vào P ta có
P=x\(\sqrt{\dfrac{\left(xy+yz+zx+z^2\right)\left(zx+xy+yz+x^2\right)}{xy+yz+zx+x^2}}\)
+y\(\sqrt{\dfrac{\left(zx+zy+xy+z^2\right)\left(zx+zy+xy+x^2\right)}{xy+yz+xz+y^2}}\)
+\(\sqrt{\dfrac{\left(zx+xy+zy+x^2\right)\left(xz+xy+zy+y^2\right)}{xz+xy+zy+z^2}}\)
=x\(\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}\)
+y\(\sqrt{\dfrac{\left(x+z\right)\left(z+y\right)\left(x+y\right)\left(z+x\right)}{\left(y+z\right)\left(x+y\right)}}\)
+z\(\sqrt{\dfrac{\left(y+z\right)\left(x+y\right)\left(z+x\right)\left(x+y\right)}{\left(z+x\right)\left(z+y\right)}}\)
=x\(\sqrt{\left(y+z\right)^2}\)+y\(\sqrt{\left(x+z\right)^2}\)+z\(\sqrt{\left(x+y\right)^2}\)=x(z+y)+y(x+z)+z(x+y)
=2(xy+zx+zy)=3992
*có gì ko hiểu thì hỏi
x+y+z=0
nên x+y=-z; y+z=-x; x+z=-y
\(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
\(=\dfrac{x+y}{y}\cdot\dfrac{y+z}{z}\cdot\dfrac{x+z}{x}=-1\)
Đề bài sai
Ví dụ: với \(a=1;b=2;c=3,d=4\) thì \(x=\dfrac{1}{2}\) ; \(y=\dfrac{3}{4}\) ; \(z=\dfrac{2}{3}\)
Khi đó \(x< y\) nhưng \(z< y\)
\(\text{Vì }\dfrac{a}{b}< \dfrac{c}{d}\text{ nên }ad< bc\left(1\right)\)
\(\text{Xét tích}:a\left(b+d\right)=ab+ad\left(2\right)\)
\(b\left(a+c\right)=ba+bc\left(3\right)\)
\(\text{Từ(1);(2);(3)}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\text{ do đó }\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(4\right)\)
\(\text{Tương tự ta có:}\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(5\right)\)
\(\text{Từ (4);(5) ta được }\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
\(\Rightarrow x< y< z\)
\(\dfrac{x}{y}=\dfrac{z}{t}\\ \Rightarrow\dfrac{x}{z}=\dfrac{y}{t}\\ \Rightarrow\dfrac{x}{z}=\dfrac{y}{t}=\dfrac{x-y}{z-t}\\ \Rightarrow\dfrac{x^{1996}}{z^{1996}}=\dfrac{y^{1996}}{t^{1996}}=\left(\dfrac{x-y}{z-t}\right)^{1996}\\ \dfrac{x^{1996}}{z^{1996}}=\dfrac{y^{1996}}{t^{1996}}=\dfrac{x^{1996}+y^{1996}}{z^{1996}+t^{1996}}\\ \Rightarrow\left(\dfrac{x-y}{z-t}\right)^{1996}=\dfrac{x^{1996}+y^{1996}}{z^{1996}+t^{1996}}\)