Chứng minh rằng nếu\(B:\frac{a^{2016}+3b^{2016}}{c^{2016}+3d^{2016}}=\left(\frac{a^2+2b^2}{c^2+2d^2}\right)^2\)
Chứng minh rằng nếu a/b=c/d\(a,\frac{a^2+2c^2}{b^2+2d^2}=\left(\frac{a+3c}{b+3d}\right)^2\) \(\frac{a^{2016}+3b^{2016}}{c^{2016}+3d^{2016}}=\left(\frac{a^2+2b^2}{c^2+2d^2}\right)^2\)
CMR nếu a/b=c/d thì a^2016+3b^2016/c^2016+3d^2016=(a^2+2b^2/c^2+2d^2)^1008
CMR nếu a/b=c/d thì a^2016+3b^2016/c^2016+3d^2016=(a^2+2b^2/c^2+2d^2)^1008
CHO A/B=C/D CHỨNG MINH RẰNG
\(\frac{\left(a-c\right)^4}{\left(b-d\right)^4}=\frac{5a^4+7c^4}{5b^4+7d^4}\)
\(\frac{a+2c}{b+2d}=\frac{a-3c}{b-3d}\)
\(\frac{a^{2016}+c^{2016}}{b^{2016}+d^{2016}}=\frac{\left(a-c\right)^{2016}}{\left(b-d\right)^{2016}}\)
AI LÀM ĐƯỢC CÂU NÀO CŨNG ĐC,GIÚP MÌNH VS GẤP LẮM,THANKS
a, \(\frac{a}{b}=\frac{c}{d}=\frac{a-c}{b-d}\Rightarrow\frac{a^4}{b^4}=\frac{c^4}{d^4}=\frac{\left(a-c\right)^4}{\left(b-d\right)^4}\) (1)
\(\frac{a^4}{b^4}=\frac{c^4}{d^4}=\frac{5a^4}{5b^4}=\frac{7c^4}{7d^4}=\frac{5a^4+7c^4}{5b^4+7d^4}\)(2)
Từ (1) và (2) => đpcm
b, \(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+2c}{b+2d}\) (3)
\(\frac{a}{b}=\frac{c}{d}=\frac{3c}{3d}=\frac{a-3c}{b-3d}\) (4)
Từ (3) và (4) => đpcm
c, làm giống câu a
a) ta có \(\frac{a}{b}=\frac{c}{d}=\frac{a+2c}{b+2d}\left(1\right)\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a-3c}{b-3d}\left(2\right)\)
(1) và (2) => \(\frac{a+2c}{b+2d}=\frac{a-3c}{b-3d}\)
Tương tự \(\left(\frac{a}{b}\right)^4=\left(\frac{c}{d}\right)^4=\left(\frac{a-c}{b-d}\right)^4\left(1\right)\)
\(\left(\frac{a}{b}\right)^4=\left(\frac{c}{d}\right)^4=\frac{5a^4+7c^4}{5b^4+7d^4}\left(2\right)\)
=> \(\left(\frac{a-c}{b-d}\right)^4=\frac{5a^4+7c^4}{5b^4+7d^4}\)
Cho a, b, c>0. Chứng minh rằng
a2016+b2016+c2016>=\(\frac{\left(b+c\right).a^{2015}}{2}\)+\(\frac{\left(c+a\right).b^{2015}}{2}\)+\(\frac{\left(a+b\right).c^{2015}}{2}\)
Cho a,b,c >0; biết \(\hept{\begin{cases}a^2=b+4032\\x+y+z=a\\x^2+y^2+z^2=b\end{cases}}\)
\(P=x\sqrt{\frac{\left(2016+y^2\right)\left(2016+z^2\right)}{2016+x^2}}+y\sqrt{\frac{\left(2016+z^2\right)\left(2016+x^2\right)}{\left(2016+y^2\right)}}+z\sqrt{\frac{\left(2016+x^2\right)\left(2016+y^2\right)}{\left(2016+z^2\right)}}\)
Chứng minh giá trị của P không phụ thuộc vào x,y,z
Bạn thêm điều kiện x,y,z lớn hơn 0 nhé :)
Từ giả thiết ta suy ra : \(a^2=b+4032\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4032\)
\(\Rightarrow xy+yz+zx=2016\)thay vào :
\(x\sqrt{\frac{\left(2016+y^2\right)\left(2016+z^2\right)}{2016+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}\)
\(=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+y\right)\left(z+x\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=x\left|y+z\right|=xy+xz\)vì x,y,z > 0
Tương tự : \(y\sqrt{\frac{\left(2016+z^2\right)\left(2016+x^2\right)}{2016+y^2}}=xy+zy\)
\(z\sqrt{\frac{\left(2016+x^2\right)\left(2016+y^2\right)}{2016+z^2}}=zx+zy\)
Suy ra \(P=2\left(xy+yz+zx\right)=2.2016=4032\)
Cho các số nguyên dương a,b,c,d và \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng: \(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}=\frac{\left(a^{2017}-b^{2017}\right)^{2016}}{\left(c^{2017}-d^{2017}\right)^{2016}}\)
Cho \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\).Chứng minh \(\frac{x^{2016}}{a^{2016}}+\frac{y^{2016}}{b^{2016}}+\frac{z^{2016}}{c^{2016}}=\frac{x^{2016}+y^{2016}+z^{2016}}{a^{2016}+b^{2016}+c^{2016}}\)
Chứng minh rằng:
a) Nếu \(\frac{a-c}{c-b}\)=\(\frac{a}{b}\)thì \(\frac{1}{c}\)=\(\frac{1}{2}\)\(\times\)(\(\frac{1}{a}\)\(+\)\(\frac{1}{b}\))
b)Nếu \(\frac{a}{b}\)=\(\frac{a}{d}\)thì\(\frac{2a^{^{2016}}}{2c^{ }^{2016}}\)\(+\)\(\frac{5b^{2016}}{5d^{ }^{2016}^{ }}\)=\(\frac{\left(a+b\right)^{2016}}{\left(c+d\right)^{2016}^{ }}\)
Giúp mình giải thêm 2 câu này với ạ!Mình cảm ơn nhiều ạ!
\(\frac{a-c}{c-b}=\frac{a}{b}\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow ba-bc=ac-ab\)
\(\Rightarrow2ab=ac+bc=c\left(a+b\right)\)
\(\Rightarrow\frac{2ab}{\left(a+b\right)}=c\Rightarrow\frac{a+b}{2ab}=\frac{1}{c}\Rightarrow\frac{1}{2}.\left(\frac{a}{ab}+\frac{b}{ab}\right)=\frac{1}{c}\Rightarrow\frac{1}{2}.\left(\frac{1}{b}+\frac{1}{a}\right)=\frac{1}{c}\)
Câu b ấy, hình như sai đề, phải bằng \(\frac{a^{2016}+b^{2016}}{c^{2016}+d^{2016}}\)có lẽ mới đúng
nếu như câu b đề như thế thì bạn có thể giải giúp mình được ko? mình cảm ơn bạn nhé!
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^{2016}=\left(\frac{b}{d}\right)^{2016}\)
\(\Rightarrow\frac{a^{2016}}{c^{2016}}=\frac{2a^{2016}}{2c^{2016}}=\frac{b^{2016}}{d^{2016}}=\frac{5b^{2016}}{5d^{2016}}=\frac{a^{2016}+b^{2016}}{c^{2016}+d^{2016}}\)