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 \(\frac{x^{2016}}{a^{2016}}+\frac{y^{2016}}{b^{2016}}+\frac{z^{2016}}{c^{2016}}=\frac{x^{2016}+y^{2016}+z^{2016}}{a^{2016}+b^{2016}+c^{2016}}\)
Cho các số \(a,b,c\ne0\). Tính \(M=x^{2016}+y^{2016}+z^{2016}\). Biết x,y,z thỏa mãn \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
Ta có:
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow\left(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}\right)+\left(\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}\right)+\left(\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}\right)=0\)
\(\Leftrightarrow x^2.\frac{b^2+c^2}{a^2+b^2+c^2}+y^2.\frac{a^2+c^2}{a^2+b^2+c^2}+z^2.\frac{a^2+b^2}{a^2+b^2+c^2}=0\)
Vì a, b, c khác 0 nên dấu bằng xảy ra khi \(x=y=z=0\)
\(\Rightarrow M=x^{2016}+y^{2016}+z^{2016}=0^{2016}+0^{2016}+0^{2016}=0\)
Cho a,b,c khác 0 . Tính giá trị biểu thức T=x2016+y2016+z2016 biết x,y,z thỏa mãn \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\frac{x^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}+\frac{y^2}{a^2+b^2+c^2}-\frac{y^2}{b^2}\)\(+\frac{z^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}=0\)
\(x^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)\)\(+y^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)\)\(=0\)
Vì \(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\ne0,\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\ne0\)\(,\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\ne0\) và \(a,b,c\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\\z^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)\(\Rightarrow T=0\)
Cho các số a,b,c khác 0. Tính giá trị của biểu thức:
T=\(x^{2016}+y^{2016}+z^{2016}\)
Biết x,y,z thỏa mãn: \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
Cô ơi em có cách khác ạ :)
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
Dấu "=" xảy ra tại x=y=z=0
Khi đó T=0
Ta có:
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
<=> \(\left(a^2+b^2+c^2\right)\)\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\left(a^2+b^2+c^2\right)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)
<=> \(x^2+y^2+z^2=\left(a^2+b^2+c^2\right)\frac{x^2}{a^2}+\left(a^2+b^2+c^2\right)\frac{y^2}{b^2}+\left(a^2+b^2+c^2\right)\frac{z^2}{c^2}\)
<=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)
vì a, b , c khác 0 nên \(\frac{\left(b^2+c^2\right)}{a^2};\frac{\left(c^2+a^2\right)}{b^2};\frac{\left(b^2+a^2\right)}{c^2}\ne0\)
\(\frac{\left(b^2+c^2\right)}{a^2}x^2\ge0;\frac{\left(a^2+c^2\right)}{b^2}y^2\ge0;\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x, y, z
=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x; y; z
Do đó: \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)
=> x = y = z = 0
Vậy T = 0
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 \(P=2\left(xy+yz+zx\right)=2.2016=4032\)
1.Cho \(\frac{a_1}{2a_2}=\frac{2a_2}{3a_3}=.......=\frac{2015a_{2015}}{2016a_{2016}}=\frac{2016a_{2016}}{a_1}\) và \(a_1+a_2+a_3+...+a_{2016}\ne0\)
CMR \(a_1=a_2=a_3...=a_{2016}\)
2.Cho\(\frac{a}{2014}=\frac{a}{2015}=\frac{a}{2016}\) CMR:\(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
3.Tìm x,y,z biết \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và \(x^2-\left(x-y\right)=0\)
4.Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) CMR \(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3+b^3}{c^3+d^3}\)
Giúp mình với ạ!Mai phải nộp rồi☹
Tính P=\(x^{2014}+y^{2015}+z^{2016}\) biết x,y,z thỏa mãn điều kiện sau:
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow\left(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}\right)+\left(\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}\right)+\left(\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}\right)=0\)
\(\Leftrightarrow\left(x^2.\frac{b^2+c^2}{a^2+b^2+c^2}\right)+\left(y^2.\frac{a^2+c^2}{a^2+b^2+c^2}\right)+\left(z^2.\frac{a^2+b^2}{a^2+b^2+c^2}\right)=0\)
Vì a,b,c khác
=>Dấu bằng xảy ra khi x=y=z=0
\(\Rightarrow x^{2014}+y^{2015}+z^{2016}=0^{2014}+0^{2015}+0^{2016}=0\)
cho x^2/a^2 + y^2/b^2 + z^2/c^2 =x^2+y^2+z^2/a^2+b^2+c^2
CMR x^2016/a^2016 + y^2016/b^2016 +z^2016/c^2016 = x^2016+y^2016+z^2016/a^2016+b^2016+c^2016
a, Cho các số a,b,c,d nguyên dương đôi một khác nhau thoả mãn:
\(\frac{2a+b}{a+b}+\frac{2b+c}{b+c}+\frac{2c+d}{c+d}+\frac{2d+a}{d+a}=6\). CMR: A = abcd là số chính phương
b, Giải phương trình: \(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-3\left(\frac{2x-4}{x-4}\right)^2=0\)
c, Cho x,y,z dương và x + y + z = 1. CMR: \(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge9\)
d, Tìm nghiệm nguyên dương của phương trình: \(\frac{2016}{x+y}+\frac{x}{y+2015}+\frac{y}{4031}+\frac{2015}{x+2016}=2\)
a) Ta có:
\(\frac{2a+b}{a+b}+\frac{2b+c}{b+c}+\frac{2c+d}{c+d}+\frac{2d+a}{d+a}=6\)
\(\Leftrightarrow\left[\left(\frac{2a+b}{a+b}-1\right)+\left(\frac{2b+c}{b+c}-1\right)-1\right]+\left[\left(\frac{2c+d}{c+d}-1\right)+\left(\frac{2d+a}{d+a}-1\right)-1\right]=0\)
\(\Leftrightarrow\left(\frac{a}{a+b}+\frac{b}{b+c}-1\right)+\left(\frac{c}{c+d}+\frac{d}{d+a}-1\right)=0\)
\(\Leftrightarrow\left(\frac{a.\left(b+c\right)}{\left(a+b\right).\left(b+c\right)}+\frac{b.\left(a+b\right)}{\left(a+b\right).\left(b+c\right)}-\frac{\left(a+b\right).\left(b+c\right)}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{c.\left(d+a\right)}{\left(c+d\right).\left(d+a\right)}+\frac{d.\left(c+d\right)}{\left(c+d\right).\left(d+a\right)}-\frac{\left(c+d\right).\left(d+a\right)}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\left(\frac{ab+ac}{\left(a+b\right).\left(b+c\right)}+\frac{ab+b^2}{\left(a+b\right).\left(b+c\right)}-\frac{ab+ac+b^2+bc}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{cd+ac}{\left(c+d\right).\left(d+a\right)}+\frac{cd+d^2}{\left(c+d\right).\left(d+a\right)}-\frac{cd+ac+d^2+ad}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\left(\frac{ab+ac+ab+b^2-ab-ac-b^2-bc}{\left(a+b\right).\left(b+c\right)}\right)+\left(\frac{cd+ac+cd+d^2-cd-ac-d^2-ad}{\left(c+d\right).\left(d+a\right)}\right)=0\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}+\frac{cd-ad}{\left(c+d\right).\left(d+a\right)}=0\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}=-\frac{cd-ad}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{ab-bc}{\left(a+b\right).\left(b+c\right)}=\frac{ad-cd}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{b.\left(a-c\right)}{\left(a+b\right).\left(b+c\right)}=\frac{d.\left(a-c\right)}{\left(c+d\right).\left(d+a\right)}\)
\(\Leftrightarrow\frac{b}{\left(a+b\right).\left(b+c\right)}=\frac{d}{\left(c+d\right).\left(d+a\right)}\) (vì \(a;b;c;d\) là số nguyên dương).
\(\Leftrightarrow b\left(c+d\right).\left(d+a\right)=d\left(a+b\right).\left(b+c\right)\)
\(\Leftrightarrow\left(bc+bd\right).\left(d+a\right)=\left(ad+bd\right).\left(b+c\right)\)
\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)
\(\Leftrightarrow bd^2+abc=b^2d+acd\)
\(\Leftrightarrow bd^2-b^2d=acd-abc\)
\(\Leftrightarrow bd.\left(d-b\right)=ac.\left(d-b\right)\)
\(\Leftrightarrow bd.\left(d-b\right)-ac.\left(d-b\right)=0\)
\(\Leftrightarrow\left(d-b\right).\left(bd-ac\right)=0\)
Vì \(a;b;c;d\) là số nguyên dương.
\(\Rightarrow d-b>0\)
\(\Rightarrow d-b\ne0.\)
\(\Leftrightarrow bd-ac=0\)
\(\Leftrightarrow bd=ac.\)
Lại có:
\(A=abcd\)
\(\Rightarrow A=ac.bd\)
\(\Rightarrow A=ac.ac\)
\(\Rightarrow A=\left(ac\right)^2.\)
\(\Rightarrow A=abcd\) là số chính phương (đpcm).
Chúc bạn học tốt!
