Cho x+y+z=0
CMR \(x^5+y^5+z^5=\frac{5}{2}xyz\left(x^2+y^2+z^2\right)\)
Giải các hệ phương trình sau:
a) \(\hept{\begin{cases}x^3+y^3+x^2\left(y+z\right)=xyz+14\\y^3+z^3+y^2\left(x+z\right)=xyz-21\\z^3+x^3+z^2\left(x+y\right)=xyz+7\end{cases}}\)
b)\(\hept{\begin{cases}\frac{xyz}{x+y}=2\\\frac{xyz}{y+z}=\frac{6}{5}\\\frac{xyz}{x+z}=\frac{3}{2}\end{cases}}\)
Bài b nhé bạn!
\(\hept{\begin{cases}\frac{xyz}{x+y}=2\\\frac{xyz}{y+z}=\frac{6}{5}\\\frac{xyz}{x+z}=\frac{3}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{x+y}{xyz}=\frac{1}{2}\\\frac{y+z}{xyz}=\frac{5}{6}\\\frac{x+z}{xyz}=\frac{2}{3}\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{1}{yz}+\frac{1}{xz}=\frac{1}{2}\\\frac{1}{xz}+\frac{1}{xy}=\frac{5}{6}\\\frac{1}{xy}+\frac{1}{yz}=\frac{2}{3}\end{cases}}\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=\frac{\frac{1}{2}+\frac{5}{6}+\frac{2}{3}}{2}=1\)
Trừ lại từng phương trình trong hệ:
\(\hept{\begin{cases}\frac{1}{xy}=\frac{1}{2}\\\frac{1}{yz}=\frac{1}{6}\\\frac{1}{xz}=\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}xy=2\\yz=6\\xz=3\end{cases}\Rightarrow xyz=\sqrt{2.6.3}=6}\)
Chia lại từng phương trình trong hệ mới, được:
\(\hept{\begin{cases}z=3\\x=1\\y=2\end{cases}}\)
Vậy \(\left(x;y;z\right)=\left(1;2;3\right)\)
Xong rồi đó!!!
Tìm max
\(A=3\sqrt{2x-1}+x\sqrt{5-4x^2}\left(\frac{1}{2}\le x\le\frac{\sqrt{5}}{2}\right)\)
\(B=\frac{xyz\left(x+y+z+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}\left(x,y,z>0\right)\)
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
1) Cho x,y,z>0 thoả mãn : xyz<=1. Chứng minh rằng: \(\frac{x\left(1-y^3\right)}{y^3}\)+ \(\frac{y\left(1-z^3\right)}{z^3}\)+\(\frac{z\left(1-x^3\right)}{x^3}\)>=0
2) Cho x, y, z là các số thực dương thỏa mãn x ≥ z. CMR: xz /(y^2 + yz) + y^2 / (xz + yz) + (x + 2z)/(x + z) ≥ 5/2
tìm max:
a, \(A=3\sqrt{2x-1}+x\sqrt{5-4x^2}\) với 1/2<=x<= căn 5/2
b, \(B=\frac{xyz\left(x+y+z+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)};x,y,z>0\)
hpt
\(\left\{\begin{matrix}\frac{x+y}{xyz}=\frac{1}{2}\\\frac{y+z}{xyz}=\frac{5}{6}\\\frac{x+z}{xyz}=\frac{2}{3}\end{matrix}\right.\)
Đặt \(\left ( \frac{1}{xy},\frac{1}{yz},\frac{1}{xz} \right )=(a,b,c)\)
\(\text{HPT}\Leftrightarrow \left\{\begin{matrix} b+c=\frac{1}{2}\\ c+a=\frac{5}{6}\\ a+b=\frac{2}{3}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2b=\frac{2}{3}+\frac{1}{2}-\frac{5}{6}\\ 2c=\frac{1}{2}+\frac{5}{6}-\frac{2}{3}\\ 2a=\frac{5}{6}+\frac{2}{3}-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} b=\frac{1}{6}\\ c=\frac{1}{3}\\ a=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} yz=6\\ xz=3\\ xy=2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=1\\ y=2\\ z=3\end{matrix}\right.\)
\(\left\{\begin{matrix}\frac{x+y}{xyz}=\frac{1}{2}\\\frac{y+z}{xyz}=\frac{5}{6}\\\frac{x+z}{xyz}=\frac{2}{3}\end{matrix}\right.\).Cộng theo vế ta có:
\(\frac{x+y+y+z+x+z}{xyz}=\frac{1}{2}+\frac{5}{6}+\frac{2}{3}=2\)
\(\Leftrightarrow\frac{2\left(x+y+z\right)}{xyz}=2\Rightarrow2\left(x+y+z\right)=2xyz\)
\(\Leftrightarrow x+y+z=xyz\). Thay vào hệ đầu ta có:
\(\left\{\begin{matrix}\frac{x+y}{x+y+z}=\frac{1}{2}\\\frac{y+z}{x+y+z}=\frac{5}{6}\\\frac{x+z}{x+y+z}=\frac{2}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}2\left(x+y\right)=x+y+z\\6\left(y+z\right)=5\left(x+y+z\right)\\3\left(x+z\right)=2\left(x+y+z\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}2\left(x+y\right)=x+y+z\\\frac{6}{5}\left(y+z\right)=x+y+z\\\frac{3}{2}\left(x+z\right)=x+y+z\end{matrix}\right.\)
\(\Leftrightarrow2x+2y=\frac{6}{5}y+\frac{6}{5}z=\frac{3}{2}x+\frac{3}{2}z=x+y+z\)\(\Leftrightarrow\left\{\begin{matrix}y=2x\\z=3x\end{matrix}\right.\)
1.Cho x+y+z=0. CMR:
a) \(5\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)=6\left(x^5+y^5+z^5\right)\)
b) \(x^7+y^7+z^7=7xyz\left(x^2y^2+y^2z^2+z^2x^2\right)\)
c) \(10\left(x^7+y^7+z^7\right)=7\left(x^2+y^2+z^2\right)\left(x^5+y^5+z^5\right)\)
d) \(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
2. Tìm n∈ N để biểu thức sau là số nguyên tố
a) \(A=n^3-4n^2-4n-1\)
b) \(B=n^3-6n^2+9n-2\)
c) \(C=n^{1975}+n^{1973}+1\)
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
1c. Sử dụng kq phần a,b:
\(10(x^7+y^7+z^7)=70xyz(xy+yz+xz)^2\)
\(=-35xyz(xy+yz+xz).-2(xy+yz+xz)=-35xyz(x+y+z)(x^2+y^2+z^2)\)
\(=\frac{7}{6}.-30xyz(xy+yz+xz)(x^2+y^2+z^2)=\frac{7}{6}.6(x^5+y^5+z^5).(x^2+y^2+z^2)\)
\(=7(x^5+y^5+z^5)(x^2+y^2+z^5)\)
(đpcm)
1d. Áp dụng kq phần a
$6(x^5+y^5+z^5)=-30xyz(xy+y+xz)=15xyz.-2(xy+yz+xz)=15xyz(x^2+y^2+z^2)$
$\Rightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)$ (đpcm)
Tìm các số hữu tỉ x, y, z biết rằng:
\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}\)
\(\left(x+y\right)xyz^2+\left(y+z\right)yzx^2+\left(z+x\right)zxy^2=477120\)
Đặt \(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=k\Rightarrow x=3k;y=5k;z=7k\)
\(xy+yz+zx=3k.5k+5k.7k+7k.3k=k^2\left(15+35+21\right)=71k^2;xyz=3k.5k.7k=105k^3\)
Ta có : \(xyz\left(xz+yz+xy+xz+yz+xy\right)=477120\)
\(\Rightarrow xyz\left(xz+yz+xy\right)=238560\)\(\Rightarrow105k^3.71k^2=238560\Rightarrow k^5=32=2^5\Rightarrow k=2\)
Vậy : x= 6 ; y = 10 ; z = 14
1/ Giải phương trình: \(x^2-\sqrt{x+5}=5\)
2/Cho x,y,z là ba số thỏa mãn: \(xyz=1\)và \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\).Tính giá trị của biểu thức:
\(P=\left(x^{19}-1\right)\left(y^5-1\right)\left(z^{2013}-1\right)\)
2/ \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow x+y+z=xy+yz+zx\)
\(\Leftrightarrow x+y+z-xy-yz-zx+xyz-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=1\\z=1\end{cases}}\)
\(\Rightarrow P=0\)
\(x^2-\sqrt{x+5}=5\)
\(\Leftrightarrow x^2-5=\sqrt{x+5}\)
\(\Leftrightarrow x^4-10x^2+25=x+5\)
\(\Leftrightarrow x^4-10x^2-x+20=0\)
\(\Leftrightarrow\left(x^2-x-5\right)\left(x^2+x-4\right)=0\)
Cho x,y,z > 0 và \(xyz\ge1\)
CMR: \(\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{y^5+z^2+x^2}+\frac{z^5-z^2}{z^5+x^2+y^2}\ge0\)
dinh lam nhung thoi vi chac chan se con nguoi vao lam ho :)
Áp dụng BĐT AM-GM ta có:
\(\frac{4\left(x^5-x^2\right)}{x^5+y^2+z^2}+1=\frac{5x^5-4x^2+y^2+z^2}{x^5+y^2+z^2}=\frac{3x^5+\left(2x^5+y^2+z^2-4x^2\right)}{x^5+y^2+z^2}\)
\(\ge\frac{3x^5+4\sqrt[4]{x^{10}y^2z^2}-4x^2}{x^5+y^2+z^2}\ge\frac{3x^5}{x^5+y^2+z^2}=\frac{3x^4}{x^4+\frac{y^2+z^2}{x}}\ge\frac{3x^4}{x^4+yz\left(y^2+z^2\right)}\ge\frac{3x^4}{x^4+y^4+z^4}\)
suy ra: \(\frac{x^5-x^2}{x^5+y^2+z^2}\ge\frac{3}{4}.\frac{x^4}{x^4+y^4+z^4}-\frac{1}{4}\)
tương tự ta có: \(\frac{y^5-y^2}{y^5+z^2+x^2}\ge\frac{3}{4}.\frac{y^4}{x^4+y^4+z^4}-\frac{1}{4}\)
\(\frac{z^5-z^2}{z^5+y^2+x^2}\ge\frac{3}{4}.\frac{z^4}{x^4+y^4+z^4}-\frac{1}{4}\)
Cộng theo vế ta được:
\(VT\ge\frac{3}{4}.\frac{x^4+y^4+z^4}{x^4+y^4+z^4}-\frac{3}{4}=0\)
Vậy BĐT đc c/m
p/s: bài này mk cx k chắc (nhờ bn ktra nó kêu cứ sai sai nên mk cx k rõ) bạn tham khảo, sai đâu ib cho mk nhé
thân ái!