cho x,y,z thoa man x^2=yz,y^2=xz,z^2=xy
tinh gia tri bieu thucM=\(\frac{x^{2019}+y^{2019}+z^{2019}}{\left(x+y+z\right)^{2019}}\)
cho x,y ,z là các số dương thỏa mãn:xy+yz+zx=2019
Tính gtrị bt\(P=x\sqrt{\frac{\left(y^2+2019\right).\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right).\left(x^2+2019\right)}{y^{2^{ }}+2019}}+z\sqrt{\frac{\left(x^2+2019\right).\left(y^2+2019\right)}{z^2+2019}}\)
Có \(y^2+2019=y^2+xy+yz+zx=y\left(x+y\right)+z\left(x+y\right)=\left(y+z\right)\left(x+y\right)\)
\(x^2+2019=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)
\(z^2+2019=z^2+xy+yz+xz=z\left(z+y\right)+x\left(y+z\right)=\left(z+x\right)\left(y+z\right)\)
Có \(P=x\sqrt{\frac{\left(y^2+2019\right)\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right)\left(x^2+2019\right)}{y^2+2019}}+z\sqrt{\frac{\left(x^2+2019\right)\left(y^2+2019\right)}{z^2+2019}}\)
=\(x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(z+y\right)}{\left(x+z\right)\left(y+x\right)}}+y\sqrt{\frac{\left(z+x\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
=\(x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=\(x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)
=\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\) (vì x,y,z >0)
= xy+xz+xy+yz+xz+yz
=2(xy+xz+yz)=2.2019(vì xy+xz+yz=2019)
=4038
Vậy P=4038
cho x,y,z >0 thỏa mãn xy+yz+zx=673
CMR: \(\frac{x}{x^2-yz+2019}+\frac{y}{y^2-xz+2019}+\frac{z}{z^2-yx+2019}\ge\frac{1}{x+y+z}\)
Đk: $x\geq \frac{1}{2}$
Pt $\Leftrightarrow 4x^2+3x-7=4(\sqrt{x^3+3x^2}-2)+2(\sqrt{2x-1}-1)$
$\Leftrightarrow +4\frac{(x-1)(x+2)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-(x-1)(4x+7)=0$
$\Leftrightarrow (x-1)[\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-(4x+7)]=0$
$\Leftrightarrow x=1\vee \frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0$ $(*)$
Xét hàm số $f(x)=\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\in [\frac{1}{2};+\infty )$ thì $f(x)>0,\forall x\in [\frac{1}{2};+\infty )$
$\Rightarrow $ Pt $(*)$ vô nghiệm
Giai he phuong trinh:
a) \(\left\{{}\begin{matrix}\left(x+y\right).\left(y+z\right)=187\\\left(y+z\right).\left(z+x\right)=154\\\left(z+x\right).\left(x+y\right)=238\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+y^2+z^2=xy+yz+xz\\x^{2019}+y^{2019}+z^{2019}=3^{2020}\end{matrix}\right.\)
Cho x + y + z = 1 ; x , y , z > 0
CMR : \(\frac{3}{xy+yz+zx}+\frac{2}{x^2+y^2+z^2}\) >/ 14
Cho x , y , z thuộc Z ; x,y,z khác 0 và \(\sqrt{x+y+z-2018}+\sqrt{2018\left(xy+yz+zx-xyz\right)}=0\)
Tính S = \(\frac{1}{x^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}\)
CÁC BẠN GIẢI GIÚP MÌNH CHI TIẾT BÀI NÀY VỚI !
Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
\(\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\)
\(\text{CMR: }x^{2019}+y^{2019}+z^{2019}=\left(x+y+z\right)^{2019}\)
Bổ đề: xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)
Cm:
VT: xyz+(x+y)(y+z)(z+x)
=xyz+xyz+x2z+x2y+y2x+y2z+z2x+z2y+xyz
=xyz+x2z+x2y+xyz+y2z+y2x+xyz+z2x+z2y
=(x+y+z)(xy+yz+xz)
AD bổ đề và đề bài cho
=> (x+y)(y+z)(z+x)=0
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\x+z=0\\y+z=0\end{matrix}\right.\)
1. x+y=0
ta có x2019+y2019=(x+y)(x2018-x2017y+...+y2018)=0
=> x2019+y2019+z2019=z2019
Có (x+y+z)2019=z2019
=> x2019+y2019+z2019= (x+y+z)2019
Làm tương tự với 2 trường hợp còn lại ta được đpcm
Cho 3 số x,y,z thỏa mãn \(\hept{\begin{cases}x+y+z=2019\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2019}\end{cases}}\).Tính giá trị biểu thức \(P=\left(x^{2017}+y^{2017}\right)\left(y^{2019}+z^{2019}\right)\left(z^{2021}+x^{2021}\right)\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z}{\left(x+y+z\right).z}-\frac{x+y+z}{z.\left(x+y+z\right)}=\frac{-x-y}{z.\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{x+y}{-z.\left(x+y+z\right)}\)
TH1: x+y=0
=> x=-y => P=0
TH2: xy=-z.(x+y+z)
\(\Leftrightarrow xy=-xz-zy-z^2\Leftrightarrow xy+xz+zy+z^2=0\Leftrightarrow x.\left(y+z\right)+z.\left(y+z\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left(y+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-z\\y=-z\end{cases}\Rightarrow P=0}\)
Giải hệ phương trình:\(\left\{{}\begin{matrix}x^2+y^2+z^2=xy+yz+zx\\x^{2019}+y^{2019}+z^{2019}=3^{2020}\end{matrix}\right.\)
Ta có: x2 + y2 + z2 = xy + yz + zx
<=> [(x - y)2 + (y - z)2 + (z - x)2] . 1/2 = 0
<=> x = y = z
Thay vào pt thứ 2...
1. Cho xyz = 2019
Cm A = 2019/ 2019 +x + xy
+ 2019/ 2019 + z + xz + 2019/ 2019 + y + yz thuộc N
2. Tìm GTLN, GTNN ( nếu có )
A= 4x - 9 / |x|
3. So sánh
a) 3 × căn 2 và 7,(21)
b) 1/ căn 1 + căn 2 + 1/ căn 2 + căn 3 + ...... + 1/ căn 99 + căn 100 và 9
4. Tính S = x+ y + z biết 19/ x+ y + 19/ y + z + 19/ x + z = 14x/ y + z + 14y/ z + x + 14z/ x + y = 133/5
Với x,y,z thoả mãn x + y = z = 3xyz và x + y + z khác 0
Tính A = \(\frac{x^{2019}+y^{2019}+z^{2019}}{\left(x+y+z\right)^{2019}}\)