Cho các số thực a, b, c, x, y, z thoả mãn abc khác 0 và:
\(\dfrac{x^4+y^4+z^4}{a^4+b^4+c^4}=\dfrac{x^4}{a^4}+\dfrac{y^4}{b^4}+\dfrac{z^4}{c^4}\)
Tính: \(P=10x^{10}+100y^{100}+1000z^{1000}+10000\)
Bài 1: a;b;c > 0 và abc = 1
Chứng minh : \(\dfrac{a}{b^4+c^4+a}+\dfrac{b}{a^4+c^4+b}+\dfrac{c}{a^4+b^4+c}\le1\)
Bài 2: x;y;z > 0 và x + y + z = 2
Chứng minh : \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
1.
Ta có:
\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\left(x^2+y^2\right)xy\)
Đặt vế trái của BĐT cần chứng minh là P, áp dụng bồ đề vừa chứng minh ta có:
\(P\le\dfrac{a.abc}{bc\left(b^2+c^2\right)+a.abc}+\dfrac{b.abc}{ca\left(c^2+a^2\right)+b.abc}+\dfrac{c.abc}{ab\left(a^2+b^2\right)+c.abc}\)
\(P\le\dfrac{a^2.bc}{bc\left(a^2+b^2+c^2\right)}+\dfrac{b^2.ac}{ca\left(a^2+b^2+c^2\right)}+\dfrac{c^2.ab}{ab\left(a^2+b^2+c^2\right)}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
2.
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)
Cho x, y, z là các số thoả mãn:
\(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{12}-\dfrac{z}{4}=1\\\dfrac{x}{10}+\dfrac{y}{5}+\dfrac{z}{3}=1\end{matrix}\right.\)
Tính \(M=x^{10}+y^{100}+z^{1000}\)
cho các số thực a,b,c,x,y,z thỏa mãn a,b,c khác 0 và ( x^4 +y^4 +z^4)/(a^4+b^4+c^4)=x^4/a^4+y^4/b^4+z^4/c^4,tính P=x^2+y^9+z^1945+2017
Cho x, y, z > 0 thoả mãn x+y+z=2. Tìm GTNN của các biểu thức:
a) \(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
b) \(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
c) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
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x5=y7=z3=x225=y249=z29" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x5=y7=z3=x225=y249=z29
x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
Cho các số thực x, y, z thỏa mãn \(x^2+y^2+z^2=5\) và x - y + z = 3 . Giá trị nhỏ nhất của biểu thức \(P=\dfrac{x+y-2}{z+2}\) bằng
A. \(\dfrac{1}{2}\) B. \(0\) C. \(\dfrac{-36}{23}\) D. \(\dfrac{-13}{4}\)
1)cho Q=\(\dfrac{a^4+a^3-a^2-2a-2}{a^4+2a^3-a^2-4a-2}\)
Tìm GTNN của Q
2)cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\) và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
CMR: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
Cho x, y, z > 0 thoả mãn x+y+z=2. Tìm GTNN của các biểu thức:
a) \(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
b) \(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
c) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
Cho các số x,y,z khác 0 thỏa mãn 2(x+y)=3(y+z)=4(x+z). Tính P = \(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\)
Bạn tham khảo tại đây:
https://hoc24.vn/cau-hoi/cho-xyz-khac-0-thoa-man-2-xy-3yz4zx-tinh-p-dfracxydfracyzdfraczx.3861996653762
a)A=\(\dfrac{5}{X}+\dfrac{Y}{5}+\dfrac{1}{Z}\) tại X=\(\dfrac{1}{2}\); Y=20; Z=\(\dfrac{-1}{4}\)
b)B=\(\dfrac{4x+7y}{x-3y}tại\dfrac{y}{x}=\dfrac{1}{4}\) (x,y khác 0)
a:\(A=5:\dfrac{1}{2}+\dfrac{20}{5}+1:\dfrac{-1}{4}=10+4-4=10\)
b: y/x=1/4
nên x=4y
\(A=\dfrac{4x+7y}{x-3y}=\dfrac{16y+7y}{4y-3y}=23\)