cho xyz la cac so thuc thoa man 2x=3y=5z va /x-2y/=5 tinh gia tri lon nhat cua 3x-2z
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1. Cho x,y la cac so thuc ko am con z la so thuc bat ki thoa man : 3x+5y-4z=23 va x-2y+6z=4 .Tap gtri cua T =2x-3y+6z co bn so nguyen duong
cho ham so y=2x^2-3x+1. gia tri lon nhat cua x thoa man f(-x)=f(2x) la x=?
cho x,y,z la cac so huu ti duong thoa man x+1/yz y +1/xz z+1/xy la cac so nguyen tim gia tri lon nhat cua bieu thuc A=x+y^2+z^3
tim cac so x,y,z ko am thoa man x+6y=12 va 4x +5z=2018 sao cho F=x+y+z co gia tri lon nhat
1. gia tri cua x thoa man (2x+3)(x+1)2-(2x+3)(2x-3)
2. so nghiem cua da thuc x3+6x2+11x+6 la ?
3. bieu thuc C = 8-5x-2x2 dat gia tri lon nhat tai x = ?
cho x,y,z la cac so thuc duong thoa man x+y+z=1 tim gia tri nho nhat cua bieu thuc M=1/16x+1/4y+1/z
\(M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\)
\(M=\frac{1}{16x}+\frac{4}{16y}+\frac{16}{16z}\)
\(M=\frac{1^2}{16x}+\frac{2^2}{16y}+\frac{4^2}{16z}\)
\(M\ge\frac{\left(1+2+4\right)^2}{16\left(x+y+z\right)}\)
\(=\frac{49}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}=\frac{1+2+4}{16\left(x+y+z\right)}=\frac{7}{16}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}\)
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Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow x+y+z\ge3\sqrt[3]{xyz}\)
\(\Rightarrow1\ge3\sqrt[3]{xyz}\)
\(\Rightarrow\frac{1}{27}\ge xyz\)
Ta có \(M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{64xyz}}\)( 1 )
Xét \(3\sqrt[3]{\frac{1}{64xyz}}\)
Ta có \(\frac{1}{27}\ge xyz\)
\(\Rightarrow\frac{64}{27}\ge64xyz\)
\(\Rightarrow\frac{27}{64}\le\frac{1}{64xyz}\)
\(\Rightarrow\frac{9}{4}\le3\sqrt[3]{\frac{1}{64xyz}}\)( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{64xyz}}\ge\frac{9}{4}\)
Vậy \(M_{min}=\frac{9}{4}\)
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\(M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}=\frac{1}{16x}+\frac{4}{16y}+\frac{16}{16z}=\frac{1^2}{16x}+\frac{2^2}{16y}+\frac{4^2}{16z}\)
Áp dụng bất đẳng thức Cauchy Schawrz dạng Engel ta được:
\(M=\frac{1^2}{16x}+\frac{2^2}{16y}+\frac{4^2}{16z}\ge\frac{\left(1+2+4\right)^2}{16x+16y+16z}=\frac{7^2}{16\left(x+y+z\right)}=\frac{49}{16.1}=\frac{49}{16}\)
Dấu "=" xảy ra khi \(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}\). Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}=\frac{1+2+4}{16x+16y+16z}=\frac{7}{16\left(x+y+z\right)}=\frac{7}{16.1}=\frac{7}{16}\)
=>\(x=\frac{1}{7};y=\frac{2}{7};z=\frac{4}{7}\)
Vậy Mmin=49/16 khi \(x=\frac{1}{7};y=\frac{2}{7};z=\frac{4}{7}\)
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1. so x0 thoa man frac{5}{x}frac{x+1}{22}2. GTLN cua x thoa man 5^{left(x-2right)left(x+3right)}3. Gia tri x0 thoa man /x^2+50//-2x^2-1/4. Gia tri nguyen cua x thoa man left(3x-4right)^5^{left(3x-4right)^7}la x5. So gia tri cua x thoa man /x-2014/+/x-2015/06. Gia tri cua x de bieu thuc Afrac{x-3+5}{3-x+2}dat gia tri lon nhat la x
Đọc tiếp
1. so x>0 thoa man \(\frac{5}{x}=\frac{x+1}{22}\)
2. GTLN cua x thoa man \(5^{\left(x-2\right)\left(x+3\right)}\)
3. Gia tri x<0 thoa man /\(x^2+50\)/=/\(-2x^2\)-1/
4. Gia tri nguyen cua x thoa man \(\left(3x-4\right)^5\)=\(^{\left(3x-4\right)^7}\)la x=
5. So gia tri cua x thoa man /x-2014/+/x-2015/=0
6. Gia tri cua x de bieu thuc A=\(\frac{x-3+5}{3-x+2}\)dat gia tri lon nhat la x=
giup to 3 bai nay
1. gia tri cua x thoa man (2x+3)(x+1)2-(2x+3)(2x-3)
2. so nghiem cua da thuc x3+6x2+11x+6 la ?
3. bieu thuc C = 8-5x-2x2 dat gia tri lon nhat tai x = ?
giup to 3 bai nay
1. gia tri cua x thoa man (2x+3)(x+1)2-(2x+3)(2x-3)
2. so nghiem cua da thuc x3+6x2+11x+6 la ?
3. bieu thuc C = 8-5x-2x2 dat gia tri lon nhat tai x = ?