tim gia tri cua bieu thuc X=2x^5-5y^3+2015 tai x,y thoa man /x-1/=(y+2)^20=0
Những câu hỏi liên quan
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 cac so thuc x , y thay doi thoa man x + y = 2 . Tim gia tri nho nhat cua bieu thuc P = ( x4 + 1 )(y4 + 1) + 2013
ap dung bunhiacopki
\(\left(x^4+1\right)\left(y^4+1\right)>=\left(x^2+y^2\right)^2>=\left[\frac{\left(x+y\right)^2}{2}\right]^2=4\)
do do P>=4+2013=2017
= xảy ra <=>x=y=1
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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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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 = ?
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=
1. tim x biet :
a, (x-2)(x+3) > 2x\(^2\) -x -5
b, x( x-5) > x-4
2. cho 2 so x va y thoa man : x+y = 7 va xy=2 . khong tinh x va y , hay tinh gia tri cua bieu thuc A= x - y ( biet x< y)
Câu 1:
a: \(\Leftrightarrow2x^2-x-5< x^2+x-6\)
\(\Leftrightarrow x^2-2x+1< 0\)
hay \(x\in\varnothing\)
b: \(\Leftrightarrow x^2-5x-x+4>0\)
\(\Leftrightarrow x^2-6x+4>0\)
\(\Leftrightarrow\left(x-3\right)^2>5\)
hay \(\left[{}\begin{matrix}x>\sqrt{5}+3\\x< -\sqrt{5}+3\end{matrix}\right.\)
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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
1. tich cac gia tri x thoa man x2-\(\frac{1}{10}\)x-\(\frac{1}{5}\)+0
2.gia tri cua x de bieu thuc A=6x2+7x-8 dat gia tri nho nhat
3.tích cac gia tri cua x thoa man (2x-1)2=(2-x)2