1)Rut gon bieu thuc:P=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
2) Cho 3 so duong a,b,c thoa man dieu kien:a+b+c=\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Chung minh rang:a=b=c
cho a,b,c la ba so thuc duong thoa man dieu kien a+b+c=1
chung minh rang P=\(\sqrt{\frac{ab}{c+ab}}+\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}\le\frac{3}{2}\)
lấy bút xóa mà xóa hết là khỏe
cho ca so a,b,c duong thoa man ab+bc+ca =1 chung minh \(P=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\le\frac{1}{4}\)
Cho 3 so duong a,b,c thoa man dieu kien : a+b+c=1. Chung minh rang
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5\)
Áp dụng BĐT Bunhia:
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{\left(1+1+1\right)\left(4a+1+4b+1+4c+1\right)}\)
\(\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{3.\left(4\left(a+b+c\right)+3\right)}=\sqrt{21}< \sqrt{25}=5\)
Vậy \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5\)
cho cac so a,b,c duong thoa man ab+bc+ca=1 chung minh : \(p=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
tim tat ca cac so duong a,b,c thoa man dieu kien \(\left\{{}\begin{matrix}\sqrt{a}+\sqrt{b}+\sqrt{c}=6\\\frac{1}{\sqrt{a}}+\frac{4}{\sqrt{b}}+\frac{9}{\sqrt{c}}=6\end{matrix}\right.\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1^2}{\sqrt{a}}+\frac{2^2}{\sqrt{b}}+\frac{3^2}{\sqrt{c}}\ge\frac{\left(1+2+3\right)^2}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\frac{36}{6}=6\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\frac{1}{\sqrt{a}}=\frac{2}{\sqrt{b}}=\frac{3}{\sqrt{c}}\\\sqrt{a}+\sqrt{b}+\sqrt{c}=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a}=1\\\sqrt{b}=2\\\sqrt{c}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=9\end{matrix}\right.\)
cho 2 bieu thuc:
A=(\(\sqrt{20}\) -\(\sqrt{45}\) +3\(\sqrt{5}\) ).\(\sqrt{5}\) va B=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}\) +\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) (Dieu kien: x>0, x khac 1
a) Rut gon bieu thuc A va B
b)Tim cac gia tri cua x de gia tri cua bieu thuc A bang 2lan gia tri B
a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
b: A=2B
=>\(10=4\sqrt{x}-2\)
=>\(4\sqrt{x}=12\)
=>x=9(nhận)
1 cho 3 so thuc duong thoa man x^2010+y^2010+z^2010=3 tim gia tri lon nhat cua x^2+y^2+z^2
2 cho a;b;c duong c/m \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>hoac=3\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)
3 tim gia tri nho nhat cua \(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ac+a^2}\) voi a+b+c=1
4 cho a;b;c;d va A;B;C;D la cac so duong thoa man \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)C/ M \(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)
5 tim gia tri lon nhat cua \(\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)
6 phan tich da thuc thanh nhan tu \(y-5x\sqrt{y}+6x^2\)
7 cho x;y;z>0 xy+yz+xz=1 tinh \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
8 cho a;b;c >0 c/m \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}
pn oi nhieu the nay ai ma giai cho het dc
bài lớp mấy mà nhìn ghê quá zật bạn..................Nhìu quá
cho bieu thuc:P=\(\frac{\sqrt{x}}{\sqrt{x}-3}\)+\(\frac{2\sqrt{x}}{\sqrt{x}-3}\)--\(\frac{3x+9}{x-9}\) voi x>= 0;x#9 .a; Rut gon bieu thuc P . b; Tinh gia tri cua bieu thuc voi \(x=4-2\sqrt{3}\)
Bai 1)Cho bieu thuc A=\(\frac{x+y-2\sqrt{xy}}{x-y}\)
a)Tim dieu kien de A co nghia
b)Rut gon A
c)Tinh A biet x=\(3+2\sqrt{2}\)va y=\(3-2\sqrt{2}\)
Bai 2) Cho bieu thuc B=\(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\)
a)Tim dieu kien de B co nghia
b)Rut gon B
c) Tinh B voi x=\(4\left(2-\sqrt{3}\right)\)
d)Tim x de B co gia tri nho nhat
a) A có nghĩa\(\Leftrightarrow x-y\ne0\Leftrightarrow x\ne y\)
b) \(A=\frac{x+y-2\sqrt{xy}}{x-y}=\frac{\left(\sqrt{x-\sqrt{y}}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)