a) tim GTNN cua A = \(\sqrt{\left(x-1\right)}\)+ \(\sqrt{\left(5+x\right)}\)
b) cho cac so dang x, y thoa man x+y>=3
CM: x+y + 1/2x+ 2/y >= 9/2
a) tim GTNN, GTLN cua A = \(\sqrt{\left(x-1\right)}\)+\(\sqrt{\left(5-x\right)}\)
b) cho cac so duong x,y thoa man x+y>=3
CM: x+y+1/2x+2/y>=9/2
a ) Tìm GTLN : Áp dụng BĐT bunhiacopski, ta có :
Dầu bằng xảy ra khi \(x-1=5-x\Leftrightarrow x=3\).
Sao ko hiện làm lại :
\(\left(\sqrt{x-1}.1+\sqrt{5-x}.1\right)^2\le\) bé hơn hoặc bằng ( 1 + 1 ) ( x - 1 + 5 -x ) = 8
tim cac so x,y,z thoa man dang thuc :
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+Ix+y+zI=0\)
Ta có: \(\sqrt{\left(x-\sqrt{2}\right)^2}\ge0;\sqrt{\left(y+\sqrt{2}\right)^2}\ge0;\left|x+y+z\right|\ge0\)
Mà theo đề: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
=> \(\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{\left(y+\sqrt{2}\right)^2}=\left|x+y+z\right|=0\)
=> \(x-\sqrt{2}=y+\sqrt{2}=x+y+z=0\)
=> \(x=\sqrt{2};y=-\sqrt{2};z=0\).
cho x,y,z nguyen duong thoa man: \(\left\{{}\begin{matrix}\left|x-2y\right|\le\dfrac{1}{\sqrt{x}}\\\left|y-2x\right|\le\dfrac{1}{\sqrt{y}}\end{matrix}\right.\)
tim Max \(A=x^2+2y^2\)
Sau vài phút cố gắng thì khẳng định đề bài của em bị sai
cho x,y,z thuc duong thoa man \(\left\{{}\begin{matrix}\left|x-2y\right|\le\dfrac{1}{\sqrt{x}}\\\left|y-2x\right|\le\dfrac{1}{\sqrt{y}}\end{matrix}\right.\)
tim Max\(A=x^2+2y\)
Đề này còn có lý, lần sau chú ý đọc kĩ đề trước khi đăng lên, tránh làm mất thời gian vô ích:
\(\left|x-2y\right|\le\dfrac{1}{\sqrt{x}}\Rightarrow1\ge\sqrt{x}\left|x-2y\right|\Rightarrow1\ge x\left(x-2y\right)^2\)
\(\Rightarrow1\ge x^3-4x^2y+4xy^2\)
Tương tự: \(\dfrac{1}{\sqrt{y}}\ge\left|y-2x\right|\Rightarrow1\ge y^3-4xy^2+4xy^2\)
Cộng vế:
\(\Rightarrow2\ge x^3+y^3=\dfrac{1}{2}\left(x^3+x^3+1\right)+\left(y^3+1+1\right)-\dfrac{5}{2}\ge\dfrac{1}{2}.3x^2+3y-\dfrac{3}{2}=\dfrac{3}{2}\left(x^2+2y\right)-\dfrac{5}{2}\)
\(\Rightarrow\dfrac{3}{2}\left(x^2+2y\right)\le\dfrac{9}{2}\Rightarrow x^2+2y\le3\)
cho cac so thuc x va y thoa man
\(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)=1\)1
chung minh x+y=0
\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+2-2\sqrt{ }x+1}=\frac{x+5}{2}\)\(\frac{x+5}{2}\)
Xét \(x^2+\sqrt{1+x^2}\)ta có:
\(x^2\ge0\)
nên \(1+x^2\ge1\)
\(\Rightarrow\sqrt{1+x^2}\ge\sqrt{1}=1\)
\(\Rightarrow x^2+\sqrt{1+x^2}\ge1\)
Tương tự ta có:
\(y^2+\sqrt{1+y^2}\ge1\)
Do đó: \(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)\ge1\)
Dấu bằng xảy ra khi \(x=0;y=0\)
Khi đó \(x+y=0\left(ĐPCM\right)\)
Cho x,y, z la cac so duong thoa man dieu kien x+y+z=a
tim GTNN : Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)
Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)
\(Q=\left(\dfrac{x+a}{x}\right)\left(\dfrac{y+a}{y}\right)\left(\dfrac{z+a}{z}\right)\)\
=\(\left(\dfrac{2x+y+z}{x}\right)\left(\dfrac{2y+x+z}{y}\right)\left(\dfrac{2z+x+y}{z}\right)\)
=\(\dfrac{\left(2x+y+z\right)\left(2y+x+z\right)\left(2z+x+y\right)}{xyz}\)
ÁP dụng BĐT cô si
\(2x+y+z=x+x+y+z\ge4\sqrt[4]{x^2yz}\)
\(2y+x+z=y+y+x+z\ge4\sqrt[4]{y^2xy}\)
\(2z+y+x=z+z+x+y\ge4\sqrt[4]{z^2xy}\)
=> Q\(\ge\dfrac{64.\sqrt[4]{x^4y^4z^4}}{xyz}=64\)
=> MinQ=64 khi x=y=z=a/3
Cho 3 so x, y, z thoa man cac he thuc: \(\left(z-1\right)x-y=1\) va \(x+zy=2\)
Chmr: \(\left(2x-y\right)\left(z^2-z+1\right)=7\) va tim tat ca cac so nguyen x, y, z thoa man cac he thuc tren.
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
tim gia tri nho nhat cua bieu thuc P=\(\left(1+x\right)\left(1+\dfrac{1}{y}\right)+\left(1+y\right)\left(1+\dfrac{1}{x}\right)\) trong do x,y la cac so duong thoa man \(x^2+y^2=1\)