cho a c la cac so thuc khong am thoa man dieu kien a+b+c=3 tim gia tri lon nhat cua bieu thucP=a\b3+1+b\c3+1+c\a3+1
1. Cho a,b,c,d la cac so nguyen thoa man \(a^2=b^2+c^2+d^2\)
chung minh rang a.b.c.d + 2015 viet duoc duoi dang hieu cua 2 so chinh phuong.
2. Cho a,b la cac so duong thoa man dieu kien a+b=1. tim gia tri nho nhat cua bieu thuc
\(P=\frac{2+a}{\sqrt{2-a}}+\frac{2+b}{\sqrt{2-b}}\)
cho a,b,c la 3 so thuc duong thoa man dieu kien (a+b-c)/c=(b+c-a)/a=(c+a-b)/b tinh gia tri bieu thuc B=(1+b/a)*(1+a/c)*(1+c/b)
(a+b-c)/c+2 =(b+c-a)/c+2 =(c+a-b)/c+2
rồi bạn tự làm tiếp nhé
xét 2 trường hợp
thay vào thôi nhé bạn
Nhớ k cho mình nhé
cho 3 so a,b,c thoa man 0<a<b<c<1. tim gia tri lon nhat cua bieu thuc B=(a+b+c+3)[1/(a+1)+1/(b+1)+1/(c+1)]
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
giúp mình với
cho a,b,c >0 thoa man dieu kien a^2 +b^2 +c^2 = 1
tinh gia tri nho nhat cua bieu thuc A= ab/c + bc/a + ca/b
cho x,y la cac so duong thay doi va thoa man dieu kien x+y\(\le\)1. tim gia tri nho nhat cua bieu thuc M=\(\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy\)
Ta có: \(\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy\)
\(=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)\(\ge4+2+1=7\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
Vậy \(\left(\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy\right)_{Min}=7\Leftrightarrow x=y=\frac{1}{2}\)
à nhầm, bạn pham trung thanh làm đúng rồi đấy mọi người ủng hộ bạn ấy nha
Cho a,b,c la cac so duong thoa man a+b+c=9.Tim gia tri nho nhat cua bieu thuc:
\(P=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+c^2+\frac{1}{c^2}\)
Ta có:\(P=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+c^2+\frac{1}{c^2}\)
\(\Rightarrow P\ge a^2+b^2+c^2+\frac{9}{a^2+b^2+c^2}\)(bđt cauchy-schwarz)
\(P\ge\frac{a^2+b^2+c^2}{81}+\frac{9}{a^2+b^2+c^2}+\frac{80\left(a^2+b^2+c^2\right)}{81}\)
\(\Rightarrow P\ge\frac{2}{3}+\frac{80\left(a^2+b^2+c^2\right)}{81}\left(AM-GM\right)\)
Sử dụng đánh giá quen thuộc:\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=27\)
\(\Rightarrow P\ge\frac{2}{3}+\frac{80\cdot27}{81}=\frac{82}{3}\)
"="<=>a=b=c=3
Cho biet a,b,c la cac so co 1 chu so . Viet tiep vao cho cham :
a, Gia tri lon nhat cua bieu thuc a+b+c= .......
b, Gia tri be nhat cua bieu thuc a+b+c = .......
a+b+c lon nhat=9+9+9=27
a+b+c be nhat =0+0+0=0
tick nha
o minh cung dang dinh hoi cau ay
giup minh voi
cho 2 bieu thuc A=x+x^2/2-x va B=2x/x+1+3/x-2-2x^2+1/x^2-x-2 a, tinh gia tri cua A khi /2x-3/=1 b,tim dieu kien xac dinh va rut gon bieu thuc B c,tim so nguyen x de P=A.B dat gia tri lon nhat
mk dang can gap
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)