1) Gia tri lon nhat cua ham so y = sin2x + cos2x la:
A. \(\dfrac{\sqrt{2}}{2}\) B. 1 C. \(\sqrt{2}\) D. 2
1) Gia tri lon nhat cua ham so: y = \(\dfrac{cosx+2sinx+3}{2cosx-sinx+4}\)
A. 0 B. 3-2\(\sqrt{3}\) C. \(2-2\sqrt{2}\) D. -1
1) Tim tat ca cac gia tri thuc cua tham so m de ham so y = \(\sqrt{m-2x}-\sqrt{x+1}\) co tap xac dinh la 1 doan tren truc so
A. m < -2 B. m > 2 C. m > \(\dfrac{-1}{2}\) D. m > -2
cho ham so y=2x^2-3x+1. gia tri lon nhat cua x thoa man f(-x)=f(2x) la x=?
1) Gia tri lon nhat cua ham so: y = sin6x + cos6x
Cho bieu thuc A=\(\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\div\dfrac{1}{\sqrt{x}-1}\)
a/ Tim dieu kien cua x de bieu thuc A co gia tri xac dinh
b/ Rut gon A
c/ Tinh gia tri cua A khi x = \(4-2\sqrt{3}\)
d/ Tim gia tri nho nhat cua A
a. ĐKXĐ : x>1.
b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)
c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:
\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)
Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).
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 lon nhat hoac be nhat cua bieu thuc
A=\(2x-2\sqrt{x}+1\)
B=\(-x+2\sqrt{x-1}+1\)
Rut gon P=\(\dfrac{3x+6\sqrt{x}}{x+\sqrt{x}-2}-\dfrac{\sqrt{x-1}}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{1-\sqrt{x}}\)
Tim x de P dat gia tri lon nhat
ĐKXĐ: \(x\ge0;x\ne1\)
Sửa lại đề chỗ \(\dfrac{\sqrt{x-1}}{\sqrt{x}+2}\) thành \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(P=\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=2-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{2\sqrt{x}+4-\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}=1+\dfrac{3}{\sqrt{x}+2}\)
Để P lớn nhất \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\) lớn nhất
Mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)
\(\Rightarrow P_{max}=1+\dfrac{3}{2}=\dfrac{5}{2}\) khi \(\sqrt{x}+2=2\Leftrightarrow x=0\)
tim gia tri lon nhat cua \(\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)
GTLN là \(\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\) Sách mình ghi thế nhưng không có lời giải li ke nha