Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(cos^215^o+cos^225^o+cos^235^o+cos^245^o+cos^255^o+cos^265^o+cos^275^o\)
=\(sin^275^o+sin^265^o+sin^255^o+cos^245^o+cos^255^o+cos^265^o+cos^275^o\)
=1+1+1+\(\dfrac{1}{2}\)=\(\dfrac{7}{2}\)
d) c o s 2 25 0 - c o s 2 35 0 + c o s 2 45 0 - c o s 2 55 0 + c o s 2 65 0
= c o s 2 25 0 - c o s 2 35 0 + c o s 2 45 0 - sin 2 35 0 + sin 2 25 0
= cos 2 25 0 + sin 2 25 0 - cos 2 35 0 + sin 2 35 0 + c o s 2 45 0
= 1 - 1 + 1/2
= 1/2
Lời giải:
a. ĐKXĐ: $x\geq 0; x\neq 1$
\(P=\frac{x\sqrt{x}+26\sqrt{x}-19}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}+\frac{(\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}\)
\(=\frac{x\sqrt{x}+26\sqrt{x}-19-(2x+6\sqrt{x})+(x-4\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-1)}\\ =\frac{x\sqrt{x}+16\sqrt{x}-x-16}{(\sqrt{x}+3)(\sqrt{x}-1)}\\ =\frac{(x+16)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{x+16}{\sqrt{x}+3}\)
b.
$P=\frac{x+16}{\sqrt{x}+3}=\frac{(x-9)+25}{\sqrt{x}+3}$
$=(\sqrt{x}-3)+\frac{25}{\sqrt{x}+3}=(\sqrt{x}+3)+\frac{25}{\sqrt{x}+3}-6$
$\geq 2\sqrt{25}-6=4$ (áp dụng BĐT Cô-si)
Vậy $P_{\min}=4$. Giá trị này đạt tại $\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow x=4$
a: \(A=\sqrt{x}+\dfrac{\sqrt{x}\left(1+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}+\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
Khi x=4 thì \(A=2+\dfrac{2\cdot2+1}{2+1}=2+\dfrac{5}{3}=\dfrac{11}{3}\)
b: Khi x=(2-căn 3)^2 thì \(A=2-\sqrt{3}+\dfrac{2\left(2-\sqrt{3}\right)+1}{2-\sqrt{3}+1}\)
\(=2-\sqrt{3}+\dfrac{4-2\sqrt{3}+1}{3-\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{5-2\sqrt{3}}{3-\sqrt{3}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\left(3-\sqrt{3}\right)+5-2\sqrt{3}}{3-\sqrt{3}}\)
\(=\dfrac{6-2\sqrt{3}-3\sqrt{3}+3+5-2\sqrt{3}}{3-\sqrt{3}}\)
\(=\dfrac{14-7\sqrt{3}}{3-\sqrt{3}}\)
d: A=2
=>\(\dfrac{x+\sqrt{x}+2\sqrt{x}+1}{\sqrt{x}+1}=2\)
=>\(x+3\sqrt{x}+1=2\left(\sqrt{x}+1\right)=2\sqrt{x}+2\)
=>\(x+\sqrt{x}-1=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{-1+\sqrt{5}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{-1-\sqrt{5}}{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{6-2\sqrt{5}}{4}=\dfrac{3-\sqrt{5}}{2}\)
x | -2,5 | -2,25 | -2 | -1,75 | -1,5 | -1,25 | -1 |
y = f(x) = 1,2x | -3 | -2,7 | -2,4 | -2,1 | -1,8 | -1,5 | -1,2 |
x | -0,75 | -0,5 | -0,25 | 0 | 0,25 | 0,5 | 0,75 |
y = f(x) = 1,2x | -0,9 | -0,6 | -0,3 | 0 | 0,3 | 0,6 | 0,9 |
x | 1 | 1,25 | 1,5 | 1,75 | 2 | 2,25 | 2,5 |
y = f(x) = 1,2x | 1,2 | 1,5 | 1,8 | 2,1 | 2,4 | 2,7 | 3 |
a: Khi x=16/9 thì \(A=\left(\dfrac{4}{3}-2\right):\left(\dfrac{4}{3}-3\right)=\dfrac{-2}{3}:\dfrac{-5}{3}=\dfrac{2}{5}\)
b: \(=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}-10}{x-4}\)
\(=\dfrac{x-4\sqrt{x}-16}{x-4}\)
giúp mình với cả nhà