tim GTNN cua bieu thuc
a) A=\(\sqrt{x}+1\)
b) B=\(\frac{1}{2}+\sqrt{x}\)
Những câu hỏi liên quan
1) Cho bieu thuc: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(x\ge0,x\ne16\right)\)
a) Cho bieu thuc A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) ; voi cac cua bieu thuc A va B da cho, hay tim cac gia tri cua x nguyen de gia tri cua bieu thuc B(A;-1) la so nguyen
1. Tim GTNN cua bieu thuc:
\(A=\dfrac{1}{2}\sqrt{x^2}+\sqrt{x^2-2x+1}\)
2. Rut gon bieu thuc:
a) \(A=\sqrt{29-4\sqrt{7}}+\sqrt{23+8\sqrt{7}}\)
b)\(B=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}},voix>=2\)
Bài 2:
a: \(A=2\sqrt{7}-1+\left(\sqrt{7}+4\right)\)
\(=2\sqrt{7}-1+\sqrt{7}+4=3\sqrt{7}+3\)
b: \(B=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
Đúng 0
Bình luận (0)
Cho bieu thuc: \(p=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a) Tim DKXD cua bieu thuc p
b) Rut gon bieu thuc p
Giúp mình nhé!!
2. Rut gon bieu thuc:
a) \(A=\sqrt{x-2+2\sqrt{x-3}}+\sqrt{x+6+6\sqrt{x-3}},voix>=3\)
3. Tim GTNN cua bieu thuc:
a) \(A=\sqrt{4x^2-12x+9}+\sqrt{x^2-10x+25}+\sqrt{9x^2-6x+1}+\sqrt{16x^2-72x+81}\)
b) \(B=\dfrac{1}{2}\sqrt{x^2}+\sqrt{x^2-2x+1}\)
cho bieu thuc A = \(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
a. tim x de bieu thuc A co nghia ?rut gon A ?
b. tinh gia tri cua bieu thuc A tai x=7+4√3
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
a\Rut gon bieu thuc
b/Tim gia tri nho nhat cua 4A
A=\(\frac{5\sqrt{x}+4}{x+\sqrt{x}-2}+\frac{\sqrt{x}-1}{\sqrt{x}+2}-\)\(\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(a.A=\frac{5\sqrt{x}+4}{x+\sqrt{x}-2}+\frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}.\)
\(=\frac{5\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)\(+\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)\(-\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{5\sqrt{x}+4+x-2\sqrt{x}+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=-\frac{1}{\sqrt{x}+2}\)
\(b,4A_{min}\Leftrightarrow A_{min}\Rightarrow\frac{-1}{\sqrt{x}+2}\)nhỏ nhất
\(\frac{\Rightarrow1}{\sqrt{x}+2}\)lớn nhất \(\Leftrightarrow\sqrt{x}+2\)nhỏ nhất
\(\sqrt{x}+2\ge2\Leftrightarrow\sqrt{x}=0\Rightarrow x=0\)
\(\Rightarrow A_{min}=\frac{-1}{0+2}=-\frac{1}{2}\Rightarrow4A_{min}=-1\Leftrightarrow x=0\)
Đúng 0
Bình luận (0)
\(\frac{2017-2015x}{\sqrt{1-x^2}}\)
tim GTNN cua bieu thuc tren
\(M=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)N=\frac{\sqrt{x}+1}{\sqrt{x}}\) voi x>0 va x khac 1
a) tim gia tri bieu thuc cua N khi x = 25
b) rut gon S = M.N
c) tim m de S<-1
a) Với x = 25 thì \(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)
b) Ta có \(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\)
\(M=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\)
Suy ra \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
Đúng 0
Bình luận (0)
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)
Đúng 0
Bình luận (0)