Rut gon da thuc sau: X=\(\sqrt{a^2+1+\left(1-\frac{1}{a+1}\right)^2}+\frac{a}{a+1}\)
voi a>0
Rut gon bieu thuc:
P=\(\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\) voi \(x=\frac{1}{2}.\left(a+\frac{1}{a}\right)\); y=\(\frac{1}{2}.\left(b+\frac{1}{b}\right)\) va \(a\ge1;b\ge1\)
\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}-2\right)=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
Tương tự \(y^2-1=\frac{1}{4}\left(b-\frac{1}{b}\right)^2\)
\(P=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}\)
\(=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab+\frac{a}{b}+\frac{b}{a}-\frac{1}{ab}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab-\frac{a}{b}-\frac{b}{a}+\frac{1}{ab}}=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)
rut gon cac bieu thuc sau:\(\left(1+\frac{a=\sqrt{a}}{\sqrt{a+1}}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a-1}}\right)\)
Cho bieu thuc p=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
a/tim dkxd va rut gon p
b/ tìm giá trị của a để p<0
a) ĐKXĐ: \(x\ge0;x\ne1\)
P=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a-1}{2\sqrt{a}}\right)^2.\left(\frac{-1-3\sqrt{a}}{a-1}\right)\)
=\(\frac{\left(a-1\right)^2}{4a}.\frac{-1-3\sqrt{a}}{a-1}\)
=\(\frac{\left(a-1\right)\left(-1-3\sqrt{a}\right)}{4a}\)
cho bieu thuc A =\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
(x≥0;x≠1)
a. tim x de bieu thuc A co nghia ?rut gon A ?
b. tinh gia tri bieu thuc A tai x=7+4√3
a/ Ta có: A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right):\left(\sqrt{x}\right)=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b/ Ta có :\(x=7+4\sqrt{3}=3+4\sqrt{3}+4=\left(\sqrt{3}+2\right)^2
\)
\(\Rightarrow\sqrt{x}=|\sqrt{3}+2|=\sqrt{3}+2\)
Thay x vào A ta có:
A\(=\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}+2+1}{\sqrt{3}+2}=\frac{\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{3-\sqrt{3}}{1}=3-\sqrt{3}\)
rut gon bieu thuc A=\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\) tìm giá trị của x để A>0
\(A=\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)(DK : \(x\ge0;x\ne4\))
\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{x-4+10-x}{\sqrt{x}+2}\)
\(=\frac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{6}=\frac{1}{2-\sqrt{x}}\)
Để A > 0 thì \(2-\sqrt{x}>0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Vậy để A < 0 thì x < 4
Bảo Ngọc kết luận hơi sai một chút nhé. Để A > 0 thì x < 4 nhé :)
\(M=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)N=\frac{\sqrt{x}+1}{\sqrt{x}}\) 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 \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
cho bieu thuc \(P=\left(\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
a) rut gon P
b) tìm x để P <= 0
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\left(\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
\(\Leftrightarrow P=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{x+\sqrt{x}+1}{x+1}\right)\)
\(\Leftrightarrow P=\frac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\frac{x+1}{x+\sqrt{x}+1}\)
\(\Leftrightarrow P=\frac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\frac{-\sqrt{x}+1}{x+\sqrt{x}+1}\)
b) Ta có : \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Để \(P\le0\Leftrightarrow-\sqrt{x}+1\le0\)
\(\Leftrightarrow-\sqrt{x}\le-1\)
\(\Leftrightarrow\sqrt{x}\ge1\)
\(\Leftrightarrow x\ge1\)
Vì đkxđ : \(x\ne1\)
Vậy để \(P\le0\Leftrightarrow x>1\)
cho bieu thuc A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
(x≥0;x≠1)
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
rut gon bieu thuc
a) A= \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
b) B=\(\frac{\sqrt{a-1-2\sqrt{a-2}}}{\sqrt{a-2}-1}\)
a) \(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{6}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(A=\frac{\left(-\sqrt{a}+1\right)^2}{\left(-a+1\right)^2}.\left(\sqrt{a}+\frac{-a\sqrt{a}+1}{-\sqrt{a}+1}\right)\)
\(A=\frac{\left(1-\sqrt{a}\right)^2\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(A=\frac{\frac{-a\sqrt{a}+\sqrt{a}.\left(-\sqrt{a}+1\right)+1}{-\sqrt{a}+1}.\left(-\sqrt{a}+1\right)^2}{\left(1-a\right)^2}\)
\(A=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(A=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)
\(A=1\)