M=\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\)
Rút gon biểu thức M
Tính x để M=2
Cho biểu thức:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x+3}}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a)Rút gọn biểu thức P
b)Tìm x để \(p< -\frac{1}{2}\)
c)Tìm x để \(P.\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
d)Tìm m để \(P.\left(\sqrt{x}+3\right)+x\left(\sqrt{x}-m\right)=x-\sqrt{x}\left(3+m\right)\)
P/s : sửa đề
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}-3x}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}\left(1+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{-3\sqrt{x}}{\sqrt{x}+3}\)
b) \(P< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6\sqrt{x}+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\frac{-5\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
Mà \(2\left(\sqrt{x}+3\right)>0\)
\(\Rightarrow-5\sqrt{x}+3< 0\)
\(\Leftrightarrow-5\sqrt{x}< -3\)
\(\Leftrightarrow\sqrt{x}>\frac{3}{5}\)
\(\Leftrightarrow x>\frac{9}{25}\)
Vấy .................
c) \(P.\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}-2-2+x=0\)
\(\Leftrightarrow-\sqrt{x}-4+x=0\)
\(\Leftrightarrow-\sqrt{x}\left(1-\sqrt{x}\right)=4\)
Còn lại lập bảng tự tìm giá trị của x là ra .( Chú ý : đối chiếu ĐKXĐ )
d)
\(P.\left(\sqrt{x}+3\right)+x\left(\sqrt{x}-m\right)=x-\sqrt{x}\left(3+m\right)\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+x\sqrt{x}-xm=x-3\sqrt{x}-m\sqrt{x}\)
\(\Leftrightarrow-3\sqrt{x}+x\sqrt{x}-xm-x+3\sqrt{x}+m\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(x+m\right)-x\left(m+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left[x+m-m\sqrt{x}-\sqrt{x}\right]=0\)
\(\Leftrightarrow\sqrt{x}\left[m\left(1-\sqrt{x}\right)-\sqrt{x}\left(1-\sqrt{x}\right)\right]=0\)
\(\Leftrightarrow\sqrt{x}=0;m-\sqrt{x}=0;1-\sqrt{x}=0\)
+) \(\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
+) \(1-\sqrt{x}=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
+) \(m-\sqrt{x}=0\)
\(\Leftrightarrow\orbr{\begin{cases}m-\sqrt{0}=0\\m-\sqrt{1}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=0\\m=1\end{cases}}}\)
Vậy ..................
a,Cho biểu thức:\(M=\left(\frac{x+2\sqrt{x}+4}{x\sqrt{x}-8}+\frac{x+2\sqrt{x}+1}{x-1}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-2}+\frac{2\sqrt{x}+10}{x+6\sqrt{x}5}\right)\)
Rút gọn M và tìm x để M>1
Rút gọn biểu thức
M=\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
a.Rút gọn M
b.Tìm x để 3M=4-x
a, ĐKXĐ: \(x>0;x\ne1;x\ne4\)
\(M=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+2}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}}\)
Rút gon biểu thức:
\(D=\left(\frac{2+\sqrt{x}}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\left(\frac{x\sqrt{x}+x-\sqrt{x}-1}{\sqrt{x}}\right)\)
cho biểu thức
M=\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\)
a. tìm ĐK của x để M có nghĩa
b. Rút gọn M
rút gon biểu thức
M=\(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)
ĐKXĐ: \(x\ge0,x\ne1\)
\(M=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-\left(x+\sqrt{x}-2\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)\sqrt{x}}\)
\(=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2}{x-1}\)
@Phạm Mỹ Dung Sao lúc nãy không gửi luôn?
Để \(M< -1\Leftrightarrow\frac{2}{x-1}< -1\)\(\Leftrightarrow\frac{2}{x-1}+1< 0\)
\(\Leftrightarrow\frac{2+x-1}{x-1}< 0\)\(\Leftrightarrow\frac{x+1}{x-1}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -1\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x>-1\\x< 1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow-1< x< 1\)
(vì x < -1, x>1 không xảy ra)
Vậy để \(M< -1\Leftrightarrow-1< x< 1\)
1/ Rút gọn biểu thức:\(G=\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}}\right)\div\frac{\sqrt{x}+1}{x}\)
2/ Cho biểu thức: \(M=x-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)
a. Tìm ĐKXĐ
b. Rút gọn M
c. Tìm giá trị nhỏ nhất của M
3/ Chứng minh: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}|\)với \(a\ne0,b\ne0,a+b\ne0\)
4/ Biết a,b,c là số dương và ab + bc + ac =1. Hãy tính tổng:
\(M=a\sqrt{\frac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}+b\sqrt{\frac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}+c\sqrt{\frac{\left(1+a^2\right)\left(1+b^2\right)}{1+c^2}}\)
Ai giải giúp mình bài 1 với bài 4 trước đi
Rút gọn biểu thức: \(M=\left(\frac{1}{2}-\frac{1}{2x}\right)\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
ĐK:x>1
M=\(\frac{x-1}{2x}\) .\(\frac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{x-1}\)
=\(\frac{x-1}{2x}\).\(\frac{x\sqrt{x}-x-x+\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{x-1}\)=\(\frac{x-1}{2x}\).\(\frac{-4x}{x-1}\)=-2
Vậy M=-2
rút gon biểu thức \(Q=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)