Rút gọn các biểu thức :
\(\sqrt{x-2\sqrt{x-1}}\) \(\left(x\ge1\right)\)
Rút gọn các biểu thức sau:
a, \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
b, \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) với \(x\ge1\)
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)( do \(x\ge1\))
a: Ta có: \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
\(=\sqrt{xy}\)
b: Ta có: \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\dfrac{ \left|\sqrt{x}-1\right|}{\left|\sqrt{x}+1\right|}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Rút gọn các biểu thức :
\(\sqrt{x-2\sqrt{x-1}}\) \(\left(x\ge1\right)\)
\(\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}\right)^2-2\cdot\sqrt{x-1}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|\)
\(A=\dfrac{\sqrt{X}-2}{\sqrt{X}-1};B=\dfrac{\sqrt{X}}{\sqrt{X}+1}-\dfrac{\sqrt{X}-4}{1-X}\left(X\ge1;X\ne1\right)\)
a) Tính giá trị của biểu thức A khi x = 25
b) Rút gọn biểu thức B
c) Tìm x để A: B <1/2
a: Khi x=25 thì \(A=\dfrac{5-2}{5-1}=\dfrac{3}{4}\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{x-1}=\dfrac{x-4}{x-1}\)
c: \(P=\dfrac{A}{B}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
P<1/2
=>P-1/2<0
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{1}{2}< 0\)
=>\(\dfrac{2\sqrt{x}+2-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}< 0\)
=>\(\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}< 0\)
=>\(x\in\varnothing\)
Cho biểu thức M = \(\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\)
a/ Rút gọn biểu thức M
b/ Tìm giá trị của x để M=2
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Rút gọn biểu thức
P= \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}vs\left(x\ge1\right)\)
a, \(M=\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\) (ĐK : \(\forall x\in R\))
\(=\sqrt{\left(x-2\right)^2}-\sqrt{\left(x+2\right)^2}\)
* Nếu x\(\ge2\Rightarrow M=x-2-x-2=-4\)
*Nếu x<2 => M=2-x-x-2=-2x
b,Để M=2\(\ne-4\)
=>M=-2x
=>-2x=-4
=>x=2
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P=\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
* Nếu \(x\ge2\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
* Nếu x<2 =>P=\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
VẬY.......
Tk nha!
Rút gọn biểu thức :
\(A=\left(\sqrt{x-1}-1\right)^2+\sqrt{4x-3+4\sqrt{x-1}}\) với \(x\ge1\)
Xét : \(\sqrt{4x-3+4\sqrt{x-1}}=\sqrt{4\left(x-1\right)+4\sqrt{x-1}+1}=\sqrt{\left(2\sqrt{x-1}+1\right)^2}=2\sqrt{x-1}+1\)
Khi đó : \(A=\left(\sqrt{x-1}-1\right)^2+2\sqrt{x-1}=x-1-2\sqrt{x-1}+1+2\sqrt{x-1}+1=x+1\)
Rút gọn biểu thức :
\(B=\dfrac{\sqrt{1+\sqrt{1-x^2}}\left[\left(1+x\right)\sqrt{1+x}-\left(1-x\right)\sqrt{1-x}\right]}{x\left(2+\sqrt{1-x^2}\right)}\)
Giúp mình với các cao nhân
Rút gọn các biểu thức sau:
A= \(3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
B= \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)
C= \(3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
D= \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
E= \(\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)
\(A=3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x+6\sqrt{x}-\left(x-1\right)\)
\(=3x+6\sqrt{x}-x+1\)
\(=2x+6\sqrt{x}+1\)
\(B=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)
\(=x+3\sqrt{x}+\sqrt{x}+3-2\left(x-2\sqrt{x}+1\right)\)
\(=x+4\sqrt{x}+3-2x+4\sqrt{x}-2\)
\(=-x+8\sqrt{x}+1\)
\(C=3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x-3\sqrt{x}-2+\left(\sqrt{x^2}-1\right)\)
\(=3x-3\sqrt{x}-2+x-1\)
\(=4x-3\sqrt{x}-3\)
\(D=\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(=x-9-\left(2x-3\sqrt{x}-2\right)\)
\(=x-9-2x+3\sqrt{x}+2\)
\(=-x+3\sqrt{x}-7\)
\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)
\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)
\(=x-4-2\left(2x+3\sqrt{x}-2\right)\)
\(=x-4-4x-6\sqrt{x}+4\)
\(=-3-6\sqrt{x}\)
Câu 1: Rút gọn biểu thức: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{2}}+\dfrac{6}{x+3\sqrt{x}}\right)\) với x > 0
Câu 2: Rút gọn biểu thức:
\(P=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\) với x > 0; x \(\ne\) 2
Câu 3: Rút gọn biểu thức:
\(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\) với a > 0; a \(\ne\) 4
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
Cho biểu thức \(A=x-\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{1}{\sqrt{x}+\sqrt{x-1}}\right)\)với \(x\ge1\). Rút gọn A và chứng tỏ rằng giá trị của biểu thức A luôn không âm
Ta có:
\(A=x-\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{1}{\sqrt{x}+\sqrt{x-1}}\right)\)
\(A=x-\frac{\sqrt{x}+\sqrt{x-1}-\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}\)
\(A=x-\frac{2\sqrt{x-1}}{x-x+1}\)
\(A=x-2\sqrt{x-1}\)
\(A=\left(x-1\right)-2\sqrt{x-1}+1\)
\(A=\left(\sqrt{x-1}-1\right)^2\ge0\left(\forall x\ge1\right)\)
=> đpcm
cho biểu thức P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\) với x>0
1.rút gọn biểu thức P
2.tìm các soosnguyeen x thả mãn P>0
1. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\left(x>0\right)\)
\(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
2. Để \(P>0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\sqrt{2}\\x>\sqrt{-1}\left(L\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< \sqrt{2}\\x< \sqrt{-1}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< \sqrt{2}\end{matrix}\right.\)
Vậy \(P>0\Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< \sqrt{2}\end{matrix}\right.\)