Cho x\(\ge\)0, x\(\ne1\).Tìm x biết:\(\frac{1+\sqrt{x}}{2-2\sqrt{x}}-\frac{1-\sqrt{x}}{2+2\sqrt{x}}-\frac{2x}{x-1}=2\)
\(N=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}+\frac{x+\sqrt{x}}{x-1}\right)\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
với x\(\ge\)0 ;x\(\ne1\)
A = \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) (\(x\ge0;x\ne4\))
B = \(\frac{x\sqrt{x}-1}{x-1}\) (\(x\ge0;x\ne1\))
C = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) ( \(x>0;x\ne1\))
D = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) (\(x\ge2\))
E = \(\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2}-2x}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2}-2x}\)
rút gọn: \(\frac{\sqrt{x}}{\sqrt{x-1}}+\frac{\sqrt{x}}{\sqrt{x}-1}:\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\)
Đk: x \(\ge\)0 , x\(\ne1\)
A= \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) \(\left(x\ge0;x\ne4\right)\)
B= \(\frac{x\sqrt{x}-1}{x-1}\) \(\left(x\ge0;x\ne1\right)\)
C= \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) \(\left(x>0;x\ne1\right)\)
D= \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) \(\left(x\ge2\right)\)
E= \(\frac{x+\sqrt{x^2}-2x}{x-\sqrt{x^2-2x}}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2-2x}}\)
B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(B=\left(\frac{a\sqrt{a}+1}{\sqrt{a}+1}\right):\left(a-1\right)+\frac{2a+\sqrt{a}+1}{\sqrt{a}+1}-\frac{\sqrt{a}}{a-1}vớia>1\)
\(C=\left(\frac{X-1}{\sqrt{X}-1}+\frac{\sqrt{X^3}-1}{1-X}\right)-\left(\frac{\left(X-1\right)^2+\sqrt{X}}{\sqrt{X}+1}\right)vớiX>0,X\ne1\)
\(D=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}vớix>0,x\ne1\)
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
Mấy cái này chỉ có nhân lên rồi rút gọn thôi ah. Nên mình cho bạn đáp án để kiểm tra lại thôi ah
Cho biểu thức: \(A=\frac{x-2\sqrt{x}}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\frac{1+2x-2\sqrt{x}}{x^2-\sqrt{x}}\) với \(x>0,x\ne1\)
Rút gọn biểu thức A
\(\frac{4+\sqrt{X}}{7}\)
Rút gọn các biểu thức sau:
a)\(\frac{\sqrt{108x^3}}{\sqrt{12x}}\left(x>0\right)\)
b)\(\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}\left(x< 0;y\ne0\right)\)
c)\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
d) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge\right)\)
e)\(\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\left(y>0;x\ne1;y\ne1\right)\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
Linh ơi, câu a,b,c bạn làm đều đúng hết kết quả cách làm đều đúng nhưng mà ở chỗ câu c): \(\sqrt{x}^3+\sqrt{y}^3\)
không phải vậy đâu, mặc dù mình biết bạn hiểu, hay do sơ suất, nhưng mà chỗ đó là \(\sqrt{x^3}+\sqrt{y^3}\)nha! Dù sao cũng cảm ơn bạn nha!
Rút gọn:
A= (\(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\)). \(\frac{\sqrt{x}+1}{\sqrt{x}}\)với x>0 và x\(\ne1\)
B= (\(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\)) : \(\frac{\sqrt{x}+1}{x^2-x}\)với x>0 và x\(\ne1\)
C= ( \(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\)) : \(\frac{1}{\sqrt{a}.\left(\sqrt{a}-1\right)}\)với a>0 và a \(\ne1\)
D= (\(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\)) : \(\frac{2.\left(x-2\sqrt{x}+1\right)}{x-1}\)với x>0 và x\(\ne1\)
E= ( \(\frac{a\sqrt{a}+1}{a-\sqrt{a}-2}+\frac{a}{2\sqrt{a}-a}\)) :\(\frac{1-\sqrt{a}}{2-\sqrt{a}}\)với a>0, a\(\ne4\),a\(\ne1\) F= ( \(\frac{2\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}+1}+\frac{1}{\sqrt{a}+1}\)): (\(1+\frac{\sqrt{a}}{a+1}\)) với a>0 giúp mình vs mình tick cho nhiều lắm ạ!!! Mình đang cần gấp mn ơi!?!Cho biểu thức : \(\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right).\)với x\(\ge\)0; x \(\ne1\)
a/ Rút gọn P
b/ Tìm x để P \(\frac{1}{2}\)
a) Ta có:
\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\sqrt{x}-x-2}{\sqrt{x}+1}\div\frac{\left(\sqrt{x}-1\right)\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-\sqrt{x}+\sqrt{x}-4}\)
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b) Đề đánh kia ai hiểu được đây, lm đại 3 TH ra nè:
Nếu \(P=\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x}+2=2\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Rightarrow x=16\)
Nếu \(P>\frac{1}{2}\) mà \(\sqrt{x}+2>0\left(\forall x\right)\)
\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Rightarrow x>1\)
Nếu \(P< \frac{1}{2}\) mà \(\sqrt{x}+2>0\left(\forall x\right)\)
\(\Rightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Rightarrow x< 1\)