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1)đặt nhân tử chung quy đồng là xong
2)phân tích x+2cănx-3=(1-cănx)(3+cănx)
3)2a+căn a đặt căn a ra r rút gọn
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
b. \(Q=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
c. Để \(Q< 1\Rightarrow Q-1< 0\Leftrightarrow\frac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\Leftrightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)
Vậy \(0\le x< 9\)thì \(Q< 1\)
giải giúp mình bài này ới ạ mình đng cần gấp
Cho biểu thức
c=(căng x-2/căng x+2+căng x+2/căng x-2)nhân căng x+2/2 - 4 căng x/căng x-2
a)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{a-9}\)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{a-3\sqrt{a}+3+3\sqrt{a}-3a-9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{a-9}\)
b) Để \(P=\frac{1}{3}\Rightarrow\frac{-2a-3}{a-9}=\frac{1}{3}\)
\(\Rightarrow3\left(-2a-3\right)=a-9\)
\(\Rightarrow-6a-9=a-9\)
\(\Rightarrow-6a-a=-9+9\)
\(\Rightarrow-7a=0\left(L\right)\)
Vậy ko có gt của a để P=1/3 ( mk ko chắc.....)
a ) \(ĐKXĐ\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2+\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x-3}\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b ) \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}< 1\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{4}{\sqrt{x}-3}< 0\)
\(\sqrt{x}-3< 0\)
\(\Leftrightarrow x< 9\)
Vậy với \(0\le x\le9;x\ne4\) thì ...
Chúc bạn học tốt !!!
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(M=\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{3-11\sqrt{a}}{9-a}\)
ĐK : \(\hept{\begin{cases}a\ge0\\a\ne9\end{cases}}\)
\(=\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{11\sqrt{a}-3}{a-9}\)
\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{11\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\frac{2a-6\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{a+4\sqrt{a}+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{11\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\frac{2a-6\sqrt{a}+a+4\sqrt{a}+3+11\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\frac{3a+9\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\frac{3\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\frac{3\sqrt{a}}{\sqrt{a}-3}\)
b) Thiếu đề
pt \(\Leftrightarrow M=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3-11\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(\Leftrightarrow M=\frac{2a-6\sqrt{a}+a+3+4\sqrt{a}-3+11\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(\Leftrightarrow M=\frac{3a-9\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(\Leftrightarrow M=\frac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(\Leftrightarrow M=\frac{3\sqrt{a}}{\sqrt{a}+3}\)
ĐKXĐ: \(a\ge0,a\ne9\)
a)\(P=\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{3+7\sqrt{a}}{9-a}.\)
\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3+7\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)+\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)-\left(3+7\sqrt{a}\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{2a-6\sqrt{a}+a+3\sqrt{a}+\sqrt{a}+3-3-7\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{3a-9\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}=\frac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{3\sqrt{a}}{\sqrt{a}+3}\)
b) Để P<1 hay \(\frac{3\sqrt{a}}{\sqrt{a}+3}< 1\Leftrightarrow3\sqrt{a}< \sqrt{a}+3\Leftrightarrow\sqrt{a}< \frac{3}{2}\Leftrightarrow0\le a< \frac{9}{4}\)(thỏa mãn ĐKXĐ)
Vậy với \(0\le a< \frac{9}{4}\)thì P<1.
(p/s đừng ti ck cho câu trả lời này)