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a. ĐKXĐ : \(\orbr{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\end{cases}}\)<=> \(\orbr{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b. \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(\Leftrightarrow P=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(\Leftrightarrow P=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
a) Với \(x\ge0;x\ne1\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{15\sqrt{x}-11-\left(3x-9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Vậy : \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b) \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)
\(A_{max}\Leftrightarrow\left(\frac{17}{\sqrt{x}+3}\right)_{max}\)
Vì \(x\ge0;x\ne1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\\frac{17}{\sqrt{x}+3}>0\end{cases}A_{max}\Leftrightarrow}\left(\sqrt{x}+3\right)_{min}\Leftrightarrow\sqrt{x}_{min}\Leftrightarrow x=0\)
Vậy : \(A_{max}=\frac{17}{3}\Leftrightarrow x=0\)
c,d chưa làm được .-.
c) Để \(A=\frac{1}{2}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{1}{2}\)
<=> \(-10\sqrt{x}+4=\sqrt{x}+3\)
<=> \(-11\sqrt{x}=-1\)
<=> \(\sqrt{x}=\frac{1}{11}\)
<=> \(x=\frac{1}{121}\left(tm\right)\)
Vậy ...
d) \(A\le\frac{2}{3}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\le\frac{2}{3}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}-\frac{2}{3}\le0\)
<=> \(\frac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\sqrt{x}+9}\le0\)
<=> \(\frac{-17\sqrt{x}}{3\sqrt{x}+9}\le0\)
Vì \(\hept{\begin{cases}-17\sqrt{x}\le0\\3\sqrt{x}+9>0\end{cases}}\) \(\Rightarrow\frac{-17\sqrt{x}}{3\sqrt{x}+9}\le0\)(luôn đúng)
=> Ta có ĐPCM
a) \(B=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
ĐKXĐ: \(x\ge0,x\ne1\)
\(B=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{2}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{\sqrt{x+3}}\)
b) Để \(B=\frac{1}{2}\Rightarrow\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)\(\Rightarrow\sqrt{x}+3=4-10\sqrt{x}\Rightarrow11\sqrt{x}=1\Rightarrow\sqrt{x}=\frac{1}{12}\Rightarrow x=\frac{1}{121}\)(Thoả mãn ĐKXĐ)
Vậy x=1/121 thì B =1/2
Mk làm như này, k biết có sai chỗ nào k. Nếu sai thì bạn sửa nhé.
A=\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{3+\sqrt{x}}\)
A=\(\frac{15\sqrt{x}-11-\left(3x-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
A=\(\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
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