b1: cho bt p= \(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a) rg p
b) tìm các gt của x sao cho p=\(\frac{1}{2}\)
c) chứng minh \(p\le\frac{2}{3}\)
K
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23 tháng 7 2016
Điều kiện xác định : \(x\ge0,x\ne1\)
Rút gọn : \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
Câu a) bạn tự tính nhé :)
b) \(P=\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{6-15\sqrt{x}}{3\left(\sqrt{x}+3\right)}=\frac{2\left(\sqrt{x}+3\right)-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}=-\frac{17\sqrt{x}}{3\left(\sqrt{x}+3\right)}+\frac{2}{3}\le\frac{2}{3}\)
Dấu "=" xảy ra khi x = 0
B
28 tháng 7 2019
\(a,E=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\left(Đk:x\ge0;x\ne\pm1\right)\)(Đề như này mới đúng!)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2x-2\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{7\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{5\sqrt{x}+2\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(5\sqrt{x}-5x\right)+\left(2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
Vậy...
\(b,\)Ta có:\(\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{-15+17-5\sqrt{x}}{\sqrt{x}+3}=\frac{\left(-15-5\sqrt{x}\right)+17}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)
Vì \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+3\ge3\forall x\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\Rightarrow-5+\frac{17}{\sqrt{x}+3}\le\frac{2}{3}\Rightarrow E\le\frac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
MN
2 tháng 3 2020
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 \(P=3\Leftrightarrow x=4\)
MN
2 tháng 3 2020
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 \(A< 2\Leftrightarrow x>2,25\)
22 tháng 7 2016
1) \(x^2+y=y^2+x\Leftrightarrow x^2-y^2-\left(x-y\right)=0\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=x\\y=1-x\end{cases}}\). Vì x,y là hai số khác nhau nên ta loại trường hợp x = y. Vậy ta có y = x-1.
\(P=\frac{x^2+\left(1-x\right)^2+x\left(1-x\right)}{x\left(1-x\right)-1}=\frac{x^2+x^2-2x+1-x^2+x}{-x^2+x-1}\)
\(=\frac{x^2-x+1}{-\left(x^2-x+1\right)}=-1\)
13 tháng 8 2019
ĐKXĐ: \(x>1\)
a) \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(P=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(2-3\sqrt{x}\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{15\sqrt{x}-11-3x+6-7\sqrt{x}-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{-5x-2+7\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b) \(P=\frac{1}{2}\Leftrightarrow\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{1}{2}\)
\(\Leftrightarrow2\left(-5\sqrt{x}+2\right)=\sqrt{x}+3\)
\(\Leftrightarrow-10\sqrt{x}+2-\sqrt{x}-3=0\)
\(\Leftrightarrow-11\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=\frac{-1}{11}\)( vô lý )
Vậy không có giá trị của x thỏa mãn.
c) \(P\le\frac{2}{3}\)
\(\Leftrightarrow\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\le\frac{2}{3}\)
\(\Leftrightarrow3\left(-5\sqrt{x}+2\right)\le2\left(\sqrt{x}+3\right)\)
\(\Leftrightarrow-15\sqrt{x}+6\le2\sqrt{x}+6\)
\(\Leftrightarrow-17\sqrt{x}\le0\) ( luôn đúng )
Ta có đpcm.
a/ Đk: x\(\ge\)0
Khi đó ta có:
P =\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{2-3\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
=\(\frac{15\sqrt{x}-11+\left(2-3\sqrt{x}\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\frac{\left(-5x+5\sqrt{x}\right)+\left(2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\frac{5\sqrt{x}-2}{\sqrt{x}+3}\)
b/ Với x\(\ge\)0
Để P=\(\frac{1}{2}\)\(\Leftrightarrow\)\(\frac{5\sqrt{x}-2}{\sqrt{x}+3}=\frac{1}{2}\)\(\Rightarrow\)\(\frac{2\left(5\sqrt{x}-2\right)-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}=0\)
\(\Rightarrow\)\(10\sqrt{x}-4-\sqrt{x}-3=0\)
\(\Rightarrow\)\(9\sqrt{x}=7\)
\(\Rightarrow\)\(\sqrt{x}=\frac{7}{9}\)
\(\Rightarrow\)\(x=\frac{49}{81}\) (thỏa mãn đk)
Vậy .....
a/ p=\(\frac{5\sqrt{x}-2}{\sqrt{x}+3}\)
b/ x=\(\frac{49}{81}\)