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a) ĐKXĐ:\(x\ge\frac{1}{3};x\ne1\)
b)\(P=\frac{3a+\sqrt{9a-3}-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{a+6+\sqrt{9a-3}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
Ta có \(\left(\sqrt{a}+2\right)\left(1-\sqrt{a}\right)=a+\sqrt{a}-2\)
\(=\frac{3\text{a}+3\sqrt{a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)
\(=\frac{3\text{a}+3\sqrt{a}-3-a+1+a-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{3\text{a}+3\sqrt{a}-6}{a+\sqrt{a}-2}\)
\(=\frac{3\left(a+\sqrt{a}-2\right)}{a+\sqrt{a}-2}\)
\(=3\)
b/ Ta có 3 là số nguyên nên biểu thức P luôn nguyên với mọi x
TICK CHO MÌNH NHA
Bài 1
a) \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\) (ĐK : x\(\ge0\) ; x\(\ne\) 1)
\(=\frac{3a+\sqrt{9a}-3}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)
\(=\frac{3a+\sqrt{9a}-3-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{3a+\sqrt{9a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
b) \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Vậy để P là số nguyên thì: \(\sqrt{a}-1\inƯ\left(2\right)\)
Mà Ư(2)={-1;1;2;-1}
=> \(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| \(\sqrt{a}-1\) | 1 | -1 | 2 | -2 |
| a | 4 | 0 | 9 | \(\sqrt{a}=-1\) (ktm) |
vậy a={0;4;9} thì P nguyên
Bài 2
\(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)(ĐK:a\(\ge\)8)
\(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)
\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)
\(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)
\(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)
\(=\frac{2a}{\sqrt{a-4}}\)
ĐKXĐ: ...
\(P=\frac{3a+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}+\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\frac{a+\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{3a+3\sqrt{a}-3-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
\(\left|P\right|=1\Rightarrow\left[{}\begin{matrix}P=1\\P=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{\sqrt{a}+1}{\sqrt{a}-1}=1\\\frac{\sqrt{a}+1}{\sqrt{a}-1}=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}+1=\sqrt{a}-1\left(vn\right)\\\sqrt{a}+1=-\sqrt{a}+1\end{matrix}\right.\) \(\Rightarrow a=0\)
\(P=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
\(P\in N\Rightarrow\sqrt{a}-1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow\sqrt{a}=\left\{-1\left(l\right);0;2;3\right\}\)
\(\Rightarrow a=\left\{0;4;9\right\}\)
Thay vào P chỉ thấy \(a=\left\{4;9\right\}\) thỏa mãn
Đk : \(a\ge0,a\ne1\)
a,Rút gọn được P= \(\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
b, Có P<1 <=> \(\frac{\sqrt{a}+1}{\sqrt{a}-1}< 1\) <=> \(\frac{\sqrt{a}+1}{\sqrt{a}-1}-1< 0\)
<=> \(\frac{\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\) <=> \(\frac{2}{\sqrt{a}-1}< 0\) => \(\sqrt{a}-1< 0\)
<=> \(\sqrt{a}< 1\) <=> a<1 ,k/hợp với đk của a
=> \(0\le a< 1\)
Vậy để P<1 <=> 0\(\le\)a<1
c,Có P= \(\frac{\sqrt{a}+1}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Để \(P\in Z\) <=> \(\frac{2}{\sqrt{a}-1}\in Z\)
Với mọi a t/m đk có:\(\left[{}\begin{matrix}\sqrt{a}\in N\\\sqrt{a}\notin N\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\sqrt{a}-1\in Z\\\sqrt{a}-1\notin Z\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\frac{2}{\sqrt{a}-1}\in Z\left(tm\right)\\\frac{2}{\sqrt{a}-1}\notin Z\left(ktm\right)\end{matrix}\right.\)
=> \(\sqrt{a}-1\in\) Ư(2)\(=\left\{1,-1,2,-2\right\}\)
<=> \(\sqrt{a}\in\left\{2,0,3,-1\right\}\)
mà \(\sqrt{a}\ge0\) => \(\sqrt{a}\in\left\{2,0,3\right\}\) <=> \(a\in\left\{4,0,9\right\}\)
Tại a=0 => P=-1
Tại a=4=>P=3
Tại a=9 => P=2
Vậy a=4 thì P đạt GT nguyên lớn nhất
a/ Điều kiện \(\hept{\begin{cases}a\ge0\\a\ne\frac{1}{9}\end{cases}}\) \(\Rightarrow0\le a\ne\frac{1}{9}\)
b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(a-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}\left(1-3\sqrt{a}\right)+\left(\sqrt{a}-2\right)\left(1+3\sqrt{a}\right)+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}:\left(\frac{3a\sqrt{a}-2\sqrt{a}+6-a}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}-6a+\sqrt{a}+3a-2-6\sqrt{a}+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}.\left(\frac{3\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\right)\)
\(=\frac{3a-2\sqrt{a}-1}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
\(=\frac{\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
Hình như đề sai rồi bạn :(
a/ Điều kiện xác định : \(\hept{\begin{cases}a\ge0\\a\ne9\end{cases}\Leftrightarrow}0\le a\ne9\)
b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(1-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}\left(3\sqrt{a}-1\right)+\left(2-\sqrt{a}\right)\left(3\sqrt{a}+1\right)-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}:\frac{\sqrt{a}+5}{3\sqrt{a}-1}\)
\(=\frac{6a-2\sqrt{a}+6\sqrt{a}+2-3a-\sqrt{a}-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}.\frac{3\sqrt{a}-1}{\sqrt{a}+5}\)
\(=\frac{3a-2\sqrt{a}-1}{3\sqrt{a}+1}.\frac{1}{\sqrt{a}+5}\)
\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(3\sqrt{a}+1\right)\left(\sqrt{a}+5\right)}=\frac{\sqrt{a}-1}{\sqrt{a}+5}\)
c/ \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\) thay vào M được
\(\frac{\sqrt{5}-2-1}{\sqrt{5}-2+5}=\frac{\sqrt{5}-3}{\sqrt{5}+3}=\frac{-7+3\sqrt{5}}{2}\)
d/ \(M=\frac{\sqrt{a}-1}{\sqrt{a}+5}=\frac{\sqrt{a}+5-6}{\sqrt{a}+5}=1-\frac{6}{\sqrt{a}+5}\)
Với mọi \(0\le a\ne9\) thì ta luôn có \(\sqrt{a}+5\ge5\Leftrightarrow\frac{6}{\sqrt{a}+5}\le\frac{6}{5}\Leftrightarrow-\frac{6}{\sqrt{a}+5}\ge-\frac{6}{5}\Leftrightarrow1-\frac{6}{\sqrt{a}+5}\ge1-\frac{6}{5}\)
\(\Rightarrow M\ge-\frac{1}{5}\)
Đẳng thức xảy ra khi a = 0
Vậy giá trị nhỏ nhất của M bằng \(-\frac{1}{5}\) khi a = 0
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.....)
ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
a. Ta có \(P=\frac{3a+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-2}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+2}-1\)
\(=\frac{3a+3\sqrt{a}-3-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{3a+3\sqrt{a}-3-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\)
b. Để \(\left|P\right|=2\Rightarrow\orbr{\begin{cases}P=2\\P=-2\end{cases}}\)
Với \(P=2\Rightarrow\sqrt{a}+1=2\sqrt{a}-2\Rightarrow\sqrt{a}=3\Rightarrow a=9\)
Với \(P=-2\Rightarrow\sqrt{a}+1=2-2\sqrt{a}\Rightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)
c. Ta có \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Để \(P\in N\Rightarrow P\in Z\Rightarrow\sqrt{a}-1\in\left\{-2;-1;1;2\right\}\)
\(\left(tm\right)\)
Vậy \(x\in\left\{0;4;9\right\}\)thì \(P\in N\)