Rút gọn
A = \(\frac{\sqrt{a}-2}{1-\sqrt{a}}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{3a-3+\sqrt{9a}}{a+\sqrt{a}-2}\)
Rút gọn các biểu thức sau:
a) $\sqrt{9a^4}$
b) 2$\sqrt{a^{2}}$- 5a (với a<0)
c) $\sqrt{16(1+4x+4x^2)}$ với x $\geq$ $\frac{1}{2}$
d) $\frac{1}{a-3}$$\sqrt{9(a^2-3a+9)}$ với a<3
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
Cho biểu thức P=\(\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\)
a) Rút gọn P
b) Tìm các giá trị nguyên của a để P nguyên
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
p=\(\frac{\sqrt{a}-2}{1-\sqrt{a}}-\frac{1+\sqrt{a}}{2+\sqrt{a}}+\frac{3a-3+\sqrt{9a}}{a+\sqrt{a}-2}\)
a) rút gọn p
b)tìm giá trị a nguyên để phương trình tương ứng là số nguyên
Cho biểu thức:
\(P=\frac{3a+\sqrt{9a-3}}{a+\sqrt{a}-2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+2}-1\)
a) Tìm a để P có nghĩa
b) Rút gọn P
a) ĐKXĐ:\(x\ge\frac{1}{3};x\ne1\)
b)\(P=\frac{3a+\sqrt{9a-3}-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{a+6+\sqrt{9a-3}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
bài 1: cho biểu thức: P=\(\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\)
a) Rút gọn P
b) Tìm các giá trị nguyên của a để P nguyên
bài 2: cho biểu thức: P=\(\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)
a) Rút gọn P
b) Tìm các giá trị nguyên của a (a>8) để P nguyên
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)
\(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)
\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)
\(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)
\(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)
\(=\frac{2a}{\sqrt{a-4}}\)
\(\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1+\sqrt{a}}\)
P = (\(\frac{\sqrt{a}-1}{3\sqrt{a}-1}-\frac{1}{1+3\sqrt{a}}+\frac{8\sqrt{a}}{9a-1}\)) : ( \(1-\frac{3\sqrt{a}-2}{3\sqrt{a}+1}\))
a, Rút gọn P
\(ĐKXĐ:\hept{\begin{cases}a\ge0\\a\ne\frac{1}{9}\end{cases}}\)
\(P=\left(\frac{\sqrt{a}-1}{3\sqrt{a}-1}-\frac{1}{1+3\sqrt{a}}+\frac{8\sqrt{a}}{9a-1}\right)\div\left(1-\frac{3\sqrt{a}-2}{3\sqrt{a}+1}\right)\)
\(\Leftrightarrow P=\frac{\left(\sqrt{a}-1\right)\left(1+3\sqrt{a}\right)-3\sqrt{a}+1+8\sqrt{a}}{9a-1}:\frac{3\sqrt{a}+1-3\sqrt{a}+2}{3\sqrt{a}+1}\)
\(\Leftrightarrow P=\frac{\sqrt{a}+3a-1-3\sqrt{a}-3\sqrt{a}+1+8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}:\frac{3}{3\sqrt{a}+1}\)
\(\Leftrightarrow P=\frac{\left(3a+3\sqrt{a}\right)\left(3\sqrt{a}+1\right)}{3\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(\Leftrightarrow P=\frac{a+\sqrt{a}}{3\sqrt{a}-1}\)
bài 1: rút gọn
a, \(\sqrt{\frac{2}{3}}-\sqrt{24}+2\sqrt{\frac{3}{8}}+\sqrt{\frac{1}{6}}\)
b, \(\sqrt{\frac{2}{2-\sqrt{3}}}-\sqrt{\frac{2}{2+\sqrt{3}}}\)
c, \(2\sqrt{a}-\frac{5}{a}\sqrt{9a^3}+a\sqrt{\frac{4}{a}}-\frac{2}{a^2}\sqrt{25a^5}\left(vớia>0\right)\)
Cho \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+2}-1\)
a, Rút gọn P
b, Tìm a để /P/=2
c, Tìm\(a\in N\)sao cho \(P\in N\)
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 \(P\in N\Rightarrow P\in Z\Rightarrow\sqrt{a}-1\in\left\{-2;-1;1;2\right\}\)
\(\sqrt{a}-1\) | \(-2\) | \(-1\) | \(1\) | \(2\) |
\(\sqrt{a}\) | \(-1\) | \(0\) | \(2\) | \(3\) |
\(a\) | \(0\) | \(4\) | \(9\) | |
\(\left(l\right)\) | \(\left(tm\right)\) | \(\left(tm\right)\) | \(\left(tm\right)\) |
Vậy \(x\in\left\{0;4;9\right\}\)thì \(P\in N\)