rút gon:\(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
CÂU HỎI KHÓ NHẤT QUẢ ĐẤT
A)RÚT GON P
\(P=\left(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right)\):\(\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\)
\(P=\left(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\)
\(=\left(\frac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{3a}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\left(a+\sqrt{ab}+b\right)}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}\right).\frac{2\left(a+\sqrt{ab}+b\right)}{\cdot\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{2\left(a-2\sqrt{ab}+b\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)\left(a+\sqrt{ab}+b\right)}\)
\(=\frac{2}{a-1}\)
B=1+\(\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right).\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)
a)Rút gon B
b)Tìm a để B=\(\frac{\sqrt{6}}{1+\sqrt{6}}\)
c)CMR B>\(\frac{2}{3}\)
Rút gọn:
A= \(1+\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right).\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)
Rút gọn biểu thức:
A= \(1+\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right).\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(A=1+"\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{2\sqrt{a}-1}=\)
\(A="\frac{1a+\sqrt{a}-1}{1-a}-\frac{1a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{1\sqrt{a}-1}\)
P/s: Ko chắc đâu nhé
Rút gọn \(1+\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{2\sqrt{a}-1}\right)\)
Cho biểu thức
A= \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a, Rút gon A
b, Tìm a để A> \(\frac{1}{6}\)
a) \(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(ĐK:a>0;a\ne1;a\ne4\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để \(A>\frac{1}{6}\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Leftrightarrow\)\(\frac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}>0\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
\(\Leftrightarrow\sqrt{a}-4>0\Leftrightarrow a>16\left(tm\right)\)
Vậy a>16 thì \(A>\frac{1}{6}\)
ĐKXĐ : \(a>0,a\ne4,a\ne1\)
a) \(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\frac{a-1-\left(a-4\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
b) \(A>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow-\frac{2}{3\sqrt{a}}+\frac{1}{3}>\frac{1}{6}\Rightarrow\frac{2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{1}{\sqrt{a}}>\frac{1}{4}\Rightarrow a< 16\)
Kết hợp với điều kiện xác định.
Rút gọn biểu thức:
a) \(\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab+\sqrt{a}}}{\sqrt{ab}-1}+1\right)\)
b) \(1+\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{2\sqrt{a}-1}\right)\)
Rút gọn biểu thức:
\(1+\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{2\sqrt{a}-1}\right)\)
Điều kiện: x \(\ne\) 1; 1/4 ; x \(\ge\) 0
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\frac{\left(2a+\sqrt{a}-1\right).\sqrt{a}}{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)-\left(2a+\sqrt{a}-1\right)\left(1+\sqrt{a}\right).\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)\left(a+\sqrt{a}+1-a-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2a+\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\right)\)
\(A=1+\left(\frac{\left(2\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{2\sqrt{a}-1}\right)=1+\frac{-\sqrt{a}}{a+\sqrt{a}+1}=\frac{a+1}{a+\sqrt{a}+1}\)
Các bài tập dạng này hoàn toàn làm tương tự!!!
Cho A=\(1+\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\times\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)
a> Rút gọc A
b> tìm a để a=\(\frac{6}{1+\sqrt{6}}\)
ok b ơi b làm nhanh hộ mình với mình đang cần gấp