\(A=\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\cdot\left(\frac{1-\sqrt{a}}{1-a}\right)\)
Rút gọn A
Rút gọn biểu thức:
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\cdot\left(x-1\right)}{\sqrt{x}-1}\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
ĐKXĐ: Bạn tự làm nha
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+1\)
\(=\frac{x^2-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x^2+x+1}{x+\sqrt{x}+1}\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{1\left(\sqrt{a}-1\right)-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1-2}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
Rút gọn biểu thức sau :\(\left(\frac{1}{2+2\sqrt{a}}+\frac{1}{2-2\sqrt{a}}-\frac{a^2+1}{1-a^2}\right)\cdot\left(1+\frac{1}{a}\right)\)
\(\left(\frac{1}{2+2.\sqrt{a}}+\frac{1}{2-2.\sqrt{a}}-\frac{a^2+1}{1-a^2}\right).\left(1+\frac{1}{a}\right)\)
\(=\left(\frac{2-2.\sqrt{a}+2+2.\sqrt{a}}{\left(2+2.\sqrt{a}\right)\left(2-2.\sqrt{a}\right)}-\frac{a^2+1}{\left(1-a\right).\left(1+a\right)}\right).\left(\frac{a+1}{a}\right)\)
\(=\left(\frac{4}{4-4a}-\frac{a^2+1}{\left(1-a\right).\left(1+a\right)}\right).\left(\frac{a+1}{a}\right)=\frac{\left(1+a\right)}{\left(1-a\right).\left(1+a\right)}\cdot\frac{a+1}{a}=\frac{1+a}{\left(1-a\right).a}=\frac{a+1}{a-a^2}\)
Rút gọn biểu thức:
1) \(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\cdot\left(x-1\right)}{\sqrt{x}-1}\)
2) \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
3) \(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
4) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
rút gọn A = \(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(A=\)\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+\sqrt{a}^3}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}\)\(:\)\(\left[\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\right]\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\)\(\left(1+a+2\sqrt{a}\right)\left(1+a-2\sqrt{a}\right)\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(1+a\right)\left[\left(1+a\right)^2-\left(2\sqrt{a}\right)^2\right]}\)\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1+2a+a^2-4a\right)}\)
\(=\frac{\sqrt{a}\left(1-a\right)^2}{\left(a+1\right)\left(1-a\right)^2}=\frac{\sqrt{q}}{a+1}\)
Rút gọn biểu thức:
\(A=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left[\sqrt{a}-\frac{1}{\sqrt{a}}\right]\left[\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right]\)
Điều kiện : a> 0 ; a khác 1
\(A=\frac{\left(\sqrt{a}\right)^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}\right)^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{2a+2}{a-1}\right)\)
\(A=\frac{2\sqrt{a}}{\sqrt{a}}+\frac{2\left(a+1\right)}{\sqrt{a}}=2+\frac{2\sqrt{a}\left(a+1\right)}{a}\)
Rút gọn :\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
rút gọn biểu thức \(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a+1}}{\sqrt{a}}+\left(\sqrt{a-}\frac{1}{\sqrt{a}}\right).\left(\frac{3\sqrt{a}}{\sqrt{a-1}}-\frac{2+\sqrt{a}}{\sqrt{a+1}}\right)\)
1. Rút gọn biểu thức:
a) \(\sqrt{27\cdot48\cdot\left(1-a\right)^2}\)với a>1
b) \(\frac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}\) với a>b
c) \(\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}\)với \(a\ge0\)
d) \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}\)với a>0
e) \(\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)
giải hộ mk con này vs mk 2 like cho: rút gọn
:\(P1=\frac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}+\frac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}\)
\(P2=\left(\frac{1-A\sqrt{A}}{1-\sqrt{A}}+\sqrt{A}\right)\cdot\left(\frac{1-\sqrt{A}}{1-A}\right)^2\)
P2\(=\left(\frac{1-A\sqrt{A}}{1-\sqrt{A}}+\sqrt{A}\right).\left(\frac{1-\sqrt{A}}{1-A}\right)^2\)\(=\left(\frac{1-A\sqrt{A}+\sqrt{A}-A}{1-\sqrt{A}}\right).\frac{\left(1-\sqrt{A}\right)^2}{\left(1-A\right)^2}\)\(=\frac{\left(\sqrt{A}+1\right)\left(1-A\right)}{1-\sqrt{A}}.\frac{\left(1-\sqrt{A}\right)^2}{\left(1-\sqrt{A}\right)^2\left(1+\sqrt{A}\right)^2}\)
\(=\left(\sqrt{A}+1\right)^2.\frac{1}{\left(1+\sqrt{A}\right)^2}=1\)
Rút gọn các biểu thức
\(A=\left(1+\frac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\frac{a+\sqrt{a}}{a-1}\frac{\sqrt{a}}{a-\sqrt{a}}\right)\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
\(C=\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{1}{a-1}\right):\frac{a}{2+2\sqrt{a}}\)