BÀI1 : Rút gọn \(\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}-1\right).\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\) (ĐKXĐa>0;a khác 1)
BÀI2 Cho hpt\(\hept{\begin{cases}x+my=1\\x+2y=3\end{cases}}\)Tìm m để hpt có no(x,y)thỏa mãn x-y=1
Giups mk nha
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: \(A = \frac{{{{\left( {{a^{\sqrt 2 - 1}}} \right)}^{1 + \sqrt 2 }}}}{{{a^{\sqrt 5 - 1}}.{a^{3 - \sqrt 5 }}}}\,\,\,\left( {a > 0} \right).\)
\(=\dfrac{a^{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}{a^{\left(\sqrt{5}-1\right)+\left(3-\sqrt{5}\right)}}=\dfrac{a}{a^{\sqrt{5}-1+3-\sqrt{5}}}=\dfrac{a}{a^2}=\dfrac{1}{a}\)
\(ChoQ=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
a, rút gọn
b, chứng minh nếu 0<x<1 thì Q>0
c, tìm GTLN của Q
\(ChoA=\frac{1}{2\left(1+\sqrt{x}+2\right)}+\frac{1}{2\left(1-\sqrt{x}+2\right)}\)
a, tìm x để a có nghĩa
b, rút gon A
c, tìm X nguyên để A nguyên
\(ChoA=\left(\frac{\sqrt{a}}{\sqrt{a-1}}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2}{a-1}\right)\)
a, Rút gọn A
b, tính A Khi a=3+\(2\sqrt{2}\)
rút gọn biểu thức:
\(Q=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(A=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)với a lớn hơn hoặc bằng 0; a khác 1
rút gọn :
\(C=\left(\frac{\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}}{1+\sqrt{a}}\right)\div\frac{\sqrt{a}}{a-1}\left(a>0;a\ne0\right)\)
\(=\left(\frac{\sqrt{a}+a+\sqrt{a}-a}{1-a}\right)\div\frac{\sqrt{a}}{a-1}\)
\(=\frac{2\sqrt{a}}{1-a}\div\frac{\sqrt{a}}{a-1}\)
\(=\frac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\frac{2\left(\sqrt{a}-1\right)}{\left(1-\sqrt{a}\right)}\)\(=-2\)
\(Với:a>0;a\ne0\)
\(\left(\frac{1}{\sqrt{a}+\sqrt{a+1}}+\frac{1}{\sqrt{a}-\sqrt{a-1}}\right):\left(1+\sqrt{\frac{a+1}{a-1}}\right)\)
\(\left(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a^3}+3\right)\left(a-b\right)}\right)\)
Rút gọn 2 biểu thức trên?
Ai giúp mình với, tks nhiều
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
Rút gọn các biểu thức
\(A=\left(1+\frac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\frac{a+\sqrt{a}}{a-1}\frac{\sqrt{a}}{a-\sqrt{a}}\right)\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
\(C=\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{1}{a-1}\right):\frac{a}{2+2\sqrt{a}}\)
Rút gọn biểu thức P=\(\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1+\sqrt{a}}\right):\left(1+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)với a>0 và a khác 1
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)....\)
a) Rút gọn A, b) tìm a để A<0.c) Tìm A để A=-2
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\).\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
= \(\left[\left(\frac{\sqrt{a}}{2}\right)^2-2\frac{\sqrt{a}}{2}\frac{1}{2\sqrt{a}}+\left(\frac{1}{2\sqrt{a}}\right)^2\right]\).\(\left[\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{a-1}\right]\)
=\(\left(\frac{a}{4}-\frac{1}{2}+\frac{1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1\right)^2}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)^2}{a-1}\right]\)
=\(\left(\frac{a^2}{4a}-\frac{2a}{4a}+\frac{1}{4a}\right)\).\(\left[\frac{\left[\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)\right]\cdot\left[\left(\sqrt{a}-1\right)+\left(\sqrt{a}+1\right)\right]}{a-1}\right]\)
=\(\left(\frac{a^2-2a+1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1-\sqrt{a}+1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right]\)
=\(\frac{\left(a-1\right)^2}{1}\).\(\frac{-4\sqrt{a}}{a-1}\)
=\(\frac{-\left(a-1\right)}{1}\)= - a + 1
hok tốt