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\(A=\sqrt{\left(\sqrt{a-1}+1\right)^2}+\sqrt{\left(\sqrt{a-1}-1\right)^2}\)
\(A=\sqrt{a-1}+1+1-\sqrt{a-1}\) ( DO: a < 2 - gt => \(1>\sqrt{a-1}\))
\(A=2\)
Vậy A = 2.
c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}.\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}-1}{a}\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(1+a\right)-\left(1-a\right)}.\frac{\left(\sqrt{1-a^2}-1\right)}{a}=-1\)
M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu
a,
\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
b,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)
a, \(\sqrt{\left(2-\sqrt{5}\right)^2}=\sqrt{5}-2\left(\sqrt{5}>2\right)\)
b, \(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\left(3>\sqrt{2}\right)\)
c, Với a < 3
\(\sqrt{\left(a-3\right)^2}+\left(a-9\right)=3-a+a-9=-6\)
d, \(A=\sqrt{\left(2a+5\right)^2}-\left(2a-7\right)\)
\(=\left|2a+5\right|-2a+7\)
+) Xét \(x\ge\dfrac{-5}{2}\) có:
\(A=2a+5-2a+7=12\)
+) Xét \(x< \dfrac{-5}{2}\) có:
\(A=-2a-5-2a+7=-4a+2\)
Vậy...
Bài 2 :
b) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\) (1)
ĐKXĐ : \(x\ge1\)
Pt(1) tương đương :
\(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (*)
Xét \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\)
\(\Rightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+\sqrt{x-1}-1=2\)
\(\Leftrightarrow2\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\) ( Thỏa mãn )
Xét \(1\le x< 2\) thì \(x\ge2\Rightarrow\sqrt{x-1}-1< 0\)
Nên : \(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\). Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\Leftrightarrow2=2\) ( Luôn đúng )
Vậy tập nghiệm của phương trình đã cho là \(S=\left\{x|1\le x\le2\right\}\)
Bài 1 :
a) ĐKXĐ : \(-1\le a\le1\)
Ta có : \(Q=\left(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}\right)\)
\(=\left(\frac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right)\cdot\frac{\sqrt{1-a^2}}{3}\)
\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3}\)
\(=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\)
Vậy \(Q=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\) với \(-1\le a\le1\)
b) Với \(a=\frac{\sqrt{3}}{2}\) thỏa mãn ĐKXĐ \(-1\le a\le1\)nên ta có :
\(\hept{\begin{cases}1-a=1-\frac{\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{2^2}\\1-a^2=1-\frac{3}{4}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\sqrt{1-a}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2^2}}=\left|\frac{\sqrt{3}-1}{2}\right|=\frac{\sqrt{3}-1}{2}\\\sqrt{1-a^2}=\frac{1}{2}\end{cases}}\)
Do đó : \(Q=\frac{\left(3+\frac{1}{2}\right)\cdot\frac{\sqrt{3}-1}{2}}{3}=\frac{5\sqrt{3}-5}{12}\)
b)
)\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
= \(\frac{2}{2-\sqrt{5}}-\frac{2}{2+\sqrt{5}}\)
=\(\frac{2\left(2+\sqrt{5}\right)-2\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}\)
=\(\frac{4+2\sqrt{5}-4+2\sqrt{5}}{2^2-\sqrt{5}^2}\)
=\(\frac{4\sqrt{5}}{4-5}\)
=\(\frac{4\sqrt{5}}{-1}\)
\(-4\sqrt{5}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a, =\(\sqrt{\left[\left(\sqrt{3}\right)^2+2.\sqrt{3}.2+2^2\right]\left(a-1\right)^2}\)
=\(\sqrt{\left(\sqrt{3}+2\right)^2\left(a-1\right)^2}\)
=\(\left(\sqrt{3}+2\right)\left|a-1\right|\)
A=\(\sqrt{\left(a-1\right)+2\sqrt{a-1}+1}+\sqrt{\left(a-1\right)-2\sqrt{a-1}+1}\)
A=\(\sqrt{\left(\sqrt{a-1}+1\right)^2}+\sqrt{\left(\sqrt{a-1}+1\right)^2}\)
A=\(\left|\sqrt{a-1}+1\right|+\left|\sqrt{a-1}-1\right|\)
+Với a<2 thì a-1<1 => \(\sqrt{a-1}< 1\)
khi đó A=\(\left(\sqrt{a-1}+1\right)-\left(\sqrt{a-1}-1\right)\)= 2