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Mọi ngươi giúp em với ạ chứ em làm câu a Bài 1 và 2 ra kết quả dài quá :(
Bài 1:
a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)
=>căn a-2>0
=>a>4
a: \(P=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{4x}\cdot\dfrac{-4\sqrt{x}}{x-1}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{-\left(4+2\sqrt{3}-1\right)}{\sqrt{3}+1}=\dfrac{-\left(3+2\sqrt{3}\right)}{\sqrt{3}+1}=\dfrac{-3-\sqrt{3}}{2}\)
c: Để P<0 thì -(x-1)<0
=>x-1>0
=>x>1
a, A\(=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x-1}{\sqrt{x}}\) ĐK x>0 ;\(x\ne1;x\ne-1\)
\(A=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}}{x-1}\)
\(A=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}\)=\(\frac{4x^2}{\left(x-1\right)^2}\)
b, Để A =2 \(\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Rightarrow4x^2=2\left(x-1\right)^2\)
<=> \(4x^2=2x^2-4x+2\)
<=> \(2x^2+4x-2=0\)
<=> \(x^2+2x-1=0\)
\(\Delta=1^2-1.\left(-1\right)\) = 2
=> \(\orbr{\begin{cases}x_1=-1-\sqrt{2}\left(loại\right)\\x_2=-1+\sqrt{2}\left(nhận\right)\end{cases}}\)
Vậy x=\(-1+\sqrt{2}\)thì A =2
c, Thay x =\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)=2
=>A = \(\frac{4.2^2}{\left(2-1\right)^2}=16\)
Vậy A=16 thì x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a: \(A=\dfrac{-x+\sqrt{x}-3-\sqrt{x}-1}{x-1}:\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{x-1}\)
\(=\dfrac{-x-4}{x-1}\cdot\dfrac{x-1}{-4\sqrt{x}}=\dfrac{x+4}{4\sqrt{x}}\)
b: Khi x=4-2 căn 3 thì \(A=\dfrac{4-2\sqrt{3}+4}{4\left(\sqrt{3}-1\right)}=\dfrac{8-2\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\dfrac{3\sqrt{3}+1}{4}\)
c: \(A-1=\dfrac{x+4-4\sqrt{x}}{4\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{4\sqrt{x}}>0\)
=>A>1
ĐKXĐ: \(x\ge0;x\ne9\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
\(P=\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right)\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{-3\sqrt{x}-3}{x-3}\right)\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(P=\dfrac{-3}{\sqrt{x}+3}\)
b/ Do \(-3< 0\Rightarrow P_{min}\) khi \(\sqrt{x}+3\) nhỏ nhất
Mà \(\sqrt{x}+3\ge3\Rightarrow P_{min}=\dfrac{-3}{3}=-1\) khi \(\sqrt{x}+3=3\Leftrightarrow x=0\)
Vậy với \(x=0\) thì P đạt GTNN
a) \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}=\left[\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
b) Ta có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\)
Dấu bằng xảy ra khi x=0
Vậy x=0 thì P đạt GTNN là -1
Bài 2:
a: ĐKXĐ: a>0 và b>0
b: \(P=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}=\sqrt{a}-\sqrt{b}\)
c: Khi a=4 và b=1 thì P=2-1=1
a,ĐKXĐ \(x\ge0;x\ne1\)
Ta có A=\(\dfrac{x+2\sqrt{x}+1+2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2\sqrt{x}+1-x+\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
A=\(\dfrac{4\sqrt{x}}{x-1}.\dfrac{x-1}{2\sqrt{x+1}}\)
A=\(\dfrac{4\sqrt{x}}{2\sqrt{x}+1}\)
b, Thay x=\(1-\dfrac{\sqrt{3}}{2}\) vào biểu thức A ta có
A=\(\dfrac{4\sqrt{1-\dfrac{\sqrt{3}}{2}}}{2\sqrt{1-\dfrac{\sqrt{3}}{2}}+1}=\dfrac{\sqrt{16-8\sqrt{3}}}{\sqrt{4-2\sqrt{3}}+1}=\dfrac{6-2\sqrt{3}}{3}\)
a) \(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{x\sqrt{x}-x+x-\sqrt{x}+\sqrt{x}-x}{\sqrt{x}\left(x+\sqrt{x}\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}.\dfrac{\sqrt{x}\left(x+\sqrt{x}\right)}{x\sqrt{x}-x}\)
\(=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right).\dfrac{x+\sqrt{x}}{x}\)
\(=\left(\sqrt{x}+1\right).\dfrac{x+\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}\right)}{x}\)
b) Thay \(x=4-2\sqrt{3}\) vào A có:
\(A=\dfrac{\left(4-2\sqrt{3}+1\right)\left(4-2\sqrt{3}+\sqrt{4-2\sqrt{3}}\right)}{4-2\sqrt{3}}\)
\(=\dfrac{\left(5-2\sqrt{3}\right)\left(4-2\sqrt{3}+\sqrt{\left(1-\sqrt{3}\right)^2}\right)}{4-2\sqrt{3}}\)
\(=\dfrac{\left(5-2\sqrt{3}\right)\left(4-2\sqrt{3}+\sqrt{3}-1\right)}{4-2\sqrt{3}}=\dfrac{\left(5-2\sqrt{3}\right)\left(3-\sqrt{3}\right)}{4-2\sqrt{3}}\)
\(=\dfrac{15-5\sqrt{3}-6\sqrt{3}+6}{4-2\sqrt{3}}=\dfrac{21-11\sqrt{3}}{4-2\sqrt{3}}=\dfrac{\left(21-11\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}=\dfrac{\left(21-11\sqrt{3}\right)2\left(2+\sqrt{3}\right)}{16-12}\)
\(=\dfrac{\left(21-11\sqrt{3}\right)2\left(2+\sqrt{3}\right)}{4}=\dfrac{42+21\sqrt{3}-22\sqrt{3}-33}{2}=\dfrac{9-\sqrt{3}}{2}\)