Cho A = \(\left(\frac{\sqrt{a}-2}{\sqrt{a}+2}-\frac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}\frac{4}{\sqrt{a}}\right)\)
a) Tìm a để A có nghĩa
b) CMR A không phụ thuộc vào a
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) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{2}{a-4}\right)\cdot\left(\sqrt{a}-1+\frac{\sqrt{a}-4}{\sqrt{a}}\right)\)
=\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right).\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}}+\frac{\sqrt{a}-4}{\sqrt{a}}\right)\) =\(\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right).\left(\frac{a-\sqrt{a}+\sqrt{a}-4}{\sqrt{a}}\right)\) =\(\left(\frac{a+2\sqrt{a}+\sqrt{a}+2-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right).\left(\frac{a-4}{\sqrt{a}}\right)\)=\(\frac{a+3\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}.\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}}\) =\(\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}.\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}}\) =\(\sqrt{a}+3\)
Cho \(A=\left(\frac{\sqrt{a}}{2}-\frac{1}{2 \sqrt{a}}\right)^{2} \cdot\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
a) Rút gọn A
b) Tìm a để A<0
c) Tìm a để A=-2
Cho \(A=\left(\frac{\sqrt{a}}{2}-\frac{1}{2 \sqrt{a}}\right)^{2} \cdot\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
a) Rút gọn A
b) Tìm a để A<0
c) Tìm a để A=-2
bài 1: Cho biểu thức R = \(\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{x-2\sqrt{x}}\right)\cdot\left(\frac{1}{\sqrt{x+2}}+\frac{4}{x-4}\right)\)
a/ rút gọn R
b/ Tính giá trị R khi x = 4 + \(2\sqrt{3}\)
c/ Tìm giá trị của x để R >0
bài 2 : Cho A = 6 + 2\(\sqrt{2}\), B = 9 . So sánh A,B .
bài 3 : Chứng minh:
a/ \(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}\)= a - b (với a >0, b>0, \(a\ne b\))
b/ \(\left(2+\frac{a-\sqrt{a}}{\sqrt{a-1}}\right)\cdot\left(2-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)=4-a\)(với a >0, a\(\ne1\))
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
1,Cho \(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
a,Tìm điều kiện để A có nghĩa
b,Khi A có nghĩa. Cmr: A không phụ thuộc vào a
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
\(A=\frac{a+b+2\sqrt{ab}-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}+\sqrt{b}\right)\)
\(A=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
\(B=\left(\frac{6}{a-1}+\frac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\frac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\) (với a>0;a≠1)
Đặt \(C=B\cdot\left(a-\sqrt{a}+1\right)\) . So sánh C và 1
uses crt;
var B,C: real;
a:real;
begin
clrscr;
writeln('a = '); read(a);
B:=(6/(a-1)+(10-2*sqrt(a))/(a*sqrt(a)-a-sqrt(a)+1))*(((sqrt(a)-1)*(sqrt(a)-1))/(4*sqrt(a)));
C:=B*(a-sqrt(a)+1);
if C > 1 then writeln('C > 1');
if C < 1 then writeln('C < 1');
if C = 1 then writeln('C = 1');
readln
end.
\(ChoQ=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
a. Tìm điều kiện b,d để Q có ngjiax
b. Cm: Giá trị của Q không thuộc vào giá trị a
\(Q=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
DK: \(a,b\ge0\)do \(Q=2\sqrt{b}\)nên Q ko phụ thuộc vào giá trị của a