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1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
Bài 2:
a: ĐKXĐ: x>0; x<>1
b: \(P=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+1}\)
c: Khi x=1/4 thì \(P=2:\left(\dfrac{1}{2}+1\right)=2:\dfrac{3}{2}=\dfrac{4}{3}\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
b: \(B=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\cdot\left(2-\dfrac{\sqrt{a}\left(5-\sqrt{b}\right)}{-\left(5-\sqrt{b}\right)}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\)
c: \(C=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+2\right)\left(2-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\)
=4-x
a) ĐKXĐ: : phải là 1 biểu thức có nghĩa. b) ko có x nên ko phải tìm
Ô xin lỗi bạn, do lúc trước mình ko thấy đề nên bấm bậy, xin lỗi nhiều
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
\[\begin{array}{l}
Q = {\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)^2}\left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}} - \frac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right)\\
Q = {\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)^2}.\frac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
Q = {\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)^2}.\frac{{4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
Q = \frac{{4\sqrt x {{\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
Q = \frac{{4\sqrt x {{\left( {\frac{{x - 1}}{{2\sqrt x }}} \right)}^2}}}{{x - 1}}\\
Q = \frac{{\sqrt x .\frac{{{{\left( {x - 1} \right)}^2}}}{x}}}{{x - 1}}\\
Q = \frac{{x\sqrt x - \sqrt x }}{x}
\end{array}\]
hẹn ngày mai, giao hàng hôm nay, không lỡ hẹn nhé
a) ĐK: \(x\ne1\)
\(P=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{-2}\) \(=\dfrac{-\sqrt{x}-1}{2}\)
b) P=4
\(P=-4\Leftrightarrow\dfrac{-\sqrt{x}-1}{2}=-4\Leftrightarrow-\sqrt{x}-1=-8\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\left(N\right)\)
c) \(x=8-2\sqrt{15}\Rightarrow\sqrt{x}=\sqrt{5}-\sqrt{3}\)
Thay \(\sqrt{x}=\sqrt{5}-\sqrt{3}\) vào P, ta được:
\(P=\dfrac{-\sqrt{5}+\sqrt{3}-1}{2}\)
KL: a) ĐK: \(x\ne1\)
\(P=\dfrac{-\sqrt{x}-1}{2}\)
b) x= 81
c) \(P=\dfrac{-\sqrt{5}+\sqrt{3}-1}{2}\)
a)Đkxđ : x#1 , x > 0
Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
Q=\(\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
Q=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Q=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}X\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
Q=\(\dfrac{x-1}{\sqrt{x}}\)
b)Thay x = 2\(\sqrt{2}\)+3 vào phương trình ta được :
Q=\(\dfrac{2\sqrt{2}+3-1}{\sqrt{2\sqrt{2}+3}}\)
Q=\(\dfrac{2\sqrt{2}+2}{\sqrt{\left(\sqrt{2}+1\right)}^2}\)
Q=\(\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
Q= 2
Mysterious Person giup mk