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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
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
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) \(P=\dfrac{-\sqrt{5}+\sqrt{3}-1}{2}\)
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: \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(x=2+2\sqrt{5}+2-2\sqrt{5}=4\)
Khi x=4 thì \(P=\dfrac{4+2+1}{2}=\dfrac{7}{2}\)
1. b) \(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\left(x\sqrt{\dfrac{6x}{x^2}}+\sqrt{\dfrac{6x}{9}}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\dfrac{7}{3}\sqrt{6x}:\sqrt{6x}=\dfrac{7}{3}\)
2.
P=\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)(bn có ghi sai đề ko)
a) ĐKXĐ : \(x\ge1,x\ge2,x\ge0\)
b) P=\(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x-3\sqrt{x}-\sqrt{x}+3-2x+\sqrt{x}+4\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)
c) thay x= \(4-2\sqrt{3}\)vào P ta có :
\(\dfrac{1}{\sqrt{4-2\sqrt{3}}-2}=\dfrac{1}{\sqrt{3}-1-2}=\dfrac{1}{\sqrt{3}-3}\)
\(Q=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{2\sqrt{x}-x}\right):\left(\dfrac{2+\sqrt{x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
ĐKXĐ : \(x\ne0;x\ne4\)
\(Q=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(Q=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(Q=\left(\dfrac{\sqrt{x}-5\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(Q=\dfrac{-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-4}\)
\(Q=\dfrac{-4\left(\sqrt{x}-1\right)}{-4}\)
\(\Leftrightarrow Q=\sqrt{x}-1\)
b ) Khi \(Q=5\), ta có :
\(\sqrt{x}-1=5\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)