Cho M=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\) \(\left(x\ge0;x\ne4;x\ne9\right)\)
a, Rút gọn M
b, Tính M khi x=\(11-6\sqrt{2}\)
c, Tìm x để M<1
d, Tìm \(x\in Z\) để M\(\in Z\)
K
Khách
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KT
11 tháng 7 2018
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: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
P/s: Không chắc lắm nha!
a, M=\(\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)(ĐKXD: x>0, x#4, x#9)
=\(\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)^{ }}\)=\(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
Vậy.....
b, ta có x=11-6\(\sqrt{2}\)=\(\left(3-\sqrt{2}\right)^2\)
Thay vào M ta đươc:
M=\(\frac{\sqrt{\left(3-\sqrt{2}\right)^2}+1}{\sqrt{\left(3-\sqrt{2}\right)^2}-3}\)=\(\frac{3-\sqrt{2}+1}{3-\sqrt{2}-3}=\frac{4-\sqrt{2}}{-\sqrt{2}}=1-2\sqrt{2}\)
c,Để M<1<=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)<1 <=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)-1<0<=> \(\frac{4}{\sqrt{x}-3}\)<0<=> x<9(t/m x#9) mà x>0, x#4 => 0<x<9 và x#4
Vậy....
d, Để M∈Z <=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)∈Z<=>\(1+\frac{4}{\sqrt{x}-3}\)∈Z<=>\(\frac{4}{\sqrt{x}-3}\)∈Z<=> 4⋮\(\sqrt{x}-3\)<=>\(\sqrt{x}-3\)∈Ư(4)={\(\pm\)1,\(\pm\)2,\(\pm\)4}
<=>\(\sqrt{x}\) ∈ {2,4,5,1,7}
<=>x ∈ {4,16,25,1,49} mà x#4
=> x∈ {16,25,1,49}
vậy..