Question

Help me pls ;-;

Answer 1

Bài 2:ĐKXĐ: a>=0; a<>4

a: \(\frac{\sqrt{a}-2}{\sqrt{a}+2}+\frac{\sqrt{a}+2}{\sqrt{a}-2}-\frac{4a}{4-a}\)

\(=\frac{\left(\sqrt{a}-2\right)^2+\left(\sqrt{a}+2\right)^2+4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{a-4\sqrt{a}+4+a+4\sqrt{a}+4+4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{6a+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{2\left(3a+4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

Ta có: \(M=\left(\frac{\sqrt{a}-2}{\sqrt{a}+2}+\frac{\sqrt{a}+2}{\sqrt{a}-2}-\frac{4a}{4-a}\right):\frac{3a+4}{\sqrt{a}+2}\)

\(=\frac{2\left(3a+4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{3a+4}{\sqrt{a}+2}\)

\(=\frac{2\left(3a+4\right)}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}+2\right)}\cdot\frac{\sqrt{a}+2}{3a+4}=\frac{2}{\sqrt{a}-2}\)

b: M<-1

=>M+1<0

=>\(\frac{2+\sqrt{a}-2}{\sqrt{a}-2}<0\)

=>\(\frac{\sqrt{a}}{\sqrt{a}-2}<0\)

=>\(\sqrt{a}-2<0\)

=>\(\sqrt{a}<2\)

=>0<=a<4

c: Để M nguyên thì 2⋮\(\sqrt{a}-2\)

=>\(\sqrt{a}-2\in\left\lbrace1;-1;2;-2\right\rbrace\)

=>\(\sqrt{a}\in\left\lbrace3;1;4;0\right\rbrace\)

=>a∈{0;1;9;16}

Bài 1:

a: \(A=\frac{x\cdot\sqrt{2x}+1}{x-1}-\frac{x+\sqrt{2x}}{x-1}\)

\(=\frac{x\left(\sqrt{2x}-1\right)+1-\sqrt{2x}}{x-1}=\frac{\left(x-1\right)\left(\sqrt{2x}-1\right)}{x-1}=\sqrt{2x}-1\)

\(B=\sqrt2\cdot\sqrt{2+\sqrt3}-\frac{2}{\sqrt3+1}\)

\(=\sqrt{4+2\sqrt3}-\frac{2\left(\sqrt3-1\right)}{\left(\sqrt3+1\right)\left(\sqrt3-1\right)}\)

\(=\sqrt3+1-\left(\sqrt3-1\right)=2\)

b: Khi x=2 thì \(A=\sqrt{2\cdot2}-1=2-1=1\)

c: A=B

=>\(\sqrt{2x}-1=2\)

=>\(\sqrt{2x}=2+1=3\)

=>2x=9

=>x=9/2(nhận)