Đặt \(x=a+\sqrt{15};y=\dfrac{1}{a}-\sqrt{15}\left(x,y\in Z\right)\)

Ta có: \(y=\dfrac{1}{x-\sqrt{15}}-\sqrt{15}\Leftrightarrow xy-16=\left(y+x\right)\sqrt{15}\)

Nếu y=x thì VP là số vô tỉ còn VT là số nguyên ( vô lý)

=> x=y

=> xy-16=0 <=> x=y=\(\pm\)4 .

Thay vào tìm đc \(\left[{}\begin{matrix}a=4-\sqrt{15}\\a=-4-\sqrt{15}\end{matrix}\right.\)

1. Ta có: A = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để A \(\in\)Z <=> \(4⋮\sqrt{x}-3\) <=> \(\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Lập bảng:

Vậy ....

2. Ta có: B = \(\frac{x^2+15}{x^2+3}=\frac{\left(x^2+3\right)+12}{x^2+3}=1+\frac{12}{x^2+3}\)

Do x² + 3 \(\ge\)3  \(\forall\)x => \(\frac{12}{x^2+3}\le4\forall x\)

=> \(1+\frac{12}{x^2+3}\le5\forall x\)

Dấu "=" xảy ra <=> x = 0

Vậy Max B = 5 khi x = 0

a) ĐKXĐ: a\(\ge\)0, a\(\ne\)1

A=(\(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)).\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

A=\(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\).\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

A=\(\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}.\dfrac{\sqrt{a}+1}{\sqrt{a}}\)=\(\dfrac{2}{a-1}\)

b) Để A\(\in\)Z\(\Rightarrow\)x-1\(\in\) Ư(2)=\(\left\{-1,1,-2,2\right\}\)

vì x\(\ge\)0,x\(\ne\)1 nên x\(\in\)\(\left\{-1,0,2,3\right\}\)

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

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

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

\(A=\frac{4}{\sqrt{x}+2}\)

b, \(A=\frac{4}{\sqrt{x}+2}=\frac{2}{3}\)

=> 2cawn x + 4 = 12

=> 2.căn x = 8

=> căn x = 4

=> x = 16 (thỏa mãn)

c, có A = 4/ căn x + 2 và B  = 1/căn x - 2

=> A.B = 4/x - 4 

mà AB nguyên

=> 4 ⋮ x - 4

=> x - 4 thuộc Ư(4) 

=> x - 4 thuộc {-1;1;-2;2;-4;4}

=> x thuộc {3;5;2;6;0;8} mà x > 0 và x khác 4

=> x thuộc {3;5;2;6;8}

d, giống c thôi

a: \(B=\frac{2\sqrt{x}+13}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}+13+\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2\sqrt{x}+13+x+\sqrt{x}-6}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}=\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

b: C=A-B

\(=\frac{2\sqrt{x}-1}{\sqrt{x}+3}-\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(x+3\sqrt{x}+7\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{2x+3\sqrt{x}-2-x-3\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-3}{\sqrt{x}+2}\)

C nguyên khi \(\sqrt{x}-3\) ⋮\(\sqrt{x}+2\)

=>\(\sqrt{x}+2-5\) ⋮\(\sqrt{x}+2\)

=>-5⋮\(\sqrt{x}+2\)

=>\(\sqrt{x}+2=5\) (Do \(\sqrt{x}+2\ge2\forall x\) thỏa mãn ĐKXĐ)

=>\(\sqrt{x}=3\)

=>x=9(nhận)

Toán lớp 9 như này á hả

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