\(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\)

\(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)

\(=\dfrac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x\left(x-3\right)}\)

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\(\dfrac{x+1}{2x+6}-\dfrac{x-6}{2x^2+6x}\)

\(=\dfrac{x+1}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)

\(=\dfrac{x\left(x+1\right)}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)

\(=\dfrac{x^2+x-x+6}{2x\left(x+3\right)}=\dfrac{x^2+6}{2x\left(x+3\right)}\)

a: Để A là phân số thì \(n+1\ne0\)

=>\(n\ne-1\)

b: \(n^2-1=0\)

=>\(n^2=1\)

=>\(\left[{}\begin{matrix}n=1\left(nhận\right)\\n=-1\left(loại\right)\end{matrix}\right.\)

Thay n=1 vào A, ta được:

\(A=\dfrac{3\cdot1-1}{1+1}=\dfrac{3-1}{2}=1\)

c: Để A là số nguyên thì \(3n-1⋮n+1\)

=>\(3n+3-4⋮n+1\)

=>\(-4⋮n+1\)

=>\(n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(n\in\left\{0;-2;1;-3;3;-5\right\}\)

d: \(A=\dfrac{3n-1}{n+1}=\dfrac{3n+3-4}{n+1}=3-\dfrac{4}{n+1}\)

Để A min thì \(\dfrac{-4}{n+1}\) min

=>n+1 là số nguyên dương nhỏ nhất

=>n+1=1

=>n=0

=>\(A=3-\dfrac{4}{0+1}=3-4=-1\)

a)Kẻ AH⊥BC

Vì ΔABC vuông cân tại A

 ⇒ AH cũng là đường trung tuyến 

⇒ AH=BH=CH

Ta có:MB² + MC² = (BH-HM)² + (CH+HM)² = (AH-HM)²+(AH+HM)²

       = AH²-2.AH.HM+HM²+AH²+2.AH.HM+HM²=2(AH²+HM²)

Áp dụng định lý Py-ta-go vào ΔAHM vuông tại A ta có:

      MA² = AH²+HM²

⇒ MB²+MC²=2MA²  

b) Ta có: MA≥AH (đường xiên và đường vuông góc)

        ⇒ MA² ≥ AH²

        ⇒ 2MA² ≥ 2AH²

        ⇒ MB²+MC² ≥ 2AH²

Dấu "=" xảy ra ⇔ MA=AH ⇔ M là trung điểm của BC

Vậy Min K = 2AH²  ⇔ M là trung điểm của BC

1 played the piano for 4 years

2 5 years since my father last went to HN

3 been reading a newspaprt for 2 hours

4 have written to her for 4 years

5 is over a years since I last went to see them

6 not rained for 2 weeks

7 kept a diary for 5 years

8 gone swimming since 2001

9 studied E since I was 10 years old

10 saw Tom when he moved to London

i) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}=\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}+\left|\sqrt{2}-2\right|\)

\(=\sqrt{\left(\sqrt{2}+1\right)^2}+2-\sqrt{2}=\left|\sqrt{2}+1\right|+2-\sqrt{2}=\sqrt{2}+1+2-\sqrt{2}=3\)

k) \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}=\sqrt{\dfrac{8-2\sqrt{15}}{2}}-\sqrt{\dfrac{8+2\sqrt{15}}{2}}+\sqrt{6}\)

\(=\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}+\sqrt{6}\)

\(=\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}+\sqrt{6}\)

\(=\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}+\sqrt{6}\)

\(=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}+\sqrt{6}=\dfrac{-2\sqrt{3}}{\sqrt{2}}+\sqrt{6}=-\sqrt{6}+\sqrt{6}=0\)

m) \(2\sqrt{56}-14\sqrt{\dfrac{2}{7}}+\left(\sqrt{7}-\sqrt{2}\right)\sqrt{7}-\dfrac{8\sqrt{2}}{\sqrt{3}-\sqrt{7}}\)

\(=2\sqrt{4.14}-2\sqrt{49.\dfrac{2}{7}}+7-\sqrt{14}+\dfrac{8\sqrt{2}.\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}\)

\(=4\sqrt{14}-2\sqrt{14}+7-\sqrt{14}+\dfrac{8.\left(\sqrt{14}+\sqrt{6}\right)}{4}\)

\(=\sqrt{14}+7+2\left(\sqrt{14}+\sqrt{6}\right)=7+3\sqrt{14}+2\sqrt{6}\)

Lời giải:
i.

\(=\sqrt{(\sqrt{2}+1)^2}+|\sqrt{2}-2|=|\sqrt{2}+1|+|\sqrt{2}-2|=\sqrt{2}+1+2-\sqrt{2}=3\)

k.

\(=\frac{1}{\sqrt{2}}(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}+\sqrt{12})\)

\(=\frac{1}{\sqrt{2}}(\sqrt{(\sqrt{3}-\sqrt{5})^2}-\sqrt{(\sqrt{3}+\sqrt{5})^2}+2\sqrt{3})\)

\(=\frac{1}{\sqrt{2}}(|\sqrt{3}-\sqrt{5}|-|\sqrt{3}+\sqrt{5}|+2\sqrt{3})=\frac{1}{\sqrt{2}}(-2\sqrt{3}+2\sqrt{3})=0\)

m.

\(=4\sqrt{14}-2\sqrt{14}+7-\sqrt{14}-\frac{8\sqrt{2}(\sqrt{3}+\sqrt{7})}{(\sqrt{3}-\sqrt{7})(\sqrt{3}+\sqrt{7})}\)

\(=\sqrt{14}+7-\frac{8(\sqrt{14}+\sqrt{6})}{-4}=\sqrt{14}+\sqrt{7}+2(\sqrt{14}+\sqrt{6})=3\sqrt{14}+\sqrt{7}+2\sqrt{6}\)