mn giúp e với ạ

a: 2+5/6=12/6+5/6=17/6

b: 5/12+3/4+1/3=5/12+9/12+4/12=18/12=3/2

c: 2/3+3/4=8/12+9/12=17/12

\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)

\(x^2-y^2-6x+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x+y-3\right)\left(x-y-3\right)\)

Bài 1:1) \(10+2\sqrt{10}=\sqrt{10}\left(2+\sqrt{10}\right)\)

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

các câu 3,4,5 bạn làm tương tự như 2 câu trên

6) \(3a\sqrt{b}+3b\sqrt{a}=3\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\)

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

8) \(b-4=\left(\sqrt{b}\right)^2-2^2=\left(\sqrt{b}-2\right)\left(\sqrt{b}+2\right)\)

Bài 2: 1) \(\dfrac{\left(2-\sqrt{a}\right)^2-\left(\sqrt{a}+3\right)^2}{2a+\sqrt{a}}\left(a>0\right)\)

\(=\dfrac{\left(2-\sqrt{a}-\sqrt{a}-3\right)\left(2-\sqrt{a}+\sqrt{a}+3\right)}{\sqrt{a}\left(2\sqrt{a}+1\right)}\)

\(=\dfrac{\left(-2\sqrt{a}-1\right).5}{\sqrt{a}\left(2\sqrt{a}+1\right)}=-\dfrac{5}{\sqrt{a}}\)

3) \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\left(a\ge0,a\ne4\right)\)

\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}=\sqrt{a}+2-\left(2+\sqrt{a}\right)=0\)

4) \(\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\dfrac{a+\sqrt{a}}{1+\sqrt{a}}\right)\left(a\ge0,a\ne1\right)\)

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

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

mấy câu còn lại bạn làm tương tự

Bài 1: 

1) \(10+2\sqrt{10}=2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)\)

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

3) \(5\sqrt{7}-7\sqrt{5}=\sqrt{35}\left(\sqrt{5}-\sqrt{7}\right)\)

4) \(4\sqrt{3}-2\sqrt{6}=2\sqrt{6}\left(\sqrt{2}-\sqrt{1}\right)\)

5) \(6\sqrt{6}-2\sqrt{12}+3\sqrt{2}\)

\(=\sqrt{216}-\sqrt{48}+\sqrt{18}\)

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

6) \(3a\sqrt{6}+36\sqrt{a}\)

\(=3\sqrt{a}\left(\sqrt{6a}+12\right)\)

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

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

8) \(b-4=\left(\sqrt{b}-2\right)\left(\sqrt{b}+2\right)\)

Bài 2: 

a) Ta có: \(\dfrac{\left(2-\sqrt{a}\right)^2-\left(\sqrt{a}+3\right)^2}{2a+\sqrt{a}}\)

\(=\dfrac{a-4\sqrt{a}+4-a-6\sqrt{a}-9}{2a+\sqrt{a}}\)

\(=\dfrac{-10\sqrt{a}-5}{\sqrt{a}\left(2\sqrt{a}+1\right)}\)

\(=\dfrac{-5\left(2\sqrt{a}+1\right)}{\sqrt{a}\left(2\sqrt{a}+1\right)}\)

\(=-\dfrac{5}{\sqrt{a}}\)

4) Ta có: \(\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\dfrac{a+\sqrt{a}}{1+\sqrt{a}}\right)\)

\(=\left(1-\sqrt{a}\right)\left(1-\sqrt{a}\right)\)

\(=a-2\sqrt{a}+1\)

câu mấy với câu mấy

ko biết lm

mới lớp 6

đề như này không cụ thể khó làm lắm em ạ