\(\sqrt{a\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}\)

\(=\sqrt{a\left(a+6\right)\left(a+1\right)\left(a+5\right)\left(a+2\right)\left(a+4\right)+36}\)

\(=\sqrt{\left(a^2+6a\right)\left(a^2+6a+5\right)\left(a^2+6a+8\right)+36}\left(1\right)\)

Đặt \(a^2+6a=x\), Ta có:

\(\left(1\right)=\sqrt{x\left(x+5\right)\left(x+8\right)+36}\)

\(=\sqrt{\left(x^2+5\right)\left(x+8\right)+36}=\sqrt{x^3+13x^2+40x+36}\)

\(=\sqrt{x^3+9x^2+4x^2+36x+4x+36}=\sqrt{\left(x+9\right)\left(x+2\right)^2}\)

Thay \(x=a^2+6a\)vào biểu thức trên ta được:

\(\sqrt{\left(a^2+6a+9\right)\left(a^2+6a+2\right)^2}=\sqrt{\left(a+3\right)^2\left(a^2+6a+2\right)^2}=\left(a+3\right)\left(a^2+6a+2\right)\)

\(\rightarrowđpcm\)

Câu hỏi của Kunzy Nguyễn - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo bài tương tự tại đây nhé.

là 3 

ko phải mình chứ

casi hử

mk làm đc rồi , m.n ko phải giải nữa

o de vay ma

\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)

\(=\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)\left(x+y\right)^2}\)

\(=\frac{10y}{15\left(x+y\right)^2}\)

\(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)

\(=\frac{\left(x^2-x\right)-\left(xy-y\right)}{\left(x^2-x\right)+\left(xy-y\right)}\)

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

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

\(=\frac{x-y}{x+y}\)

a)\(\frac{2xy}{3\left(x+y\right)^2}\)

b)=\(\frac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)

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

\(x^2+x+1=\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0voimoix\)

Ta có

a + b + c = abc

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Ta lại có

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)

Ta có:a+b+c=abc

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Ta lại có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)

jup mk nhanh vs đk ko mk đag cần gấp lắm

\(x^2+3x+3y+xy\)

\(=\left(x^2+xy\right)+\left(3x+3y\right)\)

\(=x\left(x+y\right)+3\left(x+y\right)\)

\(=\left(x+3\right)\left(x+y\right)\)

\(b,x^3+5x^2+6x\)

\(=x^3+3x^2+2x^2+6x\)

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

\(=x^2\left(x+3\right)+2x\left(x+3\right)\)

\(=\left(x^2+2x\right)\left(x+3\right)\)

\(=x\left(x+2\right)\left(x+3\right)\)

Xét \(\Delta ABC:\)N là trung điểm AC, P là trung điểm BC

\(\Rightarrow NP\)là đường trung bình \(\Delta ABC\)

\(\Rightarrow NP\text{//}AB\)

\(\Rightarrow PQ\text{//}AM\)( Vì \(M\in AB;N\in PQ\))

\(\Rightarrow\)Tứ giác PMAQ là hình thang 

Vậy...

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