A=a4−2a3+a2+a2−2a+1+1A=a4−2a3+a2+a2−2a+1+1

=a2(a2−2a+1)+a2−2a+1+1=a2(a2−2a+1)+a2−2a+1+1

=(a2+1)(a2−2a+1)+1=(a2+1)(a2−2a+1)+1

=(a2+1)(a−1)2+1≥1=(a2+1)(a−1)2+1≥1

Amin=1Amin=1 khi a=1

\(M=\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-a\right)}\)

Đánh giá đại diện: \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}-\frac{1}{a-c}\)

Tương tự: \(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}-\frac{1}{b-a}\)

                   \(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}-\frac{1}{c-b}\)

\(\Rightarrow M=\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}\)

\(\Rightarrow M=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\)

\(\Rightarrow M=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2N\left(đpcm\right)\)

fil in the correct form of the words in blackets( AS...........AS)

1. Jonh is ( tall)..........................Glen

2.Janet is ( beautiful) .............jenifer

3.You are (crazy ) .....................my sister 

4.We are run(fast)...........they can

5.my mom is (not/strict)..........your mom

6. Your mobile phone is ( not/ expensve ) ........................mine

Ai trả lời nhanh mình sẽ cho người đấy

Sửa đề :

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

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

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

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

????

Đề:........

<=> x². (2x + 7) - 16. (2x + 7) = 0

<=> (2x + 7). (x² - 16) = 0

<=> (2x+ 7). (x - 4). (x + 4) = 0

=> \(\hept{\begin{cases}2x+7=0\\x-4=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}2x=-7\\x=4\\x=-4\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{-7}{2}\\x=4\\x=-4\end{cases}}\)

Vậy...........

\(x^2\left(2x+7\right)=16\left(2x+7\right)\)

\(x^2\left(2x+7\right)-16\left(2x+7\right)=0\)

\(\left(2x+7\right)\left(x^2-16\right)=0\)

\(\left(2x+7\right)\left(x+4\right)\left(x-4\right)=0\)

\(\hept{\begin{cases}x=-\frac{7}{2}\\x=-4\\x=4\end{cases}}\)

a) \(x\left(x+4\right)-x^2-6x=10\)

\(\Leftrightarrow x^2+4x-x^2-6x=10\)

\(\Leftrightarrow-2x=10\)

\(\Leftrightarrow x=-5\)

Vậy \(x=-5\)

b) \(x\left(x-1\right)+2x-2=0\)

\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Vậy \(x=1\)hoặc \(x=-2\)

thanks nobita nha

a) Đặt \(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)

\(\Rightarrow A=\left(x+y\right)\left(x+4y\right)\left(x+2y\right)\left(x+3y\right)+y^4\)

\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

Đặt \(x^2+5xy+5y^2=t\)

\(\Rightarrow A=\left(t+y^2\right)\left(t-y^2\right)+y^4=t^2-y^4+y^4\)

         \(=t^2=\left(x^2+5xy+5y^2\right)^2\)là số chính phương ( đpcm )