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4/ a/ Ta có \(x^2-2xy+y^2+a^2=\left(x-y\right)^2+a^2\)

Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\a^2\ge0\end{cases}}\)=> \(\left(x-y\right)^2+a^2\ge0\)

=> \(x^2-2xy+y^2+a^2\ge0\)

Vậy \(x^2-2xy+y^2\)chỉ nhận những giá trị không âm.

b/ Ta có \(x^2+2xy+2y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x+y\right)^2+\left(y+1\right)^2\)

Mà \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)=> \(\left(x+y\right)^2+\left(y+1\right)^2\ge0\)

=> \(x^2+2xy+2y^2+2y+1\ge0\)

Vậy \(x^2+2xy+2y^2+2y+1\)chỉ nhận những giá trị không âm.

c/ Ta có \(9b^2-6b+4c^2+1=\left(3b-1\right)^2+4c^2\)

Mà \(\hept{\begin{cases}\left(3b-1\right)^2\ge0\\4c^2\ge0\end{cases}}\)=> \(\left(3b-1\right)^2+4c^2\ge0\)

=> \(9b^2-6b+4c^2+1\ge0\)

Vậy \(9b^2-6b+4c^2+1\)chỉ nhận những giá trị không âm.

d/ Ta có \(x^2+y^2+2x+6y+10=\left(x+1\right)^2+\left(y+3\right)^2\)

Mà \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)=> \(\left(x+1\right)^2+\left(y+3\right)^2\ge0\)

=> \(x^2+y^2+2x+6y+10\ge0\)

Vậy \(x^2+y^2+2x+6y+10\)chỉ nhận những giá trị không âm.

1/

a/ \(x^4-y^4=\left(x^2-y^2\right)\)

b/ \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

                                                  \(=2b\left[a^2+2ab+b^2-\left(a^2-b^2\right)+\left(a^2-2ab+b^2\right)\right]\)

                                                  \(=2b\left(a^2+b^2\right)\)

c/ \(\left(a^2+2ab+b^2\right)+\left(a+b\right)\)

= \(\left(a+b\right)^2+\left(a+b\right)\)

= \(\left(a+b\right)\left(a+b+1\right)\)

A = \(\frac{1}{2}.24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{32}-1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{64}-1\right)\)

\(2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

        \(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

       \(=5^{64}-1\)

=> \(A=\frac{5^{64}-1}{2}\)

Ta có; \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\Rightarrow\widehat{A}+\widehat{B}=220^o\)

Lại có: \(\left(\widehat{A}+\widehat{B}\right)+\left(\widehat{A}-\widehat{B}\right)=220^o+10^o\)

\(\Rightarrow2\widehat{A}=230^o\Rightarrow\widehat{A}=115^o\)

\(\Rightarrow B=115^o-10^o=105^o\)

Vậy..

a) x² - 2x + 5

= x² - x - x + 1 + 4

= (x² - x) - (x - 1) + 4

= x.(x-1) - (x-1) + 4

= (x-1)^2 + 4

Có: (x-1)^2 \(\ge\)0 => (x-1)^2 + 4\(\ge4\)

Dấu ''='' xảy ra khi x-1=0 => x = 1.

Vậy Min của x^2 - 2x + 5 bằng 4 khi x = 1

\(\left(a+b\right)^0=1\)

\(\left(a+b\right)^1=a+b\)

\(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

\(\left(a+b\right)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

\(\left(a+b\right)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)

Tổng quát:

\(\left(a+b\right)^n=C_0a^n+C_1a^{n-1}b+...+C_nb^n\)

Trong đó : C₀, C₁, ..., C_n là các hệ số trong tam giác cân Paxcan:

Chúc bn học tốt <3

\(\left(3x-5\right)2x+\left(3x-5\right)4=0\)

\(\left(3x-5\right)\left(2x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-5=0\\2x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-2\end{cases}}\)

Đây ạ bạn

Đây nhé bạn

đâu bạn

\(\left(2m^2-\frac{10}{2}\right)^2\)

\(=\left(2m^2\right)^2-2.2m^2.\frac{10}{2}+\left(\frac{10}{2}\right)^2\)

\(=4m^4-20m^2+25\)

Không rõ đề bài ....

a) Ta có \(a\left(b+1\right)+b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\Rightarrow2ab+a+b=a+b+ab+1\)

=> ab=1

b) Ta có \(2\left(a+1\right)\left(b+1\right)=\left(a+b\right)\left(a+b+2\right)\Leftrightarrow2ab+2a+2b+2=a^2+ab+2a+b^2+ab+2b\)

=> a^2+b^2=2

^_^

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