\(\left(x^2+1\right)\left(x^4+1\right)\left(x^{16}+x^8+1\right)=\frac{x^{24}-x+1}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=x^6-1\)

đề như này à bạn ???

sao có 2 dấu = vậy

Bạn viết cũng sai lun

Câu hỏi của Hattory Heiji - Toán lớp 8 - Học toán với OnlineMath

chào bạn. tôi nghĩ rằng bạn đủ thông minh để làm nên tích đi đã r tôi sẽ giúp @*

\(\left(2x-1\right)^2+3\ge3\Rightarrow A=\frac{5}{\left(2x-1\right)^2+3}\le\frac{5}{3}\)

\(\text{Dấu = xảy ra khi }2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

\(\text{Vậy Max}A=\frac{5}{3}\Leftrightarrow x=\frac{1}{2}\)

• GIẢI :

Ta có : \(\left(2x-1\right)^2\ge0\)

\(\Rightarrow(2x-1)^2+3\ge3\)

\(\Rightarrow\frac{1}{\left(2x-1\right)^2+3}\le\frac{1}{3}\)

\(\Rightarrow\frac{5}{\left(2x-1\right)^2+3}\le\frac{5}{3}\)

\(\Rightarrow\text{A}_{max}=\frac{5}{3}\).

Dấu "=" xảy ra khi : \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

Vậy \(\text{A}_{max}=\frac{5}{3}\) khi \(x=\frac{1}{2}\).

\(x^2+2\left(a+b\right)x+4ab\)

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

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

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

\(a,2x^2-xy-y^2=2x^2-2xy+xy-y^2.\)

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

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

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

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

\(c,x^2+2\left(a+b\right)x+4ab\)

\(=x^2+2\left(a+b\right)x+\left(a+b\right)^2+4ab-\left(a+b\right)^2\)

\(=\left(x+a+b\right)^2+4ab-a^2-2ab-b^2\)

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

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

\(=\left(x+a+b+a-b\right)\left(x+a+b-a+b\right)\)

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