\(a,x^2+2xy+y^2\)

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

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

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

\(=x\left(2x-1\right)+3\left(2x-1\right)\)

\(=\left(x+3\right)\left(2x-1\right)\)

a, Ta có: \(x^2+3x+2\)

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

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

\(=x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+2\right)\left(x+1\right)\)

\(b,2x^2-x+3\)

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

\(=-2x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(3-2x\right)\left(x+1\right)\)

Gọi giao điểm các đường phân giác trong tứ giác ABCD lần lượt là M, N, P, Q như hình vẽ bên trên.

Xét \(\Delta APB\) có: \(\widehat{APB}=180^o-\widehat{PAB}-\widehat{PBA}=\dfrac{360^o-\widehat{DAB}-\widehat{CBA}}{2}\)

Tương tự: xét \(\Delta MCD\) có:

\(\widehat{DMC}=\dfrac{360^o-\widehat{ADC}-\widehat{BCD}}{2}\)

Suy ra: \(\widehat{QMN}+\widehat{QBN}=\dfrac{360^o-\widehat{DAB}-\widehat{CBA}}{2}+\dfrac{360^o-\widehat{ADC}-\widehat{BCD}}{2}=\dfrac{720^o-360^o}{2}=180^o\)

Do tổng 4 góc trong một tứ giác bằng \(360^o\) nên ta cũng có \(\widehat{MQP}+\widehat{MNP}=360^o-180^o=180^o\)

Vậy tứ giác MNPQ có các góc đối bù nhau.

\(a,-x^2+8x+22\)

\(=-\left(x^2-8x-22\right)\)

\(=-\left(x^2-2.x.4+4^2-16-22\right)\)

\(=-\left(x-4\right)^2-38\)

Do \(-\left(x-4\right)^2\le0\) với mọi x (dấu "=" xảy ra \(\Leftrightarrow x=4\))

\(\Rightarrow-\left(x-4\right)^2-38\le-38\) hay ​​​\(-x^2+8x+22\le-38\)​ (dấu "=" xảy ra \(\Leftrightarrow x=4\))

Vậy giá trị lớn nhất của \(-x^2+8x+22\) là -38 tại x=4

\(b,9x^2-6x+16\)

\(=9\left(x^2-\dfrac{2}{3}x+\dfrac{16}{9}\right)\)

\(=9\left[x^2-2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2-\dfrac{1}{9}+\dfrac{16}{9}\right]\)

\(=9\left[\left(x-\dfrac{1}{3}\right)^2+\dfrac{5}{3}\right]\)

\(=9\left(x-\dfrac{1}{3}\right)^2+15\)

Do \(9\left(x-\dfrac{1}{3}\right)^2\ge0\) với mọi x (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{3}\))

\(\Rightarrow9\left(x-\dfrac{1}{3}\right)^2+15\ge15\) hay \(9x^2-6x+16\ge15\)(dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{3}\))

Vậy giá trị nhỏ nhất của \(9x^2-6x+16\) là 15 tại \(x=\dfrac{1}{3}\)

\(c,-2x^2-6x+8\)

\(=-\left(2x^2+6x-8\right)\)

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

\(=-\left\{2\left[x^2+2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\dfrac{9}{4}-4\right]\right\}\)

\(=-\left\{2\left[\left(x+\dfrac{3}{2}\right)^2-\dfrac{25}{4}\right]\right\}\)

\(=-\left[2\left(x+\dfrac{3}{2}\right)^2-\dfrac{25}{2}\right]\)

\(=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{25}{2}\)

Do \(-2\left(x+\dfrac{3}{2}\right)^2\le0\) với mọi x (dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{3}{2}\))

\(\Rightarrow-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\) hay \(-2x^2-6x+8\le\dfrac{25}{2}\) (dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{3}{2}\))

Vậy giá trị lớn nhất của \(-2x^2-6x+8\) là \(\dfrac{25}{2}\) tại \(x=-\dfrac{3}{2}\)

Ta có:\(x^3-3x^2+2\)

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

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

\(=x^2\left(x-1\right)-2\left(x^2-1\right)\)

\(=x^2\left(x-1\right)-2\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left[x^2-2.\left(x+1\right)\right]\)

\(=\left(x-1\right)\left(x^2-2x-2\right)\)

chúc bạn hok tốt !!!

Ta có:\(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)

\(=20^2+18^2+16^2+...+4^2+2^2-19^2-17^2-15^2-...-3^2-1^2\)

\(=(20^2-19^2)+(18^2-17^2)+...+(4^2-3^2)+(2^2-1^2)\)

\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)

\(=20+19+18+17+...+4+3+2+1\)

\(=\dfrac{\left(20+1\right).20}{2}=\dfrac{21.20}{2}=210\)