mấy bạn giải cho mình nha sáng mai mình đi học rồi

\(x^4-3x^3+5x^2-9x+6\)

\(=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\)

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

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

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

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

a)\(-x^2-x+2\)

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

\(=-\left(x^2+x+\frac{1}{4}-\frac{7}{4}\right)\)

\(=-\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\le\frac{7}{4}.Với\forall x\in Z\)

Dấu "="  xảy ra khi 

\(x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy Max = 7/4 <=> x = -1/2

a, \(-\left(x^2+x\right)+2\)

=\(-\left(x^2+2\cdot\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}\right)+2\)

=\(-\left(x+\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

Dấu = xảy ra khi : \(x+\frac{1}{2}=0\)

=> \(x=-\frac{1}{2}\) 

Vậy Max=9/4 khi x=-1/2 

Các câu khác tương tự nhé.

Ta thấy b=111...1(30 chữ số 1) chia hết cho 3

Vì tổng b = 1+1+...+1(30 số hạng 1) = 30 chia hết cho 3

Lại có a = 111...1(31 chữ số 1) chia cho 3 dư 1

=> ab chia 3 dư 1 <=> ab-2 chia hết cho 3

Bài 2:

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

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

\(=a^3+b^3=VT\)  (đpcm)

b)  \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc\)\(+a^2c+b^2c+c^3-abc-bc^2-ac^2\)

\(=a^3+b^3+c^3-3abc\)

Bài 1:

\(N=\frac{x\left|x-2\right|}{x^2+8x-20}+12x-3\)

\(=\frac{x\left|x-2\right|}{\left(x-2\right)\left(x+10\right)}+12x-3\)

Nếu  \(x\ge2\)thì:     \(N=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3\)

                                      \(=\frac{x}{x+10}+12x+3\)  (lm tiếp nhé)

Nếu  \(x< 2\) thì:     \(N=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+10\right)}+12x-3\)

                                         \(=\frac{-x}{x+10}+12x-3\)  (lm tiếp nhé)

a)  \(A=x^2+2xy+y^2-4x-4y+1\)

\(=\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=3^2-4.3+1=-2\)

b)  \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2.7+37=100\)

c)  \(C=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2.5+10=25\)

a) \(A=x^2+2xy+y^2-4x-4v+1\)

\(=\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=3^2-4.3+1=-2\)

\(\left(\frac{3}{2}x+y\right)^3=\left(\frac{3}{2}x\right)^3+3.\left(\frac{3}{2}\right)^2.y+3.\frac{3}{2}.y^2+y^3\)

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

\(=\frac{27}{8}x^3+\frac{27}{4}x^2y+3xy^2+y^3\)

\(\frac{-x^6}{125}-\frac{y^3}{64}\)

\(=\frac{-\left(x^2\right)^3}{5^3}-\frac{y^3}{4^3}\)

\(=\left(\frac{-x^2}{5}\right)^3-\left(\frac{y}{4}\right)^3\)

\(=\left(\frac{-x^2}{5}-\frac{y}{4}\right)\cdot\left(\frac{x^4}{25}-\frac{x^2y}{20}+\frac{y^2}{16}\right)\)

Tham khảo nhé~

\(-\frac{x^6}{125}-\frac{y^3}{64}\)

\(=-\left(\frac{x^6}{125}+\frac{y^3}{64}\right)\)

\(=-\left(\frac{x^2}{5}+\frac{y}{4}\right)\left(\frac{x^4}{25}-\frac{x^2y}{20}+\frac{y^2}{16}\right)\)