a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)

\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)

\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)

\(=2a^2-b^2\)

b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)

\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)

\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)

\(=-7ab+b^2\)

c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)

\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)

\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)

\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)

\(=-7bx+3b^2+2x^2\)

d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)

\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)

\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)

\(=-5ax+32x^2-30a^2\)

a: =2ab+8a^2-b^2-4ab+2ab-6a^2

=2a^2-b^2

b: =6a^2-9ab-4ab+6b^2-6a^2+6ab

=-7ab+6b^2

c: =10bx-5b^2-16bx+8b^2+2x^2-xb

=3b^2+2x^2-7xb

d: =2xa+30x^2+5ax+2x^2-30a^2-12ax

=32x^2-30a^2-5ax

a: \(x^2+6xy+9y^2=\left(x+3y\right)^2\)

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

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

b: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

\(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)

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

Mn giúp mik vs ạ

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

\(\frac{8a^2}{4a^2x-x^3}=\frac{8a^2}{x\left(4a^2-x^2\right)}=\frac{8a^2}{x\left(2a-x\right)\left(2a+x\right)}\)

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

`@` `\text {Ans}`

`\downarrow`

`a,`

`P(x)+Q(x) = (3x^4-2x^3+3x+11)+(3x^2- x^3-5x+3x+4-x+2x^4)`

`= 3x^4-2x^3+3x+11+3x^2- x^3-5x+3x+4-x+2x^4`

`= (3x^4 + 2x^4) + (-2x^3 - x^3) + 3x^2 + (3x + 3x - 5x - x) + (11+4)`

`= 5x^4 - 3x^3 + 3x^2 + 15`

`b,`

` A(x) = P(x) + B(x)`

Thay `B(x) = 2x^3 - 3x^4 - 2`

`A(x) = P(x) + B (x)`

`=> A (x) = (2x^3 - 3x^4 - 2)+(3x^4 - 2x^3 + 3x + 11)`

`= 2x^3 - 3x^4 - 2+ 3x^4 - 2x^3 + 3x + 11`

`= (2x^3 - 2x^3) + (-3x^4 + 3x^4) + 3x + (-2+11) `

`= 3x + 9`

`A(x) = 3x+9 = 0`

`=> 3x = 0-9`

`=> 3x = -9`

`=> x = -9 \div 3`

`=> x = -3`

Vậy, nghiệm của đa thức là `x = -3.`

Bài 1:

Tao có:

\(81^7mod\left(405\right)\)

\(81^3\equiv81mod\left(405\right)\)

\(81^6\equiv81^2\equiv81mod\left(405\right)\)

\(81^7\equiv81^2.81\equiv81mod\left(405\right)\)

Ta có:

\(27^9mod\left(405\right)\)

\(27^3\equiv243mod\left(405\right)\)

\(27^9\equiv243^3\equiv162mod\left(405\right)\)

Ta có:

\(9^{13}mod\left(405\right)\)

\(9\equiv9mod\left(405\right)\)

\(9^3\equiv324mod\left(405\right)\)

\(9^9\equiv324^3\equiv324mod\left(405\right)\)

\(9^{10}\equiv324.9\equiv81mod\left(405\right)\)

\(9^{13}\equiv81.324\equiv324mod\left(405\right)\)

\(81^7+27^9-9^{13}:405=81+162-324:405=-0,2\)

\(\Rightarrow81^7+27^9-9^{13}⋮405\left(đpcm\right)\)

Casio không biết có áp dụng ntn vào bài này được không nữa? Nhưng mình ôn hổm rày thấy có bài gần giống vậy, nên mình làm thử bạn tham khảo nha chúc bạn học tốt! ^^

\(A-B=\left(3x^4+x^3-x^2+3x-5\right)-\left(7x^4-x^3+2x^2-x+4\right)\)
\(=3x^4+x^3-x^2+3x-5-7x^4+x^3-2x^2+x-4\)
\(=\left(3x^4-7x^4\right)+\left(x^3+x^3\right)-\left(x^2+2x^2\right)+\left(3x+x\right)-\left(5+4\right)\)
\(=-4x^4+2x^3-3x^2+4x-9\)

`@` `\text {Ans}`

`\downarrow`

`a,`

`P(x)+Q(x) = (3x^4-2x^3+3x+11)+(3x^2- x^3-5x+3x+4-x+2x^4)`

`= 3x^4-2x^3+3x+11+3x^2- x^3-5x+3x+4-x+2x^4`

`= (3x^4 + 2x^4) + (-2x^3 - x^3) + 3x^2 + (3x + 3x - 5x - x) + (11+4)`

`= 5x^4 - 3x^3 + 3x^2 + 15`

`b,`

` A(x) = P(x) + B(x)`

`=> A(x) - B(x) = P(x)`

`=> A(x) - B(x) = 3x^4-2x^3+3x+11`

Bạn xem lại đề ;-;.

\(a,P\left(x\right)+Q\left(x\right)=\left(3x^4-2x^3+3x+11\right)+\left(3x^2-x^3-5x+3x+4-x+2x^4\right)\\ =\left(3x^4-2x^3+3x+11\right)+\left(2x^4-x^3+3x^2-3x+4\right)\\ =3x^4-2x^3+3x+11+2x^4-x^3+3x^2-3x+4\\ =\left(3x^4+2x^4\right)+\left(-2x^3-x^3\right)+\left(3x-3x\right)+\left(11+4\right)\\ =5x^4-3x^3+15\)

`B(x)` đâu cậu nhỉ?

\(1,\\ a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x+1}=\dfrac{2}{3}\Leftrightarrow2x+1=\dfrac{4}{9}\Leftrightarrow x=-\dfrac{5}{18}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=2\Leftrightarrow\left[{}\begin{matrix}x-3=2\\3-x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\ 2,\\ a,=\left|5-x\right|=x-5\\ b,=\sqrt{4a\cdot44a}=\sqrt{176a^2}=4\left|a\right|\sqrt{11}=4a\sqrt{11}\\ c,=\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)