b: =>x/23=1+3/4+4/7=65/28

=>x=23*65/28=1495/28

c: =>3/5:x=3/5-1/4-1/2=9/40

=>x=3/5:9/40=8/3

\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)

\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

1, \(2x^2+4x=2x\left(x+2\right)\)

2, \(15x^3+5x^2-10x=5x\left(3x^2+x-2\right)=5x\left(x-\dfrac{2}{3}\right)\left(x+1\right)\)

3) \(5x^2\left(x-2y\right)+15x\left(x-2y\right)=\left(5x^2+15x\right)\left(x-2y\right)=5x\left(x+3\right)\left(x-2y\right)\)

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

5) \(5x^2-10x=5x\left(x-2\right)\)

6) \(3x-6y=3\left(x-2y\right)\)

7) \(25x^2+5x^3+x^2y=x^2\left(25+5x+y\right)\)

8) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)

9) \(x\left(y-1\right)-y\left(y-1\right)=\left(x-1\right)\left(y-1\right)\)

10) \(10x\left(x-y\right)-8y\left(y-x\right)=\left(10x+8y\right)\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)

\(1,=2x\left(x+2\right)\\ 2,=5x\left(3x^2+x-2\right)\\ 3,=\left(x-2y\right)\left(5x^2+15x\right)=5x\left(x+3\right)\left(x-2y\right)\\ 4,=\left(x-y\right)\left(3-5x\right)\\ 5,=5x\left(x-2\right)\\ 6,=3\left(x-2y\right)\\ 7,=5x^2\left(5+x+y\right)\\ 8,=7xy\left(2x-3y+4xy\right)\\ 9,=\left(y-1\right)\left(x-y\right)\\ 10,=\left(x-y\right)\left(10x+8y\right)=2\left(5x+4y\right)\left(x-y\right)\)

- Định lí Bezout: Khi chia đa thức P(x) cho nhị thức \(x-a\) thì có số dư là \(P\left(a\right)\).

Áp dụng:

P(x) chia x+1 dư 4 \(\Rightarrow P\left(-1\right)=4\)

P(x) chia x+2 dư 1\(\Rightarrow P\left(-2\right)=1\)

Vì P(x) chia x²+3x+2 được thương là 5x² nên ta có:

\(P\left(x\right)=\left(x^2+3x+2\right).5x^2+ax+b\left(1\right)\) (a,b là hằng số).

Thay \(x=-1\) vào (1) ta được:

\(P\left(-1\right)=\left(1^2-3.1+2\right).5.1^2-a+b=-a+b\)

\(\Rightarrow b-a=4\left(\cdot\right)\)

Thay \(x=-2\) vào (1) ta được:

\(P\left(-2\right)=\left(2^2-3.2+2\right).5.2^2-a.2+b\)

\(\Rightarrow b-2a=1\left(\cdot\cdot\right)\)

Từ (*), (**) ta có hệ: \(\left\{{}\begin{matrix}b-a=4\\b-2a=1\end{matrix}\right.\)

Giải ra ta được \(\left\{{}\begin{matrix}a=3\\b=7\end{matrix}\right.\)

Vậy \(P\left(x\right)=\left(x^2+3x+2\right).5x^2+3x+7\)

Thay \(x=-10\) vào P(x) ta được:

\(P\left(-10\right)=\left(10^2-3.10+2\right).5.10^2-3.10+7=35977\)

Đáp án C.

\(1,\)\(\left(2x+3\right)^2=4x^2+12x+9\)

\(2,\)\(\left(3x+2y\right)^2=9x^2+12xy+4x^2\)

\(3,\)\(\left(3a-1\right)^2=9x^2-6x+1\)

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

\(5,\)\(\left(1-5a\right)^2=1-10a+25a^2\)

\(6,\)\(\left(x-4\right)^3=x^3-12a^2+48a-64.\)

\(7,\)\(\left(x^2-2y\right)^2=x^4-4x^2y-4y^2\)

\(8,\)\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)

\(9,\)\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^4-49\)

\(10,\)\(\left(x-1\right)\left(x^2+x+1\right)=x^3-1\)

\(11,\)\(\left(x^3-2\right)\left(x^6+2x^3+4\right)=x^9-8\)

\(12,\)\(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)

\(13,\)\(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)

1, ( 2x + 3 )² = 4x² + 12x + 9

2, ( 3x + 2y )² = 9x² +12xy + 4y²

3 ( 3a - 1 )² = 9a² - 6x + 1

4, ( a - 2 )² = a² - 4a + 4

5, ( 1 - 5a )² = 1 - 10a + 25a²

6,  ( x- 4 )³ = x³ - 12x² + 48x - 64

7, ( x² - 2y )² = x⁴ - 4x²y + 4y²

8, ( 5X² - 2 ).( 5X² + 2 ) = 25X² - 4

9, ( 2a² - 7 ).( 2a² + 7 ) = 4a⁴ - 49

10, ( x - 1 ).( x² + x + 1 ) = x³ - 1

các bạn ơi đừng làm tắt được kg nhờ các bạn đó