`@` `\text {Ans}`

`\downarrow`

`a,`

`(2x - 3)^2`

`= 4x^2 - 12x + 9`

`b,`

`(x + 1)^2`

`= x^2 + 2x + 1`

`c,`

`(2x + 5)(2x - 5)`

`= 4x^2 - 25`

`d,`

`(a + b - c)(a - b + c)`

`= a^2 - b^2 + bc - c^2 + cb`

`e,`

\((x + 1)^2 - 10(x + 1) + 25\)

`= x^2 + 2x + 1 - 10x - 10 + 25`

`= x^2 - 8x +16`

`@` `\text {Kaizuu lv uuu}`

`@` CT:

Bình phương của `1` tổng: `(A + B)^2 = A^2 + 2AB + B^2`

Bình phương của `1` hiệu: `(A - B)^2 = A^2 - 2AB + B^2`

`A^2 - B^2 = (A-B)(A+B)`

a) A = (a - 2b + c) - (a - 2b - c)

       = a - 2b + c - a + 2b + c

       = (a - a) - (2b - 2b) + (c + c)

       = 2c

b) tương tự trên

c) C = 2(3a + b - 1) - 3(2a + b - 2)

       = 6a + 2b - 2 - 6a - 3b + 3

      = (6a - 6a) + (2b - 3b) - (2 - 3)

      = 0  - b + 1

      = -b + 1

d) D = 4(x - 1) - (3x + 2)

        = 4x - 4 - 3x - 2

       = (4x - 3x) - (4 + 2)

       = x - 6

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

1.

\(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c+2a+c}{2a+c}=\dfrac{a+b+c+2b}{2b}=\dfrac{a+b+c+b+c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}+1=\dfrac{a+b+c}{2b}+1=\dfrac{a+b+c}{b+c}+1\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}=\dfrac{a+b+c}{2b}=\dfrac{a+b+c}{b+c}\)

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)

TH2: \(a+b+c\ne0\)

\(\Rightarrow\dfrac{1}{2a+c}=\dfrac{1}{2b}=\dfrac{1}{b+c}\)

\(\Rightarrow\left\{{}\begin{matrix}2a+c=b+c\\2b=b+c\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a=b\\b=c\end{matrix}\right.\) \(\Rightarrow2a=b=c\)

\(\Rightarrow P=\dfrac{\left(a+2a\right)\left(2a+2a\right)\left(2a+a\right)}{a.2a.2a}=9\)

Bài 2 đề sai

Ở phân thức thứ 2 không thể là \(\dfrac{y+3x-x}{x}\)

Bài 2:

\(P=\dfrac{x+3y}{y}\cdot\dfrac{y+3z}{z}\cdot\dfrac{z+3x}{x}=\dfrac{\left(x+3y\right)\left(y+3z\right)\left(z+3x\right)}{xyz}\)

Với \(x+y+z=0\)

\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}\\ \Leftrightarrow\dfrac{x+3y+x+y}{z}=\dfrac{y+3z+y+z}{x}=\dfrac{z+3x+x+z}{y}\\ \Leftrightarrow\dfrac{2\left(x+2y\right)}{z}=\dfrac{2\left(y+2z\right)}{x}=\dfrac{2\left(z+2x\right)}{y}\\ \Leftrightarrow\dfrac{2\left(y-z\right)}{z}=\dfrac{2\left(z-x\right)}{x}=\dfrac{2\left(x-y\right)}{y}\\ \Leftrightarrow\dfrac{2y-2z}{z}=\dfrac{2z-2x}{x}=\dfrac{2x-2y}{y}\\ \Leftrightarrow\dfrac{2y}{z}-2=\dfrac{2z}{x}-2=\dfrac{2x}{y}-2\\ \Leftrightarrow\dfrac{2y}{z}=\dfrac{2z}{x}=\dfrac{2x}{y}\\ \Leftrightarrow\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x}{y}\Leftrightarrow x=y=z=0\left(\text{trái với GT}\right)\)

Với \(x+y+z\ne0\)

\(\Leftrightarrow\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x+3y-z=3z\\y+3z-x=3x\\z+3x-y=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=4z\\y+3z=4x\\z+3x=4y\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{4x\cdot4y\cdot4z}{xyz}=64\)

ta có : a) \(a.\left(-b\right)=-a.b=-\left(2,3\right)=-2,3\)

b) \(\left(-a\right).\left(-b\right)=a.b=2,3\)

c) \(a.\left(-2b\right)\Leftrightarrow-2ab=-2\left(2,3\right)=-4,6\)

d) \(\left(-3a\right).\left(2b\right)=-6ab=-6.\left(2,3\right)=-13,8\)

VT = (3a + 2b - 1)(a + 5) - 2b(a - 2)

= 3a² + 2ab - a + 15a + 10b - 5 - 2ab + 4b

= 3a² + 14a + 14b - 5 

= 3a² + 9a + 5a + 15 + 14b - 20

= 3a(a + 3) + 5(a + 3) + 2(7b - 10)

= (3a + 5)(a + 3) + 2(7b - 10)

= VP (đpcm)

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