C

C

C

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)

\(=4x^2-12x+9-4x^2+1+6x-9\)

\(=-6x+1\)

c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)

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

=1

a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)

b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)

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

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

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

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

\(=6x^2y\)

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

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

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

    \(=3x^3+3.3x^2.y+3.3x.y^2+y^3\)\(-2x^2-y^2\)\(+x^3-3.x^2.3y+3.x.3y^2-y^3\)

    \(=\left(3x^3+x^3\right)\)\(+\left(9x^2y-9x^2y\right)\)\(+\left(9xy^2-9xy^2\right)\)\(+\left(y^3-y^3\right)\)\(-2x^2-y\)

   = \(4x^3-2x^2-y^2\)

Thay x=\(\dfrac{1}{3},y=-\dfrac{1}{3}\)

\(4.\left(\dfrac{1}{3}\right)^3-2.\left(\dfrac{1}{3}\right)^2-\left(\dfrac{-1}{3}\right)^2\)

=\(4.\dfrac{1}{27}-2.\dfrac{1}{9}-\dfrac{1}{9}=\dfrac{4}{27}-\dfrac{2}{9}-\dfrac{1}{9}=\dfrac{-5}{9}\)

a. \(y=\frac{2}{2x+3}\in Z\)

\(\Rightarrow2x+3\in\left\{-2;-1;1;2\right\}\)

\(\Rightarrow2x\in\left\{-5;-4;-2;-1\right\}\). Vì x thuộc Z

\(\Rightarrow x\in\left\{-2;-1\right\}\)

b. \(y=\frac{2x-1}{2x-3}=\frac{2x-3+2}{2x-3}=1+\frac{2}{2x-3}\)

Vì y thuộc Z nên 2 / 2x - 3 thuộc Z

\(\Rightarrow2x-3\in\left\{-2;-1;1;2\right\}\)

\(\Rightarrow2x\in\left\{1;2;4;5\right\}\). Vì x thuộc Z

\(\Rightarrow x\in\left\{1;2\right\}\)

c. \(y=\frac{2x^2-1}{2x-3}=\frac{x\left(2x-3\right)+2x-3-x+2}{2x-3}=x+1-\frac{x+2}{2x-3}\)

Vì y thuộc Z nên x thuộc Z ; x + 2 / 2x - 3 thuộc Z

=> 2x + 4 / 2x - 3 thuộc Z

=> 2x - 3 + 7 / 2x - 3 thuộc Z

=> 7 / 2x - 3 thuộc Z

\(\Rightarrow2x-3\in\left\{-7;-1;1;7\right\}\)

\(\Rightarrow2x\in\left\{-4;2;4;10\right\}\)

\(\Rightarrow x\in\left\{-2;1;2;5\right\}\) ( tm x thuộc Z )

d,e tương tự

lm hết hộ mik

\(log_{\sqrt{3}}\left(2x+y\right)-log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)=\left(4x^2+y^2+2xy+2\right)-3\left(2x+y\right)-2\)

\(\Leftrightarrow log_{\sqrt{3}}\left(2x+y\right)+2+3\left(2x+y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)

\(\Leftrightarrow log_{\sqrt{3}}\left(6x+3y\right)+\left(6x+3y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)

Xét hàm \(f\left(t\right)=log_{\sqrt{3}}t+t\) với \(t>0\)

\(f'\left(t\right)=\dfrac{1}{t.ln\sqrt{3}}+1>0\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow6x+3y=4x^2+y^2+2xy+2\)

\(\Leftrightarrow4x+y=\left(x+y-1\right)^2+1+3\left(x^2+1\right)-3\ge2\left(x+y-1\right)+6x-3\)

\(\Leftrightarrow4x+y\ge2\left(4x+y\right)-5\)

\(\Leftrightarrow4x+y\le5\)

\(\Rightarrow P=\dfrac{2x+y+6+\left(4x+y-5\right)}{2x+y+6}=1+\dfrac{4x+y-5}{2x+y+6}\le1\)

\(P_{max}=1\) khi \(x=y=1\)

\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)

\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)