Tham khảo

a)

-7x²(3x - 4y)

= -7x².3x + 7x^2ư.4y

= -21x² + 28x²y

b)

(x - 3)(5x - 4)

= x.5x - x.4 - 3.5x + 3.4

= 5x² - 4x - 15x + 12

= 5x² - 19x + 12

c)

(2x - 1)² = 4x² - 4x + 1

d)

(x + 3)(x - 3) = x² - 3² = x² - 9

a, = -21x³ +28x²y

b, = 5x² -4x -15x +12

= 5x² -19x +12

c, =x² -9

a)-21x³+28x²y

b)5x²-19x+12

c)4x²-4x+1

d)x²-9

\(a,=-21x^3+28x^2y\\ b,=5x^2-4x-15x+12=5x^2-19x+12\\ c,=4x^2-4x+1\\ d,=49-x^2\)

\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

\(a,=10x^3-5x^2+5x\\ b,=x^3+27\\ c,=\dfrac{5}{2}xy-1-\dfrac{1}{2}y\\ d,=\left(2x^3-10x^2-11x^2+55x+12x-60\right):\left(x-5\right)\\ =\left[2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\right]:\left(x-5\right)\\ =2x^2-11x+12\)

a: \(=10x^3-5x^2+5x\)

b: \(=x^3-27\)

a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))

\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)

\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)

\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)

\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)

\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)

b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x^2}{x\left(x+3\right)}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)

\(=1+\dfrac{3}{x}\)

\(=\dfrac{x}{x}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x}\)

a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)

\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)

b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)

a: \(=x^2-1-x^2-x+6=-x+5\)

1.

\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)

2. 

a, ĐK: \(x\in R\)

\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)

\(\Leftrightarrow\left|x-2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b, ĐK: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)

a, A(x)= (5.1)³ - (3.1)+4

           = 125-7

           =118

a) 

 tại\(x = 1 , GTBT A(x)\) là:

\(5.1 ^3 − 3.1 + 4\)

\(= 5.1 − 3.1 + 4\)

\(= 5 − 3 + 4\)

\(= 2 + 4\)

\(=6\)

Vậy tại\(x = 1 , GTBT A ( x ) là 6\)

B(y)= 4.5-7

      =13