Phân thức đại số

a: a không chia hết cho 3 nên a=3k+1 hoặc a=3k+2

TH1: a=3k+1

\(a^2-1=\left(3k+1\right)^2-1\)

\(=9k^2+6k+1-1=9k^2+6k=3\left(3k^2+2k\right)⋮3\)

=>\(a^2\) chia 3 dư 1

TH2: a=3k+2

\(a^2-1=\left(3k+2\right)^2-1\)

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

\(=9k^2+12k+3\)

\(=3\left(3k^2+4k+1\right)⋮3\)

=>a² chia 3 dư 1

b: \(x^2+y^2-6x+10y+34=0\)

=>\(x^2-6x+9+y^2+10y+25=0\)

=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)

mà \(\left\{{}\begin{matrix}\left(x-3\right)^2>=0\forall x\\\left(y+5\right)^2>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)

c: \(A=n^3+3n^2+2n\)

\(=n\cdot\left(n^2+3n+2\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

Vì n;n+1;n+2 là ba số nguyên liên tiếp

nên \(n\left(n+1\right)\left(n+2\right)⋮3!=6\)

=>A chia hết cho 6

d: xy-5x=7

=>x(y-5)=7

=>\(\left(x;y-5\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;12\right);\left(7;6\right);\left(-1;-2\right);\left(-7;4\right)\right\}\)

1) 

a) (x + 2)² - (x² - 6x + 9)

= (x + 2)² - (x - 3)²

= [(x + 2) + (x - 3)][(x + 2) - (x - 3)]

= (x + 2 + x - 3)(x + 2 - x + 3)

= 2x - 1).5

= 5(2x - 1)

b) x⁴ - 2x³y + x²y² - 9x²

= x²(x² - 2xy + y² - 9)

= x²[(x² - 2xy + y²) - 3²]

= x²[(x - 1)² - 3²]

= x²(x - 1 - 3)(x - 1 + 3)

= x²(x - 4)(x + 2)

c) x² - 4x + 4 - y²

= (x² - 4x + 4) - y²

= (x - 2)² - y²

= (x - 2 - y)(x - 2 + y)

= (x - y - 2)(x + y - 2)

d) x² - 6xy + 9y² + xz - 3yz

= (x² - 6xy + 9y²) + (xz - 3yz)

= (x - 3y)² + z(x - 3y)

= (x - 3y)(x - 3y + z)

e) Sửa đề:

x³ - 3x² + 3x - 1 - y³

= (x³ - 3x² + 3x - 1) - y³

= (x - 1)³ - y³

= (x - 1 - y)[(x - 1)² + (x - 1)y + y²]

= (x - y - 1)(x² - 2x + 1 + xy - y + y²)

g) x² - 3x - 4

= x² - 4x + x - 4

= (x² - 4x) + (x - 4)

= x(x - 4) + (x - 4)

= (x - 4)(x + 1)

2)

a) (27x⁵y⁴ - 3x⁴y³ + 18x³y²) : 3x³y²

= 27x⁵y⁴ : 3x³y² - 3x⁴y³ : 3x³y² + 18x³y² : 3x³y²

= 9x²y² - xy + 6

b) (-5x⁸y⁴ + 24x⁶y⁵ + 7x⁵y³) : (-7x⁵y²)

= [-5x⁸y⁴ : (-7x⁵y²)] - 24x⁶y⁵ : 7x⁵y² - 7x⁵y³ : 7x⁵y²

= 5/7 x³y² - 24/7 xy³ - y

c) 4x²y(3x - 2y²) - 3x²y(4x - y²)

= 12x³y - 8x²y³ - 12x³y + 3x²y³

= (2x³y - 12x³y) + (-8x²y³ + 3x²y³)

= -5x²y³

d) (x + y)² + (x - y)² - 2(x² - 3) + 2y(x - y)

= x² + 2xy + y² + x² - 2xy + y² - 2x² + 6 + 2xy - 2y²

= (x² + x² - 2x²) + (2xy - 2xy + 2xy) + (y² + y² - 2y²) + 6

= 2xy + 6

Bài 3

a) (x² - 1)(x + 1) - x²(x + 2) + x² = 21

x³ + x² - x - 1 - x³ - 2x² + x² = 21

-x + 1 = 21

x = 1 - 21

x = -20

b) (2x + 3)² - (3x - 2)² = 0

(2x + 3 - 3x + 2)(2x + 3 + 3x - 2) = 0

(-x + 5)(5x + 1) = 0

-x + 5 = 0 hoặc 5x + 1 = 0

*) -x + 5 = 0

x = 5

*) 5x + 1 = 0

5x = -1

x = -1/5

Vậy x = -1/5; x = 5

3:

a: \(\left(x^2-1\right)\left(x+1\right)-x^2\left(x+2\right)+x^2=21\)

=>\(x^3+x^2-x-1-x^3-2x^2+x^2=21\)

=>-x-1=21

=>-x=22

=>x=-22

b: \(\left(2x+3\right)^2-\left(3x-2\right)^2=0\)

=>\(\left(2x+3+3x-2\right)\left(2x+3-3x+2\right)=0\)

=>\(\left(5x+1\right)\left(5-x\right)=0\)

=>\(\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=5\end{matrix}\right.\)

4:

\(A=-x^2+x-1\)

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

\(=-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\)

Dấu = xảy ra khi x-1/2=0

=>x=1/2

Bài 4

A = x - x² - 1

= -(x² - x + 1)

= -(x² - 2.x.1/2 + 1/4 + 3/4)

= -(x - 1/2)² - 3/4

Do (x - 1/2)² ≥ 0 với mọi x ∈ R

⇒ -(x - 1/2)² ≤ 0 với mọi x ∈ R

⇒ -(x - 1/2)² - 3/4 ≤ -3/4 với mọi x  ∈ R

Vậy GTLN của A là -3/4 khi x = 1/2

Bài 3: 

a) \(6x^2-\left(2x-3\right)\left(3x+2\right)-1=0\)

\(\Leftrightarrow6x^2-\left(6x^2+4x-9x-6\right)-1=0\)

\(\Leftrightarrow6x^2-6x^2+5x+6-1=0\)

\(\Leftrightarrow5x+5=0\)

\(\Leftrightarrow5x=-5\)

\(\Leftrightarrow x=-1\)

b) \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Bài 4:

Ta có: 

\(x+y=4\)

\(\Rightarrow\left(x+y\right)^2=4^2\)

\(\Rightarrow x^2+y^2+2xy=16\)

\(\Rightarrow10+2xy=16\)

\(\Rightarrow2xy=16-10\)

\(\Rightarrow2xy=6\)

\(\Rightarrow xy=\dfrac{6}{2}\)

\(\Rightarrow xy=3\)

Từ đó ta tính được M:
\(M=x^3+y^3\)

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

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

\(M=4\cdot\left(10-3\right)\)

\(M=4\cdot7\)

\(M=28\)

Bài 2

a) 18x³ - 12x² + 2x

= 2x(9x² - 6x + 1)

= 2x(3x - 1)²

b) 16 - 4x² + 12xy - 9y²

= 16 - (4x² - 12xy + 9y²)

= 4² - (2x - 3y)²

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

Bài 1

a) A = (3x - 2)(2x - 1) - x(6x - 4)

= 6x² - 3x - 4x + 2 - 6x² + 4x

= (6x² - 6x²) + (-3x - 4x + 4x) + 2

= -3x + 2

b) B = (x - 1)³ - 2x(x - 1)(x + 1) + (x - 1)(x² + x + 1)

= x³ - 3x² + 3x - 1 - 2x³ + 2x + x³ - 1

= (x³ - 2x³ + x³) - 3x² + (3x + 2x) + (-1 - 1)

= -3x² + 5x - 2

2:

a: \(4x\left(x-2y\right)-8y\cdot\left(x-2y\right)\)

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

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

b: \(-25x^2y^2+10xy-1\)

\(=-\left(25x^2y^2-10xy+1\right)\)

\(=-\left(5xy-1\right)^2\)

c: \(x^2+x^3+xy-y^3+y^2\)

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

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

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

5:

\(a^2+b^2+c^2=ab+bc+ac\)

=>\(2a^2+2b^2+2c^2=2ab+2ac+2bc\)

=>\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

=>a=b=c

Bài `3`

\(a,B=\dfrac{2}{x^2-5x+6}:\dfrac{x+2}{x-2}\\ =\dfrac{2}{x^2-2x-3x+6}\cdot\dfrac{x-2}{x+2}\\ =\dfrac{2}{x\left(x-2\right)-3\left(x-2\right)}\cdot\dfrac{x-2}{x+2}\\ =\dfrac{2}{\left(x-2\right)\left(x-3\right)}\cdot\dfrac{x-2}{x+2}\\ =\dfrac{2}{\left(x-3\right)\left(x+2\right)}\)

`b,` Khi `x=5` ta có :

\(\dfrac{2}{\left(x-3\right)\left(x+2\right)}\Rightarrow\dfrac{2}{\left(5-3\right)\left(5+2\right)}=\dfrac{2}{2\cdot7}=\dfrac{2}{14}=\dfrac{1}{7}\)

3:

a: \(B=\dfrac{2}{x^2-5x+6}:\dfrac{x+2}{x-2}\)

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

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

b: khi x=5 thì \(B=\dfrac{2}{\left(5+2\right)\left(5-3\right)}=\dfrac{1}{7}\)

Bài 5

A lớn nhất khi B = x² - 2x + 5 nhỏ nhất

Ta có:

B = x² - 2x + 5

= x² - 2x + 1 + 4

= (x - 1)² + 4

Do (x - 1)² ≥ 0 với mọi x ∈ R

⇒ (x - 1)² + 4 ≥ 4 với mọi x ∈ R

⇒ B nhỏ nhất là 4 khi x = 1

⇒ A lớn nhất là:

4/4 = 1 khi x = 1

Bài 4

a) Diện tích đáy của khối Rubic:

44,002 . 3 : 5,88 = 22,45 (cm²)

b) Chiều cao của hình chóp tam giác đều:

12√3 . 3 : 9√3= 4 (cm)

Bài 2

a) 2x³ + 6x² - 4x

= 2x(x² + 3x - 2)

b) (2x + 5)² - 9x²

= (2x + 5 - 3x)(2x + 5 + 3x)

= (5 - 3x)(5x + 5)

= 5(5 - 3x)(x + 1)

c) 4x² - 9y² + 4x - 6y

= (4x² - 9y²) + (4x - 6y)

= (2x - 3y)(2x + 3y) + 2(2x - 3y)

= (2x - 3y)(2x + 3y + 2)

Bài 3:

a) ĐKXĐ: \(x\ne\pm6;x\ne0;x\ne3\)  

b) \(A=\left(\dfrac{x}{x^2-36}+\dfrac{6-x}{6x+x^2}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)

\(A=\left[\dfrac{x}{\left(x+6\right)\left(x-6\right)}+\dfrac{6-x}{x\left(x+6\right)}\right]:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\left[\dfrac{x^2}{x\left(x+6\right)\left(x-6\right)}+\dfrac{\left(6-x\right)\left(x-6\right)}{x\left(x+6\right)\left(x-6\right)}\right]\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{x^2-\left(x^2-36\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{x^2-x^2+36}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{36}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{36}{\left(x-6\right)\cdot2\left(x-3\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{18}{\left(x-6\right)\left(x-3\right)}-\dfrac{x\left(x-3\right)}{\left(x-6\right)\left(x-3\right)}\)

\(A=\dfrac{18-x^2+3x}{\left(x-3\right)\left(x-6\right)}\)

\(A=\dfrac{-\left(x-6\right)\left(x+3\right)}{\left(x-3\right)\left(x-6\right)}\)

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

Câu 5.

\(4x^2+12x+9=0\\\Leftrightarrow (2x)^2+2\cdot2x\cdot3+3^2=0\\\Leftrightarrow (2x+3)^2=0\\\Leftrightarrow 2x+3=0\\\Leftrightarrow 2x=-3\\\Leftrightarrow x=\dfrac{-3}{2}\\\Rightarrow Chọn.B\)

Câu 7.

\(\dfrac{2+x}{3x^2y}+\dfrac{1-y}{3xy^2}\)

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

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

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

Câu 8.

Bạn chụp lại đề nhé, đề hơi mờ.

\(Toru\)