an có 10000000 quả cam an cho mẹ gấp đôi rồi an co ba số quả lớn hơn mẹ 200 vậy an còn bao nhiêu quả cam

a) \(\frac{5x-1}{3x^2y}+\frac{x-1}{3x^2y}=\frac{5x-1+x-1}{3x^2y}=\frac{6x}{3x^2y}=\frac{2}{xy}\)

b) \(\frac{7}{12xy^2}+\frac{11}{18x^3y}=\frac{7\left(\frac{3}{2}x^2\right)}{18x^3y^2}+\frac{11y}{18x^3y^2}=\frac{10,5x^2+11y}{18x^3y^2}\)

c) \(\frac{x}{x+2}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{x\left(4x-7\right)}{\left(x+2\right)\left(4x-7\right)}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\frac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}\)

a) \(\frac{5x-1}{3x^2y}+\frac{x+1}{3x^2y}=\frac{5x-1+x+1}{3x^2y}=\frac{6x}{3x^2y}=\frac{2}{xy}\)

b) \(\frac{7}{12xy^2}+\frac{11}{18x^3y}=\frac{7x^2.18+11.12y}{12x^3y^2.18}=\frac{126x^2+132y}{216x^3y^2}=\frac{6\left(21x^2+22y\right)}{216x^3y^2}=\frac{21x^2+22y}{36x^3y^2}\)

c) \(\frac{x}{x+2}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\frac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}=\frac{4\left(x^2-4\right)}{\left(x+2\right)\left(4x-7\right)}=\frac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}=\frac{4\left(x-2\right)}{4x-7}\)

\(\dfrac{5x-1+x+1}{3x^2y}=\dfrac{6x}{3x^2y}=\dfrac{2}{xy}\)

\(\dfrac{21x^2+22y}{36x^3y^2}\)

\(\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x-8}{4x-7}=1-\dfrac{1}{4x-7}\)

bạn ơi đề là như nào vậy

\(\frac{7}{12}\)xy² + \(\frac{11}{18}\)x³y là giải phương trình hay tìm nghiệm của phương trình vậy bn

\(\frac{7}{12xy^2}+\frac{11}{18x^3y}=\frac{21x^2}{36x^3y^2}+\frac{22y}{36x^3y^2}=\frac{21x^2+22y}{36x^3y^2}\)

Bài 1: Thực hiện phép tính.

a) \(\left(x+2y\right)\left(x-2y\right)-5-x^2=x^2-4y^2-5-x^2=-4y^2-5\)

Bài 2: Phân tích đa thức thành nhân tử.

a) \(14x^3y^3-7x^2y+21x^2y^5=7x^2y\left(2xy^2-1+3y^4\right)\)

b) \(18x\left(1-x\right)-12y+12xy=18x\left(1-x\right)-12y\left(1-x\right)=6\left(1-x\right)\left(3x-2y\right)\)

c) \(9x^2-y^2+1-6x=\left(9x^2-6x+1\right)-y^2=\left(3x-1\right)^2-y^2=\left(3x-1-y\right)\left(3x-1+y\right)\)

a, bậc 6 

b, bậc 6 

c, bậc 12 

d, bậc 9 

e, bậc 8 

Bài 1.

a) ( 7x - 3 )² - 5x( 9x + 2 ) - 4x² = 18

<=> 49x² - 42x + 9 - 45x² - 10x - 4x² = 18

<=> -52x + 9 = 18

<=> -52x = 9

<=> x = -9/52 

b) ( x - 7 )² - 9( x + 4 )² = 0

<=> x² - 14x + 49 - 9( x² + 8x + 16 ) = 0

<=> x² - 14x + 49 - 9x² - 72x - 144 = 0

<=> -8x² - 86x - 95 = 0 

<=> -8x² - 10x - 76x - 95 = 0

<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0

<=> ( x + 5/4 )( -8x - 76 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)

c) ( 2x + 1 )² + ( 4x - 1 )( x + 5 ) = 36

<=> 4x² + 4x + 1 + 4x² + 19x - 5 = 36

<=> 8x² + 23x - 4 - 36 = 0

<=> 8x² + 23x - 40 = 0

=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))

Bài 2.

a) x² - 12x + 39 = ( x² - 12x + 36 ) + 3 = ( x - 6 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) 17 - 8x + x² = ( x² - 8x + 16 ) + 1 = ( x - 4 )² + 1 ≥ 1 > 0 ∀ x ( đpcm )

c) -x² + 6x - 11 = -( x² - 6x + 9 ) - 2 = -( x - 3 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

d) -x² + 18x - 83 = -( x² - 18x + 81 ) - 2 = -( x - 9 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

\(a,3x^2-6x\)

\(=3x\left(x-2\right)\)

\(b,18x^2-4x+12\)

\(=2\left(9x^2-2x+6\right)\)

\(c,4x^2\left(2x-y\right)-12x\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2-12x\right)\)

\(=4x\left(2x-y\right)\left(x-3\right)\)

\(d,7\left(x-3y\right)-2y\left(3y-x\right)\)

\(=7\left(x-3y\right)+2y\left(x-3y\right)\)

\(=\left(x-3y\right)\left(2y+7\right)\)

\(f,6\left(x-2y\right)-3\left(2y-x\right)\)

\(=6\left(x-2y\right)+3\left(x-2y\right)\)

\(=\left(x-2y\right)\left(6+3\right)=9\left(x-2y\right)\)