a) A= (5x-2).(x+1)-(x-3).(5x+1)-17(x+3)

=> A= 5x²+5x-2x-2-5x²-x+15x+3-17x-51

=> A= -50

b) B= (6x-5) × ( x+8) - (3x-1) × (2x+3) - 9(4x-3)

=> B= 6x²+48x-5x-40-6x²-9x+2x+3-36x+27

=> B= -10

c) C = x(x³ + x² - 3x -2 ) - ( x² -2 ) × ( x²+x -1 )

=> C= x⁴+x³-3x²-2x-x⁴+x³+3x²-2x-2

=> C= 2x³-4x-2

Bn ơi mik thấy câu c) của bạn sai ở chỗ x^4+ x^3+3x^2 ý 

a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)

b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)

\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)

\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)

c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)

d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

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

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

Tìm x:

a,3(x-5)-x+5=0

=>3(x-5)-(x-5)=0

=>(x-5)(3-1)=0

=>(x-5).2=0

=>x-5=0

=>x=5

Vậy x=5.

a: \(=12x^2-9x-12x^2-10x+6x+5=-13x+5\)

b: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x\)

\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)

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

c: \(=x^3-3x^2+3x-1+x^3+8+3\left(x^2-16\right)\)

\(=2x^3-3x^2+3x+7+3x^2-48=2x^3+3x-41\)

d: \(=\left(x^3+1\right)\left(x^3-1\right)=x^6-1\)

a/ \(\dfrac{x^3}{x^2+1975}\cdot\dfrac{2x+1954}{x+1}+\dfrac{x^3}{x^2+1975}\cdot\dfrac{21-x}{x+1}=\dfrac{x^3\left(2x+1954\right)+x^3\left(21-x\right)}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{2x^4+1954x^3+21x^3-x^4}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{x^4+1975x^3}{\left(x^2+1975\right)\left(x+1\right)}\)

b/ \(\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}+\dfrac{19x+8}{x^2+1945}\cdot\dfrac{x-2}{x-7}=\dfrac{\left(19x+8\right)\left(5x-9\right)+\left(19x+8\right)\left(x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{\left(19x+8\right)\left(5x-9+x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-209x+40x-88}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-169x-88}{x^2+1938x-13615}\)

c/ \(\dfrac{x+1}{x^2-2x-8}\cdot\dfrac{4-x}{x^2+x}=\dfrac{\left(x+1\right)\left(4-x\right)}{x\left[x^2-4x+2x-8\right]\left(x+1\right)}=-\dfrac{x-4}{x\left(x-4\right)+2\left(x-4\right)}=-\dfrac{x-4}{\left(x-4\right)\left(x+2\right)}=-\dfrac{1}{x+2}\)

Bài 3: Tìm x, biết:

a) \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)

\(\Leftrightarrow40x-40=0\)

\(\Leftrightarrow4x=40\)

\(\Leftrightarrow x=10\)

Vậy x = 10

b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow\left(2x+3\right)^2-4\left(x^2-1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)

\(\Leftrightarrow12x-36=0\)

\(\Leftrightarrow12x=36\)

\(\Leftrightarrow x=3\)

Vậy x = 3

c) \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18\)

\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=18\)

\(\Leftrightarrow2\left(1-2x\right)=18\)

\(\Leftrightarrow2-4x=18\)

\(\Leftrightarrow4x=-16\)

\(\Leftrightarrow x=-4\)

Vậy x =-4

d) \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(\Leftrightarrow12x-5=0\)

\(\Leftrightarrow12x=5\)

\(\Leftrightarrow x=\frac{5}{12}\)

Vậy \(x=\frac{5}{12}\)

e) \(\left(x-5\right)^2-x\left(x-4\right)=9\)

\(\Leftrightarrow x^2-10x+25-x^2+4x=9\)

\(\Leftrightarrow25-6x=9\)

\(\Leftrightarrow6x=16\)

\(\Leftrightarrow x=\frac{8}{3}\)

Vậy \(x=\frac{8}{3}\)

f) \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)

\(\Leftrightarrow21-5x=0\)

\(\Leftrightarrow5x=21\)

\(\Leftrightarrow x=\frac{21}{5}\)

Vậy \(x=\frac{21}{5}\)

bài của bạn làm sai rồi :)