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Bài 1:
a) \(25\left(x+2y\right)^2-16\left(2x-y\right)^2\)
\(=\left[5\left(x+2y\right)\right]^2-\left[4\left(2x-y\right)\right]^2\)
\(=\left[5\left(x+2y\right)-4\left(2x-y\right)\right]\left[5\left(x+2y\right)+4\left(2x-y\right)\right]\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(14y-3x\right)\left(13x+6y\right)\)
b) \(0,25\left(x-2y\right)^2-4\left(x+y\right)^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)-2\left(x+y\right)\right]\left[\dfrac{1}{2}\left(x-2y\right)+2\left(x+y\right)\right]\)
\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)
\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)
\(=-3\left(\dfrac{1}{2}x+y\right)\left(\dfrac{5}{2}x+y\right)\)
c) \(\dfrac{4}{9}\left(x-3y\right)^2-0,04\left(x+y\right)^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)\right]^2-\left[\dfrac{1}{5}\left(x+y\right)\right]^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)-\dfrac{1}{5}\left(x+y\right)\right]\left[\dfrac{2}{3}\left(x-3y\right)+\dfrac{1}{5}\left(x+y\right)\right]\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)
\(=\dfrac{1}{5}\left(\dfrac{7}{3}x-11y\right).\dfrac{1}{5}\left(\dfrac{13}{3}x-9y\right)\)
\(=\dfrac{1}{25}\left(\dfrac{7}{3}x-11y\right)\left(\dfrac{13}{3}x-9y\right)\)
d) \(-25x^2+30x-9\)
\(=-\left(25x^2-30x+9\right)\)
\(=-\left[\left(5x\right)^2-2.5x.3+3^2\right]\)
\(=-\left(5x-3\right)^2\)
Bài 2:
a) \(x^3y^2-x^2y^3-2x+2y\)
\(=x^2y^2\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2y^2-2\right)\)
Thay x = -1 và y = -2 vào ta được
\(=\left[-1-\left(-2\right)\right]\left[\left(-1\right)^2\left(-2\right)^2-2\right]\)
\(=1\left(4-2\right)\)
\(=2\)
b) \(5x^2-3x+3y-5y^2\)
\(=5\left(x^2-y^2\right)-3\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
Thay x = 3 và y = 1 vào ta được
\(=5\left(3-1\right)\left(3+1\right)-3\left(3-1\right)\)
\(=5.2.4-3.2\)
\(=34\)
1. C. \(16x^2\left(x-y\right)\)\(-10y\left(y-1\right)\)\(=-2\left(y-x\right)\)\(\left(8x^2+5y\right)\)
2. C. \(\left(x-y\right)\left(x-y-3\right)\)
3. D. \(\left(x-2\right)\left(x+1\right)\)
4. C. \(y\left(x-2\right)\)\(5x\left(x-3\right)\)
5. D. \(3\left(x-2y\right)\)
1. Trong các kết quả sau kết quả nào sai
A. -17x^3y-34x^2y^2+51xy^3=17xy(x^2+2xy-3y^2)
B. x(y-1) +3(y-1)= -(1-y)(x+3)
C. 16x^2(x-y)-10y(y-1)=-2(y-x)(8x^2+5y)
2. Đa thức (x-y)^2+3(y-x) được phân tích thành nhân tử là:
A. (x+y)(x-y+3)
B. (x-y)(2x-2y+3)
C. (x-y)(x-y-3)
D. Cả 3 câu đều sai
3. Kết quả phân tích đa thức x(x-2)+(x-2) thành nhân tử
A. (x-2)x
B. (x-2)^2.x
C. x(2x-4)
D. (x-2)(x+1)
4. Kết quả phân tích 5x^2(xy-2y)-15x(xy-2y) thành nhân tử
A. (xy-2y)(5x^2-15x^2)
B. y(x-2)(5x^2-15x^2)
C. y(x-2)5x(x-3)
D. (xy-2y)5x(x-3)
5. Kết quả phân tích đa thức 3x-6y thành nhân tử là
A. 3(x-6y)
B. 3(3x-y)
C. 3(3x-2y)
D. 3(x-2y)
a: \(=\left(5x+10y\right)^2-\left(8x-4y\right)^2\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(-3x+14y\right)\left(13x+6y\right)\)
b: \(=\left(\dfrac{1}{2}x-y\right)^2-\left(2x+2y\right)^2\)
\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)
\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)
c: \(=\left(\dfrac{2}{3}x-2y\right)^2-\left(\dfrac{1}{5}x+\dfrac{1}{5}y\right)^2\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
Bài 3:
a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)
b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
Câu 1:
a: \(=\left(5x+10y\right)^2-\left(8x-4y\right)^2\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(-3x+14y\right)\left(13x+6y\right)\)
b: \(=\left(0.5x-y\right)^2-\left(2x+2y\right)^2\)
\(=\left(0.5x-y-2x-2y\right)\left(0.5x-y+2x+2y\right)\)
\(=\left(-1.5y-3y\right)\left(2.5x+y\right)\)
c: \(=\left(\dfrac{2}{3}x-2y\right)^2-\left(0.2x+0.2y\right)^2\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x+2y+\dfrac{1}{5}y+\dfrac{1}{5}x\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x+\dfrac{11}{5}y\right)\)
d: \(=-\left(5x-3\right)^2\)