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. Mình xin chân thành cảm ơn và có một điều bất
a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
a) \(\dfrac{a^3\left(a-5\right)}{a-5}=a^3 \)
b) \(\dfrac{3\left(b+7\right)4}{8\left(b+7\right)6}=\dfrac{12\left(b+7\right)}{48\left(b+7\right)}=\dfrac{1}{4}\)
c) \(\dfrac{15x\left(x+5\right)^2}{20x^2\left(x+5\right)}=\dfrac{15x}{20x^2}=\dfrac{3}{4x}\)
d) \(\dfrac{x^3-4x^2}{y\left(x-4\right)}=\dfrac{x^2\left(x-4\right)}{y\left(x-4\right)}=\dfrac{x^2}{y}\)
e) \(\dfrac{5\left(a-2c\right)^2}{2a^2-4ac}=\dfrac{5\left(a-2c\right)^2}{2a\left(a-2c\right)}=\dfrac{5\left(a-2c\right)}{2a}=\dfrac{5a-10c}{2a}\)
Câu 4:
\(=\dfrac{a\left(a-b\right)-c\left(a-b\right)}{a\left(a+b\right)-c\left(a+b\right)}=\dfrac{a-b}{a+b}\)
Bài 1:
\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)
Bài 2:
\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)
a)\(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x^3-2^3\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
1) \(\dfrac{15-5x}{5x^2-15x}=\dfrac{5\left(3-x\right)}{5x\left(x-3\right)}=-\dfrac{5\left(x-3\right)}{5x\left(x-3\right)}=-\dfrac{1}{x}\)
Chọn A
2) \(\dfrac{x\left(x-5\right)}{x^2+25}=\dfrac{x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x}{x+5}\)
\(A=0\Leftrightarrow\dfrac{x}{x+5}=0\Leftrightarrow x=0\)
Chọn B
3) \(\dfrac{2x-5}{5-2x}=-\dfrac{5-2x}{5-2x}=-1\)
Chọn D
a: Thay x=-3 vào B, ta được:
\(B=\dfrac{2\cdot\left(-3\right)^2}{3\cdot\left(-3\right)+6}=\dfrac{2\cdot9}{-9+6}=\dfrac{18}{-3}=-6\)
b: \(A=\dfrac{2x^2+20+3x-6-7x-14}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x+2}\)
1b.=2((x+y)+(x+y)(x-y)+(x-y))=2(x2-y2+x+y+x-y)=2(x2-y2+2x)=2x2-2y2+4x
2a.=4xy+4xy+2y=8xy+2y=2y(4x+1)
b.=(3x)2+2.3x.y+y2-(2z)2=(3x+y)2-(2z)2=(3x+y-2z)(3x+y+2z)
c.=x2-x-7x+7=x(x-1)-7(x-1)=(x-1)(x-7)
\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2\)
\(=\left(2x\right)^2\)
\(=4x^2\)
hk tốt
^^
1)
a) \(\dfrac{18ab}{27bc}=\dfrac{2a}{3c}\)
b) \(\dfrac{-21b^2y^2}{-28by}=\dfrac{3by}{4}\)
c) \(\dfrac{-49a^3}{14b^3}=\dfrac{-7a^3}{2b^3}\)
d) \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{2x^2}{3y^3}\)
2)
a) \(\dfrac{a^3\left(a-5\right)}{a-5}=a^3\)
b) \(\dfrac{3\left(b+7\right)^4}{8\left(b+7\right)^6}=\dfrac{3}{8\left(b+7\right)^2}\)
c) \(\dfrac{15x\left(x+5\right)^2}{20x^2\left(x+5\right)}=\dfrac{3\left(x+5\right)}{4x}\)
d) \(\dfrac{x^3-4x^2}{y\left(x-4\right)}=\dfrac{x^2\left(x-4\right)}{y\left(x-4\right)}=\dfrac{x^2}{y}\)
e) \(\dfrac{5\left(a-2c\right)^2}{2a^2-4ac}=\dfrac{5\left(a-2c\right)^2}{2a\left(a-2c\right)}=\dfrac{5\left(a-2c\right)}{2a}\)
3)
a) \(\dfrac{ax-3a}{bx-3b}=\dfrac{a\left(x-3\right)}{b\left(x-3\right)}=\dfrac{a}{b}\) (câu này mình sửa lại đề)
b) \(\dfrac{5x+20y}{15x+60y}=\dfrac{5\left(x+4y\right)}{15\left(x+4y\right)}=\dfrac{1}{3}\)
c) \(\dfrac{3b-9c}{5b^2-15bc}=\dfrac{3\left(b-3c\right)}{5b\left(b-3c\right)}=\dfrac{3}{5b}\)
d) \(\dfrac{8a^2+40ab}{ab+5b^2}=\dfrac{8a\left(a+5b\right)}{b\left(a+5b\right)}=\dfrac{8a}{b}\)
4)
a) \(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
b) \(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
5)
a) \(\dfrac{45x\left(3-x\right)}{15\left(x-3\right)^3}=\dfrac{-45x\left(x-3\right)}{15\left(x-3\right)^3}=\dfrac{-3x}{\left(x-3\right)^2}\)
b) \(\dfrac{36\left(x-2\right)^3}{32-16x}=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=\dfrac{-9\left(x-2\right)^2}{4}\)
c) \(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)
d) \(\dfrac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\dfrac{-\left(y+x\right)\left(x-y\right)}{\left(x-y\right)^3}=\dfrac{-x-y}{\left(x-y\right)^2}\)
1.
a, \(\dfrac{18ab}{27bc}=\dfrac{18ab:9b}{27bc:9b}=\dfrac{2a}{3c}\)
b, \(\dfrac{-21b^2y^2}{-28by}=\dfrac{-21b^2y^2:\left(-7\right)by}{-28by:\left(-7\right)by}=\dfrac{3by}{4}\)
c, \(\dfrac{-49a^3}{14b^3}=\dfrac{-49a^3:7}{14b^3:7}=\dfrac{-7a^3}{2b^3}\)
d, \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{6xy^2\cdot2x^2}{6xy^2\cdot3y^3}=\dfrac{2x^2}{3y^3}\)
2.
a,\(\dfrac{a^3\cdot\left(a-5\right)}{a-5}=\dfrac{a^3}{1}=a^3\)
b,\(\dfrac{3\cdot\left(b+7\right)^4}{8\cdot\left(b+7\right)^6}=\dfrac{3}{8\cdot\left(b+7\right)^2}\)
c,\(\dfrac{15x\cdot\left(x+5\right)^2}{20x^2\cdot\left(x+5\right)}=\dfrac{3\cdot\left(x+5\right)}{4x}\)
d,\(\dfrac{x^3-4x^2}{y\cdot\left(x-4\right)}=\dfrac{x^2}{y}\)
e,\(\dfrac{5\cdot\left(a-2x\right)^2}{2a^2-4ac}=\dfrac{5\cdot\left(a-2x\right)}{2a}\)
3.
a,\(\dfrac{ax-3a}{bx-3b}=\dfrac{a\cdot\left(x-3\right)}{b\cdot\left(x-3\right)}=\dfrac{a}{b}\)
b, \(\dfrac{5x+20y}{15x+60y}=\dfrac{5\cdot\left(x+4y\right)}{15\cdot\left(x+4y\right)}=\dfrac{5}{15}=\dfrac{1}{3}\)
c, \(\dfrac{3b-9c}{5b^2-15bc}=\dfrac{3\cdot\left(b-3c\right)}{5b\cdot\left(b-3c\right)}=\dfrac{3}{5b}\)
d, \(\dfrac{8a^2+40ab}{ab+5b^2}=\dfrac{8a\cdot\left(a+5b\right)}{b\cdot\left(a+5b\right)}=\dfrac{8a}{b}\)
4.
a,\(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\cdot\left(x^2-4x+4\right)}{x\cdot\left(x^3-8\right)}=\dfrac{3\cdot\left(x-2\right)^2}{x\cdot\left(x-2\right)\cdot\left(x^2+2x+4\right)}=\dfrac{3\cdot\left(x-2\right)}{x\cdot\left(x^2+2x+4\right)}=\dfrac{3\cdot\left(x-2\right)}{x\cdot\left(x+2\right)^2}\)
b, \(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\cdot\left(x^2+2x+1\right)}{3x\cdot\left(x+1\right)}=\dfrac{7\cdot\left(x+1\right)^2}{3x\cdot\left(x+1\right)}=\dfrac{7\cdot\left(x+1\right)}{3x}\)
5.
a, \(\dfrac{45x\cdot\left(3-x\right)}{15x\cdot\left(x-3\right)^3}=\dfrac{3\cdot\left(3-x\right)}{\left(x-3\right)^3}=\dfrac{-3\cdot\left(x-3\right)}{\left(x-3\right)^3}=\dfrac{-3}{\left(x-3\right)^2}\)
b, \(\dfrac{36\cdot\left(x-2\right)^3}{36-16x}=\dfrac{36\cdot\left(x-2\right)^3}{16\cdot\left(2-x\right)}=\dfrac{36\cdot\left(-\left(x-2\right)\right)^3}{16\cdot\left(2-x\right)}=\dfrac{-36\cdot\left(2-x\right)^3}{16\cdot\left(2-x\right)}=\dfrac{-9\cdot\left(2-x\right)^2}{4}\)
c, \(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\cdot\left(x-y\right)}{5y\cdot\left(y-x\right)}=\dfrac{-x\cdot\left(y-x\right)}{5y\cdot\left(y-x\right)}=\dfrac{-x}{5y}\)
d, \(\dfrac{y^2-x^2}{x^3-3x^2y+3xy^2+y^3}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{\left(x-y\right)^3}=\dfrac{-\left(x+y\right)\cdot\left(y-x\right)}{\left(x-y\right)^3}=\dfrac{-\left(x+y\right)}{\left(x-y\right)^2}\)