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a)\(\left(3x+2\right)\left(2x-3\right)=6x^2-5x-6\)
b) viết lại đề nhin (dùng f(x) viết mới rõ ra dduocj) ko phải dùng {[(...)]} cho chuẩn vào
c) \(\left(x-2\right)^3-x^2.\left(x-6\right)=x^3-3.x.2\left(x-2\right)-8-x^3+6x^2\)
\(=x^3-6x^2+12x-8-x^3+6x^2=12x-8=4\Rightarrow x=1\)
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}\)
\(=\frac{\left(x^2-36\right).3}{\left(2x+10\right)\left(6-x\right)}\)
\(=\frac{3\left(x+6\right)\left(x-6\right)}{\left(2x+10\right)\left(6-x\right)}\)
\(=-\frac{3\left(x+6\right)\left(x-6\right)}{2\left(x+5\right)\left(x-6\right)}\)
\(=-\frac{3\left(x+6\right)}{2\left(x+5\right)}\)
[5(x + 2y)⁶ - 6(x + 2y)⁵] : [2(x + 2y)⁴]
= 5(x + 2y)⁶ : [2(x + 2y)⁴] - 6(x + 2y)⁵ : [2(x + 2y)⁴]
= 5(x + 2y)²/2 - 3(x + 2y)
\(a.\dfrac{x+1}{2x+6}+2x=\dfrac{x+1+4x^2+12x}{2x+6}=\dfrac{4x^2+13x+1}{2x+6}\) ( x # -3)
\(b.\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\) ( x # - 3)
Các câu còn lại tương tự .
\(a,\dfrac{x+1}{2x+6}+2x\)
\(=\dfrac{x+1}{2x+6}+\dfrac{2x\left(2x+6\right)}{2x+6}\)
\(=\dfrac{x+1+4x^2+12x}{2x+6}\)
\(=\dfrac{4x^2+13x+1}{2x+6}\)
\(b,\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3x}{2x^2+6x}-\dfrac{x-6}{2x^2-6x}\)
\(=\dfrac{2x-6}{2x^2+6x}=\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}=\dfrac{x-3}{x^2+3x}\)
\(c,\dfrac{x}{x-2y}+\dfrac{x}{x+2y}+\dfrac{4xy}{4y^2-x^2}\)
\(=\dfrac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\dfrac{x\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}-\dfrac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{2x}{x+2y}\)
\(d,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\)
\(=\dfrac{3x+2}{\left(3x+2\right)\left(3x-2\right)}-\dfrac{3x-2}{\left(3x+2\right)\left(3x-2\right)}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\)
a: =-4xyz^2
b: =-9x^2y
c: =16x^2y^2
d: =1/6x^2y^3
e: =13/6x^3y^2
f: =7/12x^4y
a) -xyz² - 3xz.yz
= -xyz² - 3xyz²
= -4xyz²
b) -8x²y - x.(xy)
= -8x²y - x²y
= -9x²y
c) 4xy².x - (-12x²y²)
= 4x²y² + 12x²y²
= 16x²y²
d) 1/2 x²y³ - 1/3 x²y.y²
= 1/2 x²y³ - 1/3 x²y³
= 1/6 x²y³
e) 3xy(x²y) - 5/6 x³y²
= 3x³y² - 5/6 x³y²
= 13/6 x³y²
f) 3/4 x⁴y - 1/6 xy.x³
= 3/4 x⁴y - 1/6 x⁴y
= 7/12 x⁴y