Thực hiện phép tính sau 12x+15y+x-2y,với x+y=200
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\(7a-6b+8a\)
\(=15a-6b\)
\(=3\left(5a-2b\right)\)
\(=3.73=219\)
\(12x+15y+x-2y\)
\(=13x+13y\)
\(=13\left(x+y\right)\)
\(=13.200=2600\)
a, \(\frac{x+2y}{8x^2y^5}-\frac{3x^2+2}{12x^4y^4}\)
=\(\frac{\left(x+2y\right)3x^2}{24x^4y^5}-\frac{\left(3x^2+2\right)2y}{24x^4y^5}\)
=\(\frac{3x^3+6x^2y}{24x^4y^5}-\frac{6x^2y+4y}{24x^4y^5}\)
=\(\frac{3x^3+6x^2y-6x^2y-4y}{24x^4y^5}\)
=\(\frac{3x^3-4y}{24x^4y^5}\)
b,\(\frac{y}{xy-5x^2}-\frac{15y-25x}{y^2-25x^2}\)
=\(\frac{y}{x\left(y-5x\right)}-\frac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{\left(15y-25x\right)x}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2+5xy}{x\left(y-5x\right)\left(y+5x\right)}-\frac{15xy-25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y-5x}{x\left(y+5x\right)}\)
c,\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x^3-x^2\right)+\left(2x-2\right)}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{\left(x+5\right)x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{4x-4-x^2+x}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2}{x\left(x-1\right)}\)
kha sdaif dòng mik xin phép trình bày bằng lời ạ :
a) tìm MTC rồi quy đồng lên làm bình thường ại , tử cộng tử mấu giữ nguyên
b) cx vậy ạ tách mẫu tìm MTC rồi ....
~ hok tốt ~
a) Ta có P = ( 4 x 2 − 1 ) ( 2 x + 1 ) − ( 2 x − 1 ) − ( 4 x 2 − 1 ) ( 2 x + 1 ) ( 2 x − 1 ) = 3 − 4 x 2
b) Ta có Q = 3 x ( x + 3 ) . ( x + 3 ) ( x − 3 ) − x = 9 − 3 x x + 3
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
a,\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}.\dfrac{2}{3-x}=\dfrac{x-3}{3-x}=\dfrac{-\left(3-x\right)}{3-x}=-1\)
b, \(\dfrac{2x}{x-y}-\dfrac{2y}{x-y}=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{x-y}=2\)
\(a,=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\dfrac{2}{-\left(x-3\right)}=\dfrac{x-3}{2}\cdot\dfrac{2}{-\left(x-3\right)}=-1\\ b,=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{\left(x-y\right)}=2\)