quy đồng mẫu thức sau
- 5x2+1/8x2;3x/2x+3x;-5/4x-6
2.-7/12xz2;11z/18x2;5x/6z2y
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QT
1
Những câu hỏi liên quan
NL
Nguyễn Lê Phước Thịnh
CTVHS
26 tháng 3
Bài 1:
a: \(\frac{1}{2x^3y}=\frac{1\cdot6\cdot yz^3}{2x^3y\cdot6yz^3}=\frac{6yz^3}{12x^3y^2z^3}\)
\(\frac{2}{3xy^2z^3}=\frac{2\cdot4\cdot x^2}{3xy^2z^3\cdot4x^2}=\frac{8x^2}{12x^3y^2z^3}\)
\(\frac{5}{4yz}=\frac{5\cdot3\cdot x^3\cdot y\cdot z^2}{4yz\cdot3x^3yz^2}=\frac{15x^3yz^2}{12x^3y^2z^3}\)
b: \(\frac{x+1}{10x^3-40x}=\frac{x+1}{10x\left(x^2-4\right)}=\frac{x+1}{10x\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+1\right)\cdot4\cdot x}{4x\cdot10x\cdot\left(x+2\right)\left(x-2\right)}=\frac{4x^2+4x}{40x^2\left(x+2\right)\left(x-2\right)}\)
\(\frac{5}{8x^3+16x^2}=\frac{5x}{8x^2\left(x+2\right)}\)
\(=\frac{5x\cdot5\cdot\left(x-2\right)}{8x^2\left(x+2\right)\cdot5\cdot\left(x-2\right)}=\frac{25x^2-50x}{40x^2\left(x+2\right)\left(x-2\right)}\)
Bài 2:
\(\frac{2-x}{3x-3x^2}=\frac{-\left(x-2\right)}{-\left(3x^2-3x\right)}=\frac{x-2}{3x\left(x-1\right)}\)
\(=\frac{\left(x-2\right)\cdot4x\cdot\left(x^2+x+1\right)}{3x\left(x-1\right)\cdot4x\cdot\left(x^2+x+1\right)}=\frac{\left(4x^2-8x\right)\left(x_{}^2+x+1\right)}{12x^2\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{x^2-2}{4x^5-4x^2}=\frac{x^2-2}{4x^2\left(x^3-1\right)}=\frac{x^2-2}{4x^2\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{\left(x^2-2\right)\cdot3}{4x^2\left(x-1\right)\left(x^2+x+1\right)\cdot3}=\frac{3x^2-6}{12x^2\left(x-1\right)\left(x^2+x+1\right)}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
7 tháng 12 2021
Bài 2:
a: \(\dfrac{1}{2x^3y}=\dfrac{6yz^3}{12x^3y^2z^3}\)
\(\dfrac{2}{3xy^2z^3}=\dfrac{2\cdot4x^2}{12x^3y^2z^3}=\dfrac{8x^2}{12x^3y^2z^3}\)
CM
4 tháng 12 2018
Ta có:
Suy ra: x 3 - 7 x 2 + 7 x + 15 = x 2 - 4 x - 5 x - 3
Lại có:
Suy ra: x 3 - 7 x 2 + 7 x + 15 = x 2 - 2 x - 3 x - 5
NQ
CM
1 tháng 8 2018
M T 1 : x 3 – 1 = ( x - 1 ) ( x 2 + x + 1 )
M T 2 : x 2 + x + 1
⇒ M T C : ( x - 1 ) ( x 2 + x + 1 )
⇒ NTP1: 1
⇒ NTP2: x - 1
Quy đồng:
PT
1
\(1, \frac{5x^2+1}{8x^2}=\frac{5\left(5x^2+1\right)\left(4x-6\right)}{5.8x^2\left(4x-6\right)}=\frac{5\left(5x^2+1\right)\left(4x-6\right)}{40x^2\left(4x-6\right)}\)
\(\frac{3x}{2x+3x}=\frac{3x}{5x}=\frac{3x.8x\left(4x-6\right)}{8x.5x\left(4x-6\right)}=\frac{24x^2\left(4x-6\right)}{40x^2\left(4x-6\right)}\)
\(\frac{-5}{4x-6}=\frac{-5.5x.8x}{8x.5x.\left(4x-6\right)}=\frac{-200x^2}{40x^2\left(4x-6\right)}\)