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a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)
a: \(\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(=27x^3+8\)
b: \(\left(x-2y\right)^3-\left(x^2-2xy+y^2\right)\)
\(=x^3-6x^2y+12xy^2-8y^3-x^2+2xy-y^2\)
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
Bài 2 :
a) (2x + 1)(1 - 2x) + (2x - 1)2 = 22
=> 1 - 4x2 + (4x2 - 4x + 1) = 22
=> 1 - 4x2 + 4x2 + 4x + 1 = 22
=> 4x + 2 = 22
=> 4x = 20
=> x = 5
Vậy x = 5
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
Sửa đề 2a) một chút
a) 1. ( x + y )2 + ( x - y )2
= x2 + 2xy + y + x2 - 2xy + y2
= 2x2 + 2y2 = 2( x2 + y2 )
2. ( x + y )2 - ( x - y )2
= x2 + 2xy + y2 - ( x2 - 2xy + y2 )
= x2 + 2xy + y2 - x2 + 2xy - y2
= 4xy
b) x2 + 9x2 + 27 = 10x2 + 27
c) 9x2 - 6x + 1 = 9x2 - 3x - 3x + 1
= ( 9x2 - 3x ) - ( 3x - 1 )
= 3x( 3x - 1 ) - 1( 3x - 1 )
= ( 3x - 1 )( 3x - 1 )
= ( 3x - 1 )2
d) 4x2y2( 2xy + 9 ) = 4x2y2 . 2xy + 4x2y2 . 9
= 8x3y3 + 36x2y2
xét hình thang \(ACBD\)
CÓ \(AB//DC\)
\(\Rightarrow\widehat{ABC}+\widehat{BCD}=180^o\left(tcp\right)\)
thay\(\widehat{ABC}+117^o=180^o\)
\(\Rightarrow\widehat{ABC}=180^o-117^o=63^o\)
xét hình thang \(ACBD\)
có \(\widehat{ABC}+\widehat{BCD}+\widehat{ADC}+\widehat{BAD}=360^o\left(ĐL\right)\)
THAY \(63^o+117^o+\widehat{ADC}+36^o=360^o\)
\(\Rightarrow\widehat{ADC}=360^o-63^o-36^o-117^o=144^o\)