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6 tháng 12 2022
\(=\dfrac{4\left(x+3\right)^2}{\left(x+5\right)\left(5x+5\right)}-\dfrac{x^2-25}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(3x-3-x\right)\left(3x-3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}-\dfrac{x^2-25}{\left(x-5\right)\cdot5\left(x+1\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{3\left(x+5\right)\cdot5\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(x+5\right)}{5\left(x+1\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2-\left(x+5\right)\left(x+1\right)}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-6x-5}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{3x^2+18x+31}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{3\left(x+3\right)\left(3x^2+18x+31\right)-\left(2x-3\right)\left(4x-3\right)\left(x+1\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{\left(3x+9\right)\left(3x^2+18x+31\right)-\left(8x^2-18x+9\right)\left(x+1\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{9x^3+81x^2+255x+279-\left(8x^3+8x^2-18x^2-18x+9x+9\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{9x^3+81x^2+255x+279-\left(8x^3-10x^2-9x+9\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{x^3+91x^2+264x+270}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)
T
29 tháng 6 2017
a) \(\left(3x-5\right)\left(9x^2+15x+25\right)\)
\(=\left(3x\right)^3-5^3\)
\(=27x^3-125\)
b) \(\left(2x+7\right)\left(x^2-14x+49\right)-2x\left(2x-1\right)\left(2x+1\right)\)
\(=2x^3-28x^2+98x+7x^2-98x+343-2x\left(4x^2-1\right)\)
\(=2x^3-28x^2+7x^2+343-8x^3+2x\)
\(=-6x^3-21x^2+343+2x\)
c) \(\left(4x-7\right)\left(16x^2+28x+49\right)\left(3x+1\right)\left(9x^2-3x+1\right)-9x\left(3x^2-1\right)\)
\(=\left(64x^3-343\right)\left(3x+1\right)\left(9x^2-3x+1\right)-27x^3+9x\)
\(=\left(6x^3-343\right)\left(27x^3+1\right)-27x^3+9x\)
\(=1728x^6+64x^3-9261x^3-343-27x^3+9x\)
\(=1728x^6-9224x^3-343+9x\)
TH
19 tháng 12 2018
Bài 1:
a) \(\dfrac{3x^2-5}{x^2-5x}+\dfrac{5-15x}{5x-25}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{5\left(1-3x\right)}{5\left(x-5\right)}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{1-3x}{x-5}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x-3x^2}{x\left(x-5\right)}\)
\(=\dfrac{-5+x}{x\left(x-5\right)}\)
\(=\dfrac{x-5}{x\left(x-5\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4+x^3}{x-3}-\dfrac{2x+2x^2}{x-3}+\dfrac{2x-13}{x-3}\)
\(=\dfrac{\left(4+x^3\right)-\left(2x+2x^2\right)+\left(2x-13\right)}{x-3}\)
\(=\dfrac{4+x^3-2x-2x^2+2x-13}{x-3}\)
\(=\dfrac{x^3-2x^2-9}{x-3}\)
\(=\dfrac{x^3-3x^2+x^2-9}{x-3}\)
\(=\dfrac{x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)}{x-3}\)
\(=\dfrac{\left(x-3\right)\left(x^2+x+3\right)}{x-3}\)
\(=x^2+x+3\)
c) \(\dfrac{2}{x-5}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2x+10+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3}{x+5}\)
d) Đề sai?
Bài 2:
\(A=2\left(x+1\right)+\left(3x+2\right)\left(3x-2\right)-9x^2\)
\(A=2x+2+9x^2-4-9x^2\)
\(A=2x-2\)
\(A=2\left(x-1\right)\)
Thay x = 15 vào A ta được:
\(A=2\left(15-1\right)\)
\(A=2.14=28\)
13 tháng 12 2019
3x^3-5x^2+9x-15 3x-5 x^2+3 3x^3-5x^2 9x-15 9x-15 0
Vậy \(3x^2-5x^2+9x-15=\left(3x-5\right)\left(x^2+3\right)\)
b
\(\left(x+1\right)\left(x-2\right)-x\left(x-3\right)=0\)
\(\Leftrightarrow x^2-2x+x-2-x^2+3x=0\)
\(\Leftrightarrow2x-2=0\)
\(\Leftrightarrow x=1\)
b
\(x^2+4x+3=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)-1=0\)
\(\Leftrightarrow\left(x+2\right)^2-1=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1;x=-3\)
MN
19 tháng 1 2020
a) \(4x^4-101x^2+25=0\)
\(\Leftrightarrow4x^4-100x^2-x^2+25=0\)
\(\Leftrightarrow4x^2\left(x^2-25\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x-5\right)\left(x+5\right)=0\)
Từ đây cậu suy ra đc tập nghiệm của ptr là : \(S=\left\{\frac{1}{2};-\frac{1}{2};5;-5\right\}\)
b) Tớ chịu :>
c) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left[\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-2x^2+1-x^4-x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(-3x^2\right)=0\)
Từ đây thấy rằng tập nghiệm ptr là : \(S=\left\{1;-1;0\right\}\)
Chúc cậu học tốt !
25 tháng 8 2020
a) ( x - 5 )( 2x + 3 ) + 2x( 1 - x )
= 2x2 - 7x - 15 + 2x - 2x2
= -5x - 15
= -5( x + 3 )
b) ( 3x - 5 )2 - ( x + 5 )( 5 - x ) - 5/2( -2x )2
= 9x2 - 30x + 25 + ( x + 5 )( x - 5 ) - 5/2.4x2
= 9x2 - 30x + 25 + x2 - 25 - 10x2
= -30x
c) ( 3x + 2 )( 4 - 6x + 9x2 ) - 3x( 3x - 2 )2 + 12( -2/3 - 3x2 )
= ( 3x )3 + 23 - 3x( 9x2 - 12x + 4 ) - 8 - 36x2
= 27x3 + 8 - 27x3 + 36x2 - 12x - 8 - 36x2
= -12x
QA
4
18 tháng 8 2016
a/\(x^3-9x=0\Leftrightarrow x\left(x^2-9\right)=0\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\\x=-3\end{array}\right.\)
b/ \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\Leftrightarrow\left(5-2x\right)\left(2x+7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
18 tháng 8 2016
\(x^3-9x=0\)
\(\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-3=0\\x+3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\\x=-3\end{array}\right.\)
b ) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow-\left(2x-5\right)\left(2x+7\right)-\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-2x-7-2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-4x-12\right)=0\)
\(\Leftrightarrow-4\left(2x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-5=0\\x+3=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
Ta có:
\(3\left(3x-5\right)=9x^2-25\\ \Leftrightarrow9x^2-9x-10=0\\ \Leftrightarrow\left(3x\right)^2-2.3x.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{49}{4}\\ \Leftrightarrow\left(3x-\dfrac{3}{2}\right)^2=\dfrac{49}{4}\\ \Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{3}{2}=\dfrac{7}{2}\\3x-\dfrac{3}{2}=\dfrac{-7}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
3(3x-5)= (3x-5)(3x+5)
3(3x-5)-(3x-5)(3x+5)=0
(3-3x+5)(3x+5)=0
(-3x+8)(3x+5)=0
TH1 X=8/3
TH2 X=-5/3