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14 tháng 6 2022
a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)
\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)
\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)
c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)
d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)
22 tháng 9 2020
a) Ta có: \(\left(x+y\right)^2-8\left(x+y\right)+12\)
\(=\left[\left(x+y\right)^2-8\left(x+y\right)+16\right]-4\)
\(=\left(x+y-4\right)^2-4\)
\(=\left(x+y\right)\left(x+y-8\right)\)
3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
28 tháng 6 2021
a) (7x - 8)(7x + 8) - 10(2x + 3)2 + 5x(3x - 2)2 - 4x(x - 5)2
= 49x2 - 64 - 10(4x2 + 12x + 9) + 5x(9x2 - 12x + 4) - 4x(x2 - 10x + 25)
= 49x2 - 64 - 40x2 - 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x - 100x
= 41x3 - 51x2 - 160x - 154
b) (x2 - 3)(x2 + 3) - 5x2(x + 1)2 - (x2 - 3x)(x2 - 2x) + 4x(x + 2)2
= x4 - 9 - 5x2(x2 + 2x + 1) - x4 + 5x3 - 6x2 + 4x(x2 + 4x + 4)
= 5x3 - 6x2 - 5x4 - 10x3 - 5x2 + 4x3 + 16x2 + 16x - 9
= -5x4 - x3 + 5x2 + 16x - 9
QA
28 tháng 6 2021
Trả lời:
a , ( 7x - 8 ) ( 7x + 8 ) - 10 ( 2x + 3 )2 + 5x ( 3x - 2 )2 - 4x ( x - 5 )2
= 49x2 - 64 - 10 ( 4x2 + 12x + 9 ) + 5x ( 9x2 - 12x + 4 ) - 4x ( x2 - 10x + 25 )
= 49x2 - 64 - 40x2 + 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x2 - 100x
= 41x3 - 11x2 + 40x - 154
b , ( x2 - 3 ) ( x2 + 3 ) - 5x2 ( x + 1 )2 - ( x2 - 3x ) ( x2 - 2x ) + 4x ( x + 2 )2
= x4 - 9 - 5x2 ( x2 + 2x + 1 ) - ( x4 - 2x3 - 3x3 + 6x2 ) + 4x ( x2 + 4x + 4 )
= x4 - 9 - 5x4 - 10x3 - 5x2 - x4 + 2x3 + 3x3 - 6x2 + 4x3 + 16x2 + 16x
= - 5x4 - x3 + 5x2 + 16x - 9
S
4 tháng 12 2018
a. \(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x+2\right)\left(x-2\right)\)
b. \(x^2-y^2-4x+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x+y-2\right)\left(x-y-2\right)\)
c. \(\left(x^2+9\right)^2-36x^2=\left(x^2+6x+9\right)\left(x^2-6x+9\right)=\left(x+3\right)^2\left(x-3\right)^2\)
d. \(25-x^2+2xy-y^2=25-\left(x-y\right)^2=\left(5+x-y\right)\left(5-x+y\right)\)
còn lại làm tương tự
4 tháng 12 2018
a) \(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
b) \(x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
c) \(\left(x^2+9\right)^2-36x^2=\left(x^2+9\right)^2-\left(6x\right)^2=\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)
\(=\left(x-3\right)^2\left(x+3\right)^2\)
d) \(25-x^2+2xy-y^2=5^2-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)
e) \(x^3-4x^2+4x-1=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1-4x\right)=\left(x-1\right)\left(x^2-3x+1\right)\)
f) \(3x-3y-x^2+2xy-y^2=3\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3-x+y\right)\)
g) \(2x^2-9x+10=2x^2-4x-5x+10=2x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(2x-5\right)\)
h) \(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)
i) \(x^3-3x^2+2=x^3-2x^2-x^2+2=x^2\left(x-1\right)-2\left(x^2-1\right)\)
\(=x\left(x-1\right)-2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x-2x-2\right)\)
k) \(x^4+4=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
HN
1 tháng 1 2018
a) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{1}{x^2+x+1}\)
b) \(\dfrac{9}{x^3-9x}-\dfrac{-1}{x+3}\)
\(=\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\)
c) \(\dfrac{x^3-8}{5x+10}.\dfrac{x^2+4x}{x^2+2x+4}\)
\(=\dfrac{x\left(x-2\right)\left(x^2+2x+4\right)\left(x+4\right)}{5\left(x+2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{x\left(x-2\right)\left(x+4\right)}{5\left(x+2\right)}\)
d) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}.\dfrac{2\left(2-x\right)}{x+2}\)
\(=-\dfrac{10\left(x+2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{5}{2}\)
e) \(\dfrac{\left(x-13\right)^2}{2x^5}.\dfrac{-3x^2}{x-13}\)
\(=\dfrac{x-13}{2x^3}.\dfrac{-3}{1}\)
\(=\dfrac{-3\left(x-13\right)}{2x^3}\)
g) \(\dfrac{x^2+6x+9}{1-x}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2}{x-1}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2\left(x-1\right)^2}{2\left(x-1\right)\left(x+3\right)^2}\)
\(=-\dfrac{x-1}{2}\).
GH
24 tháng 10 2017
a) x2 - 9 + (x - 3)
= (x- 3)(x + 3) + (x - 3)
= (x - 3)(x + 3 + 1)
= (x - 3)(x + 4)
b) x3 - 4x2 + 4x - xy2
= x(x2 - 4x + 4 - y2)
= \(x\left [ (x - 2)^{2} - y^{2}\right ]\)
= x(x - 2 - y)(x - 2 + y)
= x(x - y - 2)(x + y - 2)
c) x3 - 4x2 + 12x - 27
= x3 - 27 - 4x2 + 12x
= (x - 3)(x2 + 3x + 9) - 4x(x - 3)
= (x - 3)(x2 + 3x + 9 - 4x)
= (x - 3)(x2 - x + 9)
e) 5x3 - 5x2y - 10x2 + 10xy
= 5x(x2 - xy - 2x + 2y)
= \(5x\left [ x(x - y) - 2(x - y) \right ]\)
= 5x(x - y)(x - 2)
câu f pn coi lại mũ của 3x nha nếu mũ 2 thì lm như dưới
f) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= \(3\left [ (x - y)^{2} - (2z)^{2} \right ]\)
= 3(x - y - 2z)(x - y + 2z)
pn coi lại đề câu d với f nhé
