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18 tháng 7 2019
2 .tìm x
a , x ( x + 2 ) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b, x ( x-5 )= 5 -x
<=> x ( x-5 ) + x - 5 = 0
<=> x (x-5) + ( x-5)= 0
<=> (x-5)(x+1 )=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
c) ( x + 1 ) ( 6x2 + 2x ) + ( x - 1 ) ( 6x2 + 2x ) = 0
\(\Leftrightarrow\) ( 6x2 + 2x ) \([\)(x+1)(x-1)\(]\)=0
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x\left(3x+1\right)=0\\x^{2^{ }}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3x+1=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\frac{-1}{3}\\x=1\end{matrix}\right.\)
18 tháng 7 2019
1 ,a) 2a ( x - y ) - ( y - x ) = 2ax - 2ay - y + x
= x ( 2a + 1 ) - y ( 2a + 1 )
= ( 2a + 1 ) ( x - y )
b) a2 ( x - y ) - ( y - x ) = a2x - a2y - y + x
= x ( a2+ 1 ) - y ( a2 +1 )
= ( a2+1 ) - (x-y )
c) x ( x - y ) + y ( y - x ) - 3 ( x - y ) = x 2 - xy -+ y 2 - xy - 3x + 3y
= x2 - 2xy + y2 -3x + 3y
= (x-y)2 -3 ( x - y )
= ( x-y ) ( x-y+3)
PA
30 tháng 10 2016
\(B=7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
\(E=x^2+7x+12\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
\(F=x^2-9x+18\)
\(=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)\)
\(=\left(x-3\right)\left(x-6\right)\)
\(H=8x^2-2x-1\)
\(=8x^2-4x+2x-1\)
\(=4x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
10 tháng 4 2022
Bài 1:
b: \(x^3-4x^2+7x-6=0\)
\(\Leftrightarrow x^3-2x^2-2x^2+4x+3x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-2x+3\right)=0\)
=>x-2=0
hay x=2
c: \(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2-2x+2+7x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+4x+x+2\right)=0\)
=>(x+1)(x+2)(2x+1)=0
hay \(x\in\left\{-1;-2;-\dfrac{1}{2}\right\}\)
d: \(2x^3-9x+2=0\)
\(\Leftrightarrow2x^3-4x^2+4x^2-8x-x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+1-\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1+\dfrac{\sqrt{6}}{2}\right)\left(x+1-\dfrac{\sqrt{6}}{2}\right)=0\)
hay \(x\in\left\{2;-1-\dfrac{\sqrt{6}}{2};-1+\dfrac{\sqrt{6}}{2}\right\}\)
10 tháng 7 2022
1: \(\Leftrightarrow5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x\)
\(\Leftrightarrow3x^2+16x-19=3x^2-4x-18\)
=>20x=1
hay x=1/20
2: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-\left(x^2-2x\right)+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow-20x-41=-6x+27\)
=>-14x=68
hay x=-34/7
AH
Akai Haruma
Giáo viên
1 tháng 12 2019
Lời giải:
a) ĐKXĐ: $x\neq \pm 1$
\(\frac{x^4-4x^2+3}{x^4+6x^2-7}=\frac{x^2(x^2-1)-3(x^2-1)}{x^2(x^2-1)+7(x^2-1)}=\frac{(x^2-3)(x^2-1)}{(x^2-1)(x^2+7)}=\frac{x^2-3}{x^2+7}\)
b) ĐKXĐ: Với mọi $x\in\mathbb{R}$
\(\frac{x^4+x^3-x-1}{x^4+x^4+2x^2+x+1}=\frac{(x^4-x)+(x^3-1)}{(x^4+x^3+x^2)+(x^2+x+1)}=\frac{x(x^3-1)+(x^3-1)}{x^2(x^2+x+1)+(x^2+x+1)}\)
\(=\frac{(x^3-1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{(x-1)(x^2+x+1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{x^2-1}{x^2+1}\)
c) ĐK: $x\neq 1;-2$
\(\frac{x^3+3x^2-4}{x^3-3x+2}=\frac{x^2(x-1)+4(x^2-1)}{x^2(x-1)+x(x-1)-2(x-1)}=\frac{(x-1)(x^2+4x+4)}{(x-1)(x^2+x-2)}\)
\(=\frac{(x-1)(x+2)^2}{(x-1)(x-1)(x+2)}=\frac{x+2}{x-1}\)
d) ĐK: $x^2+3x-1\neq 0$
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{(x^2+3x)^2-1}{(x^2+3x)^2-2x^2-6x+1}\)
\(=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x)^2-2(x^2+3x)+1}=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x-1)^2}=\frac{x^2+3x+1}{x^2+3x-1}\)
5 tháng 7 2018
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
5 tháng 7 2018
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
Làm tính chia nha mn
a)\(\frac{6x^3-7x^2-x+2}{2x+1}=\frac{\left(2x+1\right)\left(3x^2-5x+2\right)}{2x+1}=3x^2-5x+2\)
b)\(\frac{6x^3-2x^2-9x+5}{x-1}=\frac{\left(x-1\right)\left(6x^2+4x-5\right)}{x-1}=6x^2+4x-5\)