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a: Ta có: \(x\left(2-3x\right)+\left(3x^3-x^2\right):x\)
\(=2x-3x^2+3x^2-x\)
=x
b: Ta có: \(2x\left(x-3y\right)-\left(8x^3y-12x^2y^2\right):2xy\)
\(=2x^2-6xy-4x^2+6xy\)
\(=-2x^2\)
`(2x-y)(16x^4+8x^3y+4x^2y^2+2xy^3+y^4)`
`=(2x-y)[(2x)^4+(2x)^3y+(2x)^2y^2+2xy^3+y^4)`
`=(2x)^5-y^5`
`=32x^5-y^5`
Bài 1 : 5x2 + 10y2 - 4x - 6xy - 2y + 3 > 0
= (4x2-4x+1)+(x^2-6xy+9y2)+(y^2-2y+1)+1
= (2x-1)^2+(x-3y)^2+(y-1)^2+1>0 (đpcm)
Ta có:
=11^(n+2)+12^(2n+1)
= 121.11^n + 12.144^n
= (133 -12).11^n + 12.144^n
= 133.11^n - 12.11^n + 12.144^n
=133.11^n + 12.(144^n - 11^n)
vì (144^n - 11^n) chia hết cho 133
và: 133.11^n chia hết cho 133
=> chia hết cho 133.
bạn viết rõ đề câu a;b nhé
c, \(2x\left(x-5\right)-\left(x-5\right)=0\Leftrightarrow\left(2x-1\right)\left(x-5\right)=0\Leftrightarrow x=\dfrac{1}{2};x=5\)
d, \(\left(x+3\right)\left(x+3-5+x\right)=0\Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\Leftrightarrow x=-3;x=1\)
e, \(\left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\Leftrightarrow x=-2;x=\dfrac{1}{5}\)
\(-x^2+8x-19=-\left(x^2-8x+16\right)-3=-\left(x-4\right)^2-3\le-3< 0\)
Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333
\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
a, 5\(x\)(\(x^2\) - 9) = 0
\(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -3; 0; 3}
b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0
3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0
(\(x+3\))( 3 - \(x\)) = 0
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -3; 3}
c, \(x^2\) - 9\(x\) - 10 = 0
\(x^2\) + \(x\) - 10\(x\) - 10 = 0
\(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0
(\(x+1\))(\(x-10\)) = 0
\(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -1; 10}
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
4, x^2-10x+8x-80=0
x(x-8)+10(x-8)=0
x+10=0 =)x=-10
hoặc
x-8=0 =)x=8
1, =(x+2)(x-2)=0
x+2=0 =)x=-2
hoặc
x-2=0 =)x=2
2,3(x^2-5^2)=0
=x+5=0 =)x=-5
hoặc
x-5=0 =)x=5
3,(3+2)^2=25
5^2=25
5, x^2-x-11x+11=0
x(x-1)-11(x-1)=0
x-11=0 =)x=11
hoặc
x-1=0 =)x=1
xl nheee mk làm nhầm câu 4 trc