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3 tháng 11 2021
a) \(=3\left(xy-4\right)\)
b) \(=x^2\left(x-y\right)+4\left(x-y\right)=\left(x-y\right)\left(x^2+4\right)\)
c) \(=x^2-\left(y^2-12y+36\right)=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\)
d) \(=\left(4p^2-36p+81\right)-25=\left(2p-9\right)^2-25=\left(2p-9-5\right)\left(2p-9+5\right)=4\left(p-7\right)\left(p-2\right)\)
Y
19 tháng 6 2023
Bài 2 :
\(a,\left(x+2\right)\left(x^2+3x-2\right)=2\left(x+2\right)x^2\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+3x-2\right)-2x^2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+3x-2-2x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-x^2+3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\-x^2+x+2x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\-x\left(x-1\right)+2\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left(x-1\right)\left(-x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[{}\begin{matrix}x-1=0\\-x+2=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{-2;2;1\right\}\)
\(b,9x^2-\left(6x+2\right)\left(x-5\right)=1\)
\(\Leftrightarrow9x^2-\left(6x^2-30x+2x-10\right)-1=0\)
\(\Leftrightarrow9x^2-6x^2+30x-2x+10-1=0\)
\(\Leftrightarrow3x^2+28x+9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-9\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{3};-9\right\}\)
\(c,\dfrac{x}{3x-2}-\dfrac{x}{2+3x}=\dfrac{6x^2}{9x^2-4}\left(dkxd:x\ne\pm\dfrac{2}{3}\right)\)
\(\Leftrightarrow\dfrac{x}{3x-2}-\dfrac{x}{3x+2}-\dfrac{6x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(3x+2\right)-x\left(3x-2\right)-6x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)
\(\Leftrightarrow3x^2+2x-3x^2+2x-6x^2=0\)
\(\Leftrightarrow4x-6x^2=0\)
\(\Leftrightarrow-2x\left(-2+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\-2+3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmdk\right)\\x=\dfrac{2}{3}\left(ktmdk\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
Y
19 tháng 6 2023
Bài 1 :
\(a,P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\left(dkxd:x\ne0,x\ne\pm1\right)\)
\(=\dfrac{x^2+x}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x^2+x}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\left(dpcm\right)\)
\(b,P=-\dfrac{1}{2}\Rightarrow\dfrac{x^2}{x-1}=-\dfrac{1}{2}\)
\(\Rightarrow2x^2=-\left(x-1\right)\)
\(\Rightarrow2x^2=-x+1\)
\(\Rightarrow2x^2+x-1=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy \(P=-\dfrac{1}{2}\) thì \(x=\dfrac{1}{2};x=-1\)
\(c,\) Để P nhận giá trị nguyên dương thì \(P\ge0\)
\(\Leftrightarrow\dfrac{x^2}{x-1}\ge0\Leftrightarrow x\ge0\)
23 tháng 8 2020
a, (x2 - 3)(2 + 4x)
= 2x2 + 4x3 - 6 - 12x
b, 6(3x + 5)(x - 4)
= (18x + 30)(x - 4)
= 18x2 - 72 + 30x - 120
= 18x2 + 30x - 192
NQ
1
KN
9 tháng 11 2019
Gọi số cạnh của tam giác là n
Ta có: \(\frac{\left(n-2\right).180^0}{n}=156^0\)
\(\Leftrightarrow\left(n-2\right).180^0=156^0n\)
\(\Leftrightarrow180^0n-360^0=156^0n\)
\(\Leftrightarrow180^0n-156^0n=360^0\)
\(\Leftrightarrow24^0n=360^0\)
\(\Leftrightarrow n=15\)
Vậy đa giác đó có 15 cạnh
DT
1
O
25 tháng 3 2022
a) \(\dfrac{3}{4}+\dfrac{9}{5}\div\dfrac{3}{2}-1=\dfrac{3}{4}+\dfrac{18}{15}-1=\dfrac{39}{20}-1=\dfrac{19}{20}\)
b) \(\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{6}{13}\cdot\dfrac{9}{7}-\dfrac{4}{13}\cdot\dfrac{6}{7}=\dfrac{48}{91}+\dfrac{54}{91}-\dfrac{24}{91}=\dfrac{48+51-24}{91}=\dfrac{78}{91}=\dfrac{6}{7}\)
c) \(\dfrac{-3}{7}+\left(\dfrac{3}{-7}-\dfrac{3}{-5}\right)\)\(=\dfrac{-3}{7}+\left(\dfrac{-3}{7}-\dfrac{-3}{5}\right)=\dfrac{-3}{7}+\dfrac{6}{35}=-\dfrac{9}{35}\)
TB
1
23 tháng 10 2021
\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
23 tháng 10 2021
6a.
$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$
$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$
Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$
đây bạn😀
b: \(\Leftrightarrow x^2-4-x-x=0\)
hay x=-4