giúp mình vs cần gấp
x^2-4y^2-3x+6y
a^2+2ab+b^2-ac-bc
25x^2-10x-3x
16x^2+24x-7
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HT
0
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 10 2019
\(A=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\)
\(B=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}\)
\(C=\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\)
\(D=\left(x-5\right)^2+\left(3y+1\right)^2+4\)
\(E=\left(4x+1\right)^2+\left(y-2\right)^2+1\)
\(M=-\left(x+\frac{7}{2}\right)^2-\frac{11}{4}\)
\(N=-5\left(x-\frac{3}{5}\right)^2-\frac{41}{5}\)
\(C\) đề sai ví dụ \(x=3\Rightarrow C=2>0\)
\(D=-5\left(x-\frac{7}{10}\right)^2-\frac{131}{20}\)
AH
Akai Haruma
Giáo viên
28 tháng 12 2018
a)
\(a^2+b^2+2ab+2a+2b+1\)
\(=(a^2+2ab+b^2)+(2a+2b)+1\)
\(=(a+b)^2+2(a+b)+1^2=(a+b+1)^2\)
b)
\(3x(x-2y)+6y(2y-x)\)
\(=3x(x-2y)-6y(x-2y)=(3x-6y)(x-2y)=3(x-2y)(x-2y)\)
\(=3(x-2y)^2\)
AH
Akai Haruma
Giáo viên
28 tháng 12 2018
c)
\(16xy+4y^2-9+16x^2\)
\(=(16x^2+16xy+4y^2)-9\)
\(=(4x+2y)^2-3^2=(4x+2y-3)(4x+2y+3)\)
d)
\(x^4+64y^8=(x^2)^2+(8y^4)^2=(x^2)^2+(8y^4)^2+2.x^2.8y^4-2x^2.8y^4\)
\(=(x^2+8y^4)^2-16x^2y^4=(x^2+8y^4)^2-(4xy^2)^2\)
\(=(x^2+8y^4-4xy^2)(x^2+8y^4+4xy^2)\)
e)
\(3x^2-7x+2=3x^2-6x-x+2=(3x^2-6x)-(x-2)\)
\(=3x(x-2)-(x-2)=(3x-1)(x-2)\)
28 tháng 12 2018
a, a2+b2+2ab+2a+2b+1=(a+b+1)2
b,3x(x-2y)+6y(2y-x)=3x(x-2y)-6y(x-2y)
=3(x-2y)(x-2y)=3(x-2y)2
c, 16xy +4y2-9 +16x2=(16x2+16xy+4y2)-32
=(4x-2y)2-32=(4x-2y+3)(4x-2y-3)
13 tháng 8 2021
y: Ta có: \(x^2-x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
z: Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)
j: Ta có: \(25x^2-4=0\)
\(\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
PD
3
XO
28 tháng 8 2020
x2 + y2 + 10x + 6y + 34 = 0
=> (x2 + 10x + 25) + (y2 + 6y + 9) = 0
=> (x + 5)2 + (y + 3)2 = 0
=> \(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)
Vậy x = - 5 ; y = -3
b) 25x2 + 4y2 + 10x + 4y + 2 = 0
=> (25x2 + 10x + 1) + (4y2 + 4y + 1) = 0
=> (5x + 1)2 + (2y + 1)2 = 0
=> \(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-0,2\\y=-0,5\end{cases}}\)
Vậy x = -0,2 ; y = -0,5
28 tháng 8 2020
a)
\(x^2+10x+25+y^2+6y+9=0\)
\(\left(x+5\right)^2+\left(y+3\right)^2=0\) ( 1 )
Ta có :
\(\left(x+5\right)^2\ge0\forall x\)
\(\left(y+3\right)^2\ge0\forall y\)
\(\left(1\right)=0\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\)
\(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\)
\(\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)
b)
\(25x^2+10x+1+4y^2+4y+1=0\)
\(\left(5x+1\right)^2+\left(2y+1\right)^2=0\) ( 1 )
Ta có :
\(\left(5x+1\right)^2\ge0\forall x\)
\(\left(2y+1\right)^2\ge0\forall y\)
\(\left(1\right)=0\Leftrightarrow\hept{\begin{cases}\left(5x+1\right)^2=0\\\left(2y+1\right)^2=0\end{cases}}\)
\(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\)
\(\hept{\begin{cases}x=\frac{-1}{5}\\y=\frac{-1}{2}\end{cases}}\)
a/ \(x^2-4y^2-3x+6y=\left(x^2-4y^2\right)-\left(3x-6y\right)=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x+2y-3\right)\)
b/ \(a^2+2ab+b^2-ac-bc=\left(a^2+2ab+b^2\right)-\left(ac+bc\right)=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)
c/ \(25x^2-10x-3x=25x^2-13x=x\left(25x-13\right)\)
d/ \(16x^2+24x-7=16x^2-4x+28x-7=4x\left(4x-1\right)+7\left(4x-1\right)=\left(4x-1\right)\left(4x+7\right)\)