K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
15 tháng 8 2021
a: Ta có: \(x^2-6x+9-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-y-3\right)\left(x+y-3\right)\)
b: Ta có: \(x^3+4x^2+4x\)
\(=x\left(x^2+4x+4\right)\)
\(=x\left(x+2\right)^2\)
c: Ta có: \(4xy-4x^2-y^2+9\)
\(=-\left(4x^2-4xy+y^2-9\right)\)
\(=-\left(2x-y-3\right)\left(2x-y+3\right)\)
HH
2
18 tháng 7 2021
a) \(4x^2+20x+25=\left(2x+5\right)^2\)
b) \(x^2-6x+9=\left(x-3\right)^2\)
c) \(4x^2+12x+9=\left(2x+3\right)^2\)
12 tháng 2 2021
a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)
\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)
\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)
\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)
\(\Leftrightarrow x\left(-x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
Vậy: S={0;-8}
b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)
c) Ta có: \(x^2+6x+9=4x^2\)
\(\Leftrightarrow\left(x+3\right)^2-\left(2x\right)^2=0\)
\(\Leftrightarrow\left(x+3-2x\right)\left(x+3+2x\right)=0\)
\(\Leftrightarrow\left(-x+3\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S={3;-1}
d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)
HN
0
20 tháng 7 2021
a, \(A=-x^2-2x+3=-\left(x^2+2x-3\right)=-\left(x^2+2x+1-4\right)\)
\(=-\left(x+1\right)^2+4\le4\)
Dấu ''='' xảy ra khi x = -1
Vậy GTLN là 4 khi x = -1
b, \(B=-4x^2+4x-3=-\left(4x^2-4x+3\right)=-\left(4x^2-4x+1+2\right)\)
\(=-\left(2x-1\right)^2-2\le-2\)
Dấu ''='' xảy ra khi x = 1/2
Vậy GTLN B là -2 khi x = 1/2
c, \(C=-x^2+6x-15=-\left(x^2-2x+15\right)=-\left(x^2-2x+1+14\right)\)
\(=-\left(x-1\right)^2-14\le-14\)
Vâỵ GTLN C là -14 khi x = 1
Bài 8 :
b, \(B=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\)
Dấu ''='' xảy ra khi x = 3
Vậy GTNN B là 2 khi x = 3
c, \(x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu ''='' xảy ra khi x = 1/2
Vậy ...
c, \(x^2-12x+2=x^2-12x+36-34=\left(x-6\right)^2-34\ge-34\)
Dấu ''='' xảy ra khi x = 6
Vậy ...
KK
10 tháng 10 2017
a) (x2-6xy+9y2):(3y-x)
= (x-3y)2:(3y-x)
=(3y-x)2:(3y-x)
= 3y-x
b) (8x3-1):(4x2+2x+1)
=[(2x)3-1]:(4x2+2x+1)
= (2x-1)(4x2+2x+1):(4x2+2x+1)
= 2x-1
10 tháng 10 2017
c) (4x4-9):(2x2-3)
=(2x2-3)(2x2+3):(2x2-3)
=2x2+3
d) (8x3-27):(4x2+6x+9)
=(2x-3)(4x2+6x+9):(4x2+6x+9)
=2x-3
28 tháng 7 2021
a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)
DH
8 tháng 2 2019
x2 +6xx+9=4x2-4x+1
<=>3x2-10x -8=0
<=>3x2-12x+2x-8 =0
<=>3x(x-4)+2(x-4)=0
<=>(3x+2)(x-4)=0
<=>x =-2/3 hoặc x=4
8 tháng 2 2019
x2 +6xx+9=4x2-4x+1
<=>3x2-10x -8=0
<=>3x2-12x+2x-8 =0
<=>3x(x-4)+2(x-4)=0
<=>(3x+2)(x-4)=0
<=>x =-2/3 hoặc x=4
\(C=9+6x-4x^2\)
\(=-4\left(x^2-\frac{3}{2}x\right)+9\)
\(=-4\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)+\frac{9}{4}+9\ge\frac{9}{4}+9\)