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
10 tháng 8 2017
f) x2 + 2y2 - 2xy + 2x + 2 - 4y =0
<=>x2 + y2 - 2xy+2x-2y+y2-2y+1+1=0
<=>(x-y)2+2(x-y)+1+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>y=1;x=0
Bạn học thầy Trung phải k nè~~~~
Busted :))))
14 tháng 7 2017
1) \(\left(x-3\right)\left(x-5\right)+44\)
\(=x^2-3x-5x+15+44\)
\(=x^2-8x+59\)
\(=x^2-2.x.4+4^2+43\)
\(=\left(x-4\right)^2+43\ge43>0\)
\(\rightarrowĐPCM.\)
2) \(x^2+y^2-8x+4y+31\)
\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)
\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)
\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)
\(\rightarrowĐPCM.\)
3)\(16x^2+6x+25\)
\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)
\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)
\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)
\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)
-> ĐPCM.
4) Tương tự câu 3)
5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)
\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)
-> ĐPCM.
6) Tương tự câu 5)
7) 8) 9) Tương tự câu 3).
2
25 tháng 10 2021
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
2
25 tháng 10 2021
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
DD
1
15 tháng 12 2017
\(Q=\)\(1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(Q=1+\frac{x+3}{x^2+3x+2x+6}:\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right]\)
\(Q=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left[\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\left[\frac{2x+4-2x+2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(Q=1+\frac{1}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{6}\)
\(Q=1+\frac{x-2}{6}\)
\(Q=\frac{6+x-2}{6}\)
\(Q=\frac{x+4}{6}\)
b) khi \(Q=0\)thì \(\frac{x+4}{6}=0\)
\(\Rightarrow x+4=0\)
\(\Rightarrow x=-4\)
vậy \(x=-4\)khi \(Q=0\)
c) khi \(Q>0\)thì \(\frac{x+4}{6}>0\)
\(\Rightarrow x+4>0\)
\(\Leftrightarrow x>-4\)
vậy \(x>-4\)thì \(Q>0\)
HL
18 tháng 7 2016
a) (2x+3)(4x2-6x+9)-2(4x3-1)+(8x-1)=15
<=>8x3+27-8x3+2+8x-1=15
<=>8x+28=15
<=>8x=-13
<=>x=-13/8
b) (x+3)3-(x+9)(x2+27)-(5x-216) = 3x-4
<=>x3+9x2+27x+27-x3-27x-9x2-243-5x+216=3x-4
<=>-5x=3x-4
<=>8x=4
<=>x=1/2
AT
29 tháng 6 2017
a. \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)
\(\Leftrightarrow-2x+1-x-2=8\)
\(\Leftrightarrow-2x-x=8+2-1\)
\(\Leftrightarrow-3x=9\)
\(\Leftrightarrow x=-3\)
Vậy...............
b. \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow-4x^2+6x-2+4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)
\(\Leftrightarrow-4x^2+4x^2+6x-8x=-4+2\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy...............
19 tháng 8 2018
Bài 4 : Tìm x biết:
a, 4x2 - 49 = 0
\(\Leftrightarrow\) (2x)2 - 72 = 0
\(\Leftrightarrow\) (2x - 7)(2x + 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\2x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b, x2 + 36 = 12x
\(\Leftrightarrow\) x2 + 36 - 12x = 0
\(\Leftrightarrow\) x2 - 2.x.6 + 62 = 0
\(\Leftrightarrow\) (x - 6)2 = 0
\(\Leftrightarrow\) x = 6
19 tháng 8 2018
e, (x - 2)2 - 16 = 0
\(\Leftrightarrow\) (x - 2)2 - 42 = 0
\(\Leftrightarrow\) (x - 2 - 4)(x - 2 + 4) = 0
\(\Leftrightarrow\) (x - 6)(x + 2) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
f, x2 - 5x -14 = 0
\(\Leftrightarrow\) x2 + 2x - 7x -14 = 0
\(\Leftrightarrow\) x(x + 2) - 7(x + 2) = 0
\(\Leftrightarrow\) (x + 2)(x - 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)