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30 tháng 12 2018
\(a^2-2a+b^2+4b+4c^2-4c+6=0\)'
\(\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)
\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
b tự làm nốt nhé~
30 tháng 12 2018
\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)
\(M=x^3+3^3-x^3-54+x\)
\(M=x+27-54\)
\(M=x+27-54\)
\(M=7-27\)
\(M=-20\)
VQ
1
12 tháng 3 2020
\(M=a+\frac{\left(2a+b\right)\left(2+b\right)-\left(2a-b\right)\left(2-b\right)}{4-b^2}-\frac{4a}{4-b^2}.\)
\(=a+\frac{4b\left(a+1\right)-4a}{4-b^2}\)
Ta có \(4ab+4b-4a=4\left[\frac{a^2}{a+1}+\frac{a}{a+1}-4a\right]=-12a\)
\(4-b^2=4-\frac{a^2}{\left(a+1\right)^2}=\frac{4\left(a^2+2a+1\right)-a^2}{\left(a+1\right)^2}=\frac{3a^2+8a+4}{\left(a+1\right)^2}\)
\(\Rightarrow M=a+\frac{-12a\left(a+1\right)^2}{3a^2+8a+4}\)
\(=-\frac{9a^3+16a^2+8a}{3a^2+8a+4}\)
12 tháng 3 2020
\(M=a+\frac{2a+b}{2-b}-\frac{2a-b}{2+b}+\frac{4a}{b^2-4}\)
\(=a-\frac{2a+b}{b-2}-\frac{2a-b}{2+b}+\frac{4a}{b^2-4}\)
\(=a-\frac{\left(2a+b\right)\left(2+b\right)+\left(2a-b\right)\left(b-2\right)}{\left(b-2\right)\left(b+2\right)}+\frac{4a}{b^2-4}\)
\(=a-\frac{4b\left(a+1\right)}{b^2-4}+\frac{4a}{b^2-4}\)
\(=a-\frac{4\frac{a}{a+1}\left(a+1\right)}{b^2-4}+\frac{4a}{b^2-4}\)
\(=a-\frac{4a}{b^2-4}+\frac{4a}{b^2-4}\)
\(=a\)
11 tháng 3 2020
Hắc hắc :P Cứ làm từ từ sẽ thành công em ạ :D
\(=\frac{a+b+a-b}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a\left(a^2+b^2\right)+2a\left(a^2-b^2\right)}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3\left(a^4+b^4\right)+4a^3\left(a^4-b^4\right)}{a^8-b^8}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{8a^7\left(a^8+b^8\right)+8a^7\left(a^8-b^8\right)}{\left(a^8-b^8\right)\left(a^8+b^8\right)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)
H
30 tháng 1 2019
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
H
30 tháng 1 2019
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
4 tháng 12 2018
a, Để C có nghĩa thì \(\hept{\begin{cases}2x-2\ne0\\2-2x\ne0\end{cases}\Rightarrow}x\ne1\)
b, Với x khác 1 thì
\(C=\frac{x}{2x-2}+\frac{x^2+1}{2-2x}=\frac{-x}{2-2x}+\frac{x^2+1}{2-2x}=\frac{x^2-x+1}{2-2x}\)
c, \(C=-0,5\Rightarrow\frac{x^2-x+1}{2-2x}=\frac{-1}{2}\)
\(\Rightarrow2\left(x^2-x+1\right)=\left(2-2x\right).\left(-1\right)\)
\(\Rightarrow2x^2-2x+2=-2+2x\)
\(\Rightarrow2x^2-2x+2+2-2x=0\)
\(\Rightarrow2x^2-4x+4=0\Rightarrow2\left(x^2-2x+2\right)=0\)
\(x^2-2x+2=\left(x-1\right)^2+1>0\forall x\)
Do đó: \(2\left(x^2-2x+2\right)>0\forall x\)
Vậy \(x\in\varnothing\)
MN
11 tháng 2 2020
Đề sai ! Sửa nhé :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(\Leftrightarrow A=\left(\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x-2}\right)\)
\(\Leftrightarrow A=\frac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\frac{2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow A=\frac{2x+4-4}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(\Leftrightarrow A=\frac{2x\left(x-2\right)}{-x\left(x+2\right)}\)
\(\Leftrightarrow A=-\frac{2\left(x-2\right)}{x+2}\)
b) Để \(A\le-2\)
\(\Leftrightarrow-\frac{2\left(x-2\right)}{x+2}\le-2\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{x+2}\ge2\)
\(\Leftrightarrow\frac{x-2}{x+2}\ge1\)
\(\Leftrightarrow x-2\ge x+2\)
\(\Leftrightarrow-2\ge2\)(ktm)
Vậy để \(A\le-2\Leftrightarrow x\in\varnothing\)
Q
11 tháng 2 2020
a.
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2.\left(x^2+8\right)}{\left(x+2\right).\left(x^2+8\right)}-\frac{4\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2x^2+8-4x+8}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)\left(-x\right)}\right)\)
\(A=\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^2-4x+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^3-4x-4x-4x^2+8x+16x-32\right)}{-x^3+8}\)
\(A=\frac{2x^3-4x^2+16x-32}{-x^3+8}\)
24 tháng 3 2020
\(2x^2+y^2+9=6x+2xy\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-y=0\end{cases}}\Leftrightarrow x=y=3\)
\(\Rightarrow A=x^{2019}.y^{2020}-x^{2020}.y^{2019}+\frac{1}{9xy}=\frac{1}{27}\)
\(\left(a-b\right)\left(a^2+ab+b^2\right)=a\left(a^2+ab+b^2\right)-b\left(a^2+ab+b^2\right)\)
\(=a^3+a^2b+ab^2-a^2b-ab^2-b^3\)
\(=a^3-b^3\)
\(\left(a+b\right)\left(a^2-ab+b^2\right)=a\left(a^2-ab+b^2\right)+b\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)
\(=a^3+b^3\)
(a-b)(a^2+ab+b^2)=a^3-b^3
(a+b)(a^2-ab+b^2)=a^3+b^3