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2 tháng 7 2017
a) MTC : \(\left(x+1\right)\left(x^2-x+1\right)\)
Quy đồng :
\(\frac{x-1}{x^3+1}=\frac{x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2x}{x^2-x+1}=\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x+1}=\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
b ) MTC : \(10x\left(2y-x\right)\left(2y+x\right)\)
\(\frac{7}{5x}=\frac{7.2.\left(2y-x\right)\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=\frac{-4.10x.\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}=\frac{-40x\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
c ) MTC : \(\left(x+2\right)^3\)
\(\frac{6x^2}{x^3+6x^2+12x+8}=\frac{6x^2}{\left(x+2\right)^3}\)
\(\frac{3x}{x^2+4x+4}=\frac{3x}{\left(x+2\right)^2}=\frac{3x\left(x+2\right)}{\left(x+2\right)^3}\)
\(\frac{2}{2x+4}=\frac{1}{x+2}=\frac{\left(x+2\right)^2}{\left(x+2\right)^3}\)
TT
24 tháng 4 2019
a. \(x^2-4x+3\le0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(3x-3\right)\le0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\le0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\ge0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le1\\x\ge3\end{matrix}\right.\left(Vo.li\right)\\\left\{{}\begin{matrix}x\ge1\\x\le3\end{matrix}\right.\end{matrix}\right.\)
Vậy \(1\le x\le3\)
b. \(9x^2-6x\ge0\)
\(\Leftrightarrow3x\left(3x-2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\ge0\\3x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x\le0\\3x-2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge\frac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\le\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(0\le x\le\frac{2}{3}\)
c. Câu c cậu tự làm nha, tớ đang có việc. Quy đồng lên rồi tính bình thường thôi.
VD
27 tháng 8 2017
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
1 tháng 9 2020
a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
10 tháng 12 2016
b/
\(\frac{1}{x^3-1}=\frac{a}{x-1}+\frac{6x+c}{x^2+x+1}=\frac{\left(a+6\right)x^2+\left(c+a-6\right)x-c+a}{x^3-1}\)
Đồng nhất thức 2 vế ta được
\(\hept{\begin{cases}a+6=0\\c+a-6=0\\a-c=1\end{cases}}\)
Vô nghiệm vậy không tồn tại a, c thỏa cái đó
10 tháng 12 2016
a/ Ta có
\(\frac{10x-4}{x^3-4x}=\frac{a}{x}+\frac{b}{x-2}+\frac{c}{x+2}=\frac{\left(a+b+c\right)x^2+\left(2b-2c\right)x-4a}{x^3-4x}\)
Đồng nhất thức 2 vế ta được
\(\hept{\begin{cases}a+b+c=0\\2b-2c=10\\-4a=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=-3\end{cases}}\)
17 tháng 2 2020
f/ ĐKXĐ: x khác 0
\(\Leftrightarrow\frac{1}{x}+2=2x^2+x+4+\frac{2}{x}\)
\(\Leftrightarrow2x^2+x+2+\frac{1}{x}=0\)
\(\Leftrightarrow x\left(2x+1+\frac{2}{x}+\frac{1}{x^2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\2x+1+\frac{2x+1}{x^2}=0\end{matrix}\right.\)
\(\Rightarrow\left(2x+1\right)\left(1+\frac{1}{x^2}\right)=0\Rightarrow x=-\frac{1}{2}\)( vì 1+1/x^2>0)
17 tháng 2 2020
a/\(\Leftrightarrow\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{x+4}{\left(x-1\right)\left(x-2\right)}-\frac{x+4}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\left(x+4\right)\left(\frac{1}{\left(x-1\right)\left(x-2\right)}-\frac{1}{\left(x-1\right)\left(x-3\right)}\right)=0\)
\(\Rightarrow x=-4\)
-4x + 6x2 = 0
=> -2x(3x + 2) = 0
=> x(3x + 2) = 0
=> \(\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{2}{3}\end{cases}}\)
b) \(\frac{1}{4}x^2-x=3\)
=> \(\frac{1}{4}x^2-x-3=0\)
=> \(\left(\frac{1}{2}x\right)^2-2.\frac{1}{2}x+1-4=0\)
=> \(\left(\frac{1}{2}x-1\right)^2-2^2=0\)
=> \(\left(\frac{1}{2}x-1-2\right)\left(\frac{1}{2}x-1+2\right)=0\)
=> \(\left(\frac{x}{2}-3\right)\left(\frac{x}{2}+1\right)=0\)
=> \(\orbr{\begin{cases}\frac{x}{2}-3=0\\\frac{x}{2}+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)
a, \(-4x+6x^2=0\Leftrightarrow2x\left(-2+3x\right)=0\)
TH1 : x = 0
TH2 : \(-2+3x=0\Leftrightarrow x=\frac{2}{3}\)
b, \(\frac{1}{4}x^2-x=3\Leftrightarrow\frac{1}{4}x^2-x-3=0\)
\(\Leftrightarrow\left(\frac{1}{2}x-1\right)^2-4=0\Leftrightarrow\left(\frac{1}{2}x-3\right)\left(\frac{1}{2}x+1\right)=0\)
TH1 : \(\frac{1}{2}x-3=0\Leftrightarrow\frac{1}{2}x=3\Leftrightarrow x=6\)
TH2 : \(\frac{1}{2}x+1=0\Leftrightarrow\frac{1}{2}x=-1\Leftrightarrow x=-2\)