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30 tháng 10 2020
a) \(\frac{3x+6}{x^2-4}=\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3}{x-2}\)( ĐKXĐ : x ≠ ±2 )
\(\frac{2x+6}{x^3+3x^2-9x-27}=\frac{2\left(x+3\right)}{x^2\left(x+3\right)-9\left(x+3\right)}=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x^2-9\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)( ĐKXĐ : x ≠ ±3 )
MTC : ( x - 2 )( x - 3 )( x + 3 )
=> \(\hept{\begin{cases}\frac{3}{x-2}=\frac{3\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3\left(x^2-9\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3x-27}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\\\frac{2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{4x-4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\end{cases}}\)
b) \(\frac{x^2-4x+4}{2x^2-3x+1}=\frac{\left(x-2\right)^2}{2x^2-2x-x+1}=\frac{\left(x-2\right)^2}{2x\left(x-1\right)-\left(x-1\right)}=\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}\)( ĐKXĐ : \(\hept{\begin{cases}x\ne1\\x\ne\frac{1}{2}\end{cases}}\))
\(\frac{x+4}{2x-2}=\frac{x+4}{2\left(x-1\right)}\)( ĐKXĐ : x ≠ 1 )
MTC : \(2\left(x-1\right)\left(2x-1\right)\)
=> \(\hept{\begin{cases}\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}=\frac{2\left(x^2-4x+4\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2-8x+8}{2\left(x-1\right)\left(2x-1\right)}\\\frac{x+4}{2\left(x-1\right)}=\frac{\left(x+4\right)\left(2x-1\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2+7x-4}{2\left(x-1\right)\left(2x-1\right)}\end{cases}}\)
c) \(\frac{6a}{a-b}\)( ĐKXĐ : a ≠ b ) ; \(\frac{2b}{b-a}=\frac{-2b}{a-b}\)( ĐKXĐ : a ≠ b) ; \(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)( ĐKXĐ : a ≠ ±b )
MTC : \(\left(a-b\right)\left(a+b\right)\)
=> \(\frac{6a}{a-b}=\frac{6a\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{6a^2+6ab}{\left(a-b\right)\left(a+b\right)}\)
\(\frac{-2b}{a-b}=\frac{-2b\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{-2ab-2b^2}{\left(a-b\right)\left(a+b\right)}\)
\(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)
d) \(\frac{x}{x^2+11x+30}=\frac{x}{x^2+5x+6x+30}=\frac{x}{x\left(x+5\right)+6\left(x+5\right)}=\frac{x}{\left(x+5\right)\left(x+6\right)}\)( ĐKXĐ : x ≠ -5 ; x ≠ -6 )
\(\frac{5}{x^2+9x+20}=\frac{5}{x^2+4x+5x+20}=\frac{5}{x\left(x+4\right)+5\left(x+4\right)}=\frac{5}{\left(x+4\right)\left(x+5\right)}\)( ĐKXĐ : x ≠ -4 ; x ≠ -5 )
MTC : \(\left(x+4\right)\left(x+5\right)\left(x+6\right)\)
=> \(\hept{\begin{cases}\frac{x}{\left(x+5\right)\left(x+6\right)}=\frac{x\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{x^2+4x}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\\\frac{5}{\left(x+4\right)\left(x+5\right)}=\frac{5\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{5x+30}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\end{cases}}\)
Sai chỗ nào bạn bỏ qua nhé
27 tháng 11 2018
1/ Ta có
\(x^2+9x+20=x^2+4x+5x+20=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
Tương tự
\(x^2+11x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
Đk: x khác 4, 5, 6, 7
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\frac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}+\frac{\left(x+7\right)-\left(x+6\right)}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\) EM tự làm tiếp nhé
21 tháng 10 2018
a) \(x^2-6x+9-9y^2=x^2-2\cdot3+3^2-\left(3y\right)^2=\left(x-3\right)^2-\left(3y\right)^2=\left(x-3-3y\right)\cdot\left(x-3+3y\right)\)
21 tháng 10 2018
b) \(x^3-3x^2+2x-1+2\cdot\left(x^2-x\right)=\left(x-1\right)^3+2x\cdot\left(x-1\right)=\left(x-1\right)\cdot\left[\left(x-1\right)^2+2x\right]\)
18 tháng 4 2017
a) ĐKXĐ: \(x\ne-1,x\ne0\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
<=> \(\dfrac{x\left(x+3\right)+\left(x-2\right)\left(x+1\right)-2x\left(x+1\right)}{x\left(x+1\right)}=0\)
<=> \(\dfrac{x^2+3x+x^2-x-2-2x^2-2x}{x\left(x+1\right)}=0\)
<=> \(\dfrac{-2}{x\left(x+1\right)}=0\) (vô lí)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne3,x\ne-2\)
ta có:\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\)
<=> \(\dfrac{\left(x+2\right)\left(3-x\right)+x\left(x+2\right)-5x-2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\)
<=> \(\dfrac{x-x^2+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\)
<=> \(\dfrac{0}{\left(x+2\right)\left(3-x\right)}=0\) (luôn đúng)
Vậy pt trên luôn đúng với mọi x khác 3 và -2
LM
18 tháng 4 2017
a) \(\dfrac{x+3}{x+1}\)+\(\dfrac{x-2}{x}\)=2
(đk: x\(\ne\); x\(\ne\)-1)
<=> \(x^2\)+3x + \(x^2\)-x -2 =\(2x^2\)+2x
<=> 2x -2 =2x
<=>0x=2
=>Pt vô nghiệm.
b) 1+ \(\dfrac{x}{3-x}\)= \(\dfrac{5x}{\left(x+2\right)\left(3-x\right)}\)+\(\dfrac{2}{x+2}\)
(đk:x\(\ne\)3; x\(\ne\)-2)
<=> 3x +6=3x+6
<=>0x=0
=> Pt vô số no.
c)\(\dfrac{3x+2}{3x-2}\)-\(\dfrac{6}{2+3x}\)=\(\dfrac{9x^2}{9x^2-4}\)
(đk: x\(\ne\)\(\pm\)\(\dfrac{2}{3}\))
<=>\((3x+2)^2\)-6(3x-2)=\(9x^2\)
<=>\(9x^2 \)+12x +4 -18x+12=\(9x^2\)
<=>16-6x=0
<=>6x=16
<=> x=\(\dfrac{8}{3}\)(t/m)
Vậy pt có no duy nhất là x=\(\dfrac{8}{3}\)
22 tháng 7 2018
\(A=x^2-10x+30=x^2-10x+25+5=\left(x-5\right)^2+5\ge5\)
Vậy GTNN của A là 5 khi x = 5
\(B=4x^2+4x+9=4x^2+4x+1+8=\left(2x+1\right)^2+8\ge8\)
Vậy GTNN của B là 8 khi x = \(-\dfrac{1}{2}\)
\(C=9x^2-12x+20=9x^2-12+4+16=\left(3x-2\right)^2+16\ge16\)
Vậy GTNN của C là 16 khi x = \(\dfrac{2}{3}\)
\(D=x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN của D là \(\dfrac{3}{4}\) khi x = \(-\dfrac{1}{2}\)
\(E=2x^2+3x+5=2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{31}{8}=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{31}{8}\ge\dfrac{31}{8}\)
Vậy GTNN của E là \(\dfrac{31}{8}\) khi x = \(-\dfrac{3}{4}\)
\(F=3x^2-7x+6=3\left(x^2-\dfrac{7}{3}x+\dfrac{49}{36}\right)+\dfrac{23}{12}=\left(x-\dfrac{7}{6}\right)^2\ge\dfrac{23}{12}\)Vậy GTNN của F là \(\dfrac{23}{12}\) khi x = \(\dfrac{7}{6}\)
\(\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)=112\)
\(\left(x^2+3x-4\right)\left(x^2+3x-10\right)=112\)
Đặt \(t=x^2+3x-4\)
\(t\left(t-6\right)=112\Rightarrow t^2-6t-112=0\Leftrightarrow\left(t+8\right)\left(t-14\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+3x-4=-8\\x^2+3x-4=14\end{cases}\Rightarrow\orbr{\begin{cases}\left(l\right)\\x=3,x=-6\end{cases}}}\)