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Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)
\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)
\(\Leftrightarrow15x^2-4x-4=0\)
\(\Leftrightarrow15x^2-10x+6x-4=0\)
Lỗi :vvvv
\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...
\(\left(\dfrac{2x}{3}-\dfrac{1}{3}\right)+\left(3x-2x+1\right)=8\)
\(\Leftrightarrow\dfrac{2x-1}{3}+x-7=0\Rightarrow2x-1+3x-21=0\Leftrightarrow x=\dfrac{22}{5}\)
\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)+\left[3x-2\left(x-1\right)\right]=8\)
\(\Rightarrow\dfrac{2}{3}x-\dfrac{1}{3}+3x-2x+2=8\)
\(\Rightarrow\dfrac{5}{3}x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{5}\)
\(1,\\ a,=x^2+6x+9-x^2-6x=9\\ b,=3x-1+6x-9x^2+x-10=-9x^2+10x-11\\ 2,\\ a,=4xy\left(x^2-2xy+y^2\right)=4xy\left(x-y\right)^2\)
( 3x - 1 )( x + 3 ) + 9x2 - 1 = 0
<=> 3x2 + 9x - x - 3 + 9x2 - 1 = 0
<=> 12x2 + 8x - 4 = 0
<=> 4( 3x2 + 2x - 1 ) = 0
<=> 3x2 + 2x - 1 = 0
<=> 3x2 + 3x - x - 1 = 0
<=> ( 3x2 + 3x ) - ( x + 1 ) = 0
<=> 3x( x + 1 ) - 1( x + 1 ) = 0
<=> ( 3x - 1 )( x + 1 ) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
Vậy S = { 1/3 ; -1 }
\(\frac{x+1}{3}>\frac{3x-2}{5}\)
\(\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)
\(\Leftrightarrow5x+5>9x-6\)
\(\Leftrightarrow5x-9x>-6-5\)
\(\Leftrightarrow-4x>-11\)
\(\Leftrightarrow x< \frac{11}{4}\)
Bài làm:
a) \(\left(3x-1\right)\left(x+3\right)+9x^2-1=0\)
\(\Leftrightarrow3x^2+8x-3+9x^2-1=0\)
\(\Leftrightarrow12x^2+8x-4=0\)
\(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(3x^2+3x\right)-\left(x+1\right)=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
Vậy tập nghiệm của PT \(S=\left\{-1;\frac{1}{3}\right\}\)
b) \(\frac{x+1}{3}>\frac{3x-2}{5}\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)
\(\Rightarrow5x+5>9x-6\)
\(\Leftrightarrow4x< 11\)
\(\Rightarrow x< \frac{11}{4}\)
\(\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\dfrac{36}{49}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\left(\dfrac{6}{7}\right)^2\\ \Rightarrow\dfrac{1}{2}-\dfrac{x}{3}=\pm\dfrac{6}{7}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-\dfrac{x}{3}=\dfrac{6}{7}\\\dfrac{1}{2}-\dfrac{x}{3}=-\dfrac{6}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{5}{14}\\\dfrac{x}{3}=\dfrac{19}{14}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{14}\times3\\x=\dfrac{19}{14}\times3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{14}\\x=\dfrac{57}{14}\end{matrix}\right.\)
\(\left(3-\dfrac{2}{3}x\right)^3=-\dfrac{1}{64}\\ \Rightarrow\left(3-\dfrac{2}{3}x\right)^3=\left(-\dfrac{1}{4}\right)^3\\ \Rightarrow3-\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=3-\left(-\dfrac{1}{4}\right)\\ \Rightarrow\dfrac{2}{3}x=\dfrac{13}{4}\\ \Rightarrow x=\dfrac{13}{4}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{13}{4}\times\dfrac{3}{2}\\ \Rightarrow x=\dfrac{39}{8}\)
Hic 2 câu em làm dr xong tự nhiên thử lung tung rồi lại xóa bài ;-;
\(\left(x+1\right)\left(2x-2\right)-3>-5x-\left(2x+1\right)\left(3x-x\right)\)
\(\Leftrightarrow2x^2-2x+2x-2-3>-5x-\left(6x^2-2x^2+3x-x\right)\)
\(\Leftrightarrow2x^2-2x+2x-5>-5x-6x^2+2x^2-3x+x\)
\(\Leftrightarrow2x^2-5+5x+6x^2-2x^2+3x-x>0\)
\(\Leftrightarrow6x^2-2x>5\)
\(\Leftrightarrow2x\left(3x-1\right)>5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x>5\\3x-1>5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{5}{2}\\x>2\end{matrix}\right.\)
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