tìm x
7x\(^2\)+6x\(=\)5x
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Ta có :
\(\left(1-6x\right)^2+2\left(6x-1\right)\left(2-5x\right)+\left(2-5x\right)^2\)
\(=\)\(\left(6x-1\right)^2+2\left(6x-1\right)\left(2-5x\right)+\left(2-5x\right)^2\)
\(=\)\(\left(6x-1+2-5x\right)^2\)
\(=\)\(\left(x+1\right)^2\)
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\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
\(\dfrac{6x+5}{5x-3}=\dfrac{38-6x}{21-5x}\left(đk:x\ne\dfrac{5}{3},x\ne\dfrac{21}{5}\right)\)
\(\Rightarrow\left(6x+5\right)\left(21-5x\right)=\left(5x-3\right)\left(38-6x\right)\)
\(\Rightarrow-30x^2+101x+105=-30x^2+208x-114\)
\(\Rightarrow107x=219\Rightarrow x=\dfrac{219}{107}\left(tm\right)\)
\(\dfrac{6x+5}{5x-3}=\dfrac{6x-38}{5x-21}\)
\(\Leftrightarrow30x^2-126x+25x-105=30x^2-18x-190x+114\)
\(\Leftrightarrow-101x-105=-208x+114\)
\(\Leftrightarrow107x=219\)
hay \(x=\dfrac{219}{107}\)
5x2-6x=0
\(\Leftrightarrow x\left(5x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{6}{5}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{6}{5}\)
a) \(6x^2-2x=2x\left(3x-1\right)\)
\(2x\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}2x=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{0;\dfrac{1}{3}\right\}\)
b) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+3\right)\left(x+2\right)\)
\(\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-3;-2\right\}\)
a) x^2 + 4y^2 + 6x - 12y + 18 = 0
<=>x2+6x+9+4y2-12y+9=0
<=>(x+3)2+(2y-3)2=0
<=>x+3=0 và 2y-3=0
<=>x=-3 và y=3/2
b) 5x^2 +9y^2 - 12xy - 6x +9 = 0
<=>x2-6x+9+4x2-12xy+9y2=0
<=>(x-3)2+(2x-3y)2=0
<=>x-3=0 và 2x-3y=0
<=>x=3 và 2.3-3y=0
<=>x=3 và y=2
7x2 + 6x = 5x
,=> 7x2 + 6x - 5x = 0
=> 7x2 + x = 0
=> x(7x + 1) = 0
=> x = 0 hoặc 7x + 1 =0
=> x = 0 hoặc x = -1/7
Vậy...
7x2 + 6x = 5x
=> 7x2 + 6x - 5x = 0
7x2 + x = 0
x.(7x+1) = 0
=> x = 0
7x + 1 = 0 => 7x = -1 => x = -1/7
KL: x = 0 hoặc x = -1/7