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a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b) \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2-\left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=-6\)
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b, \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x- 5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2- \left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=6\)
Giải:
1) \(\left(x-6\right)\left(x^2+6x+36\right)-\left(x+4\right)^3=\left(x-2\right)^3+\left(x+5\right)\left(x^2-10x+25\right)-\left(2x^3+6x^2\right)\)
\(\Leftrightarrow x^3-216-\left(x^3+12x^2+48x+64\right)=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow x^3-216-x^3-12x^2-48x-64=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow-280-12x^2-48x=-12x^2+12x+117\)
\(\Leftrightarrow-280-48x-12x-117=0\)
\(\Leftrightarrow-397-60x=0\)
\(\Leftrightarrow-60x=397\)
\(\Leftrightarrow x=-\dfrac{397}{60}\)
Vậy ...
2) \(\left(2x+3\right)^3-\left(2x+5\right)\left(4x^2-10x+25\right)=\left(6x-1\right)^2-\left(x-2\right)\left(x^2+2x+4\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-\left(8x^3+125\right)=36x^2-12x+1-\left(x^3-8\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3-125=36x^2-12x+1-x^3+8+x^3\)
\(\Leftrightarrow54x-98=-12x+9\)
\(\Leftrightarrow54x+12x=9+98\)
\(\Leftrightarrow66x=107\)
\(\Leftrightarrow x=\dfrac{107}{66}\)
Vậy ...
1: \(\Leftrightarrow5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x\)
\(\Leftrightarrow3x^2+16x-19=3x^2-4x-18\)
=>20x=1
hay x=1/20
2: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-\left(x^2-2x\right)+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow-20x-41=-6x+27\)
=>-14x=68
hay x=-34/7
a.(2x - 5)(3x + 4) - x(6x - 5) = 4
⇔ 6x2 +8x -15x-20-6x2+5x=4
⇔-2x=24
⇔ x=-12
vậy x=12
b.(x - 2)2 + x(x - 2) = 0
⇔(x-2)(x-2+x)=0
⇔(x-2) (2x-2)=0
⇔ (x-2)2(x-2)=0
⇔(x-2)2.2=0
⇔(x-2)2=0
⇔x-2=0
⇔x=2
vậy x=2
c.(x3 + 4x2 - x - 4) : (x + 4) = 0
⇔[(x3+4x2)-(x+4)] :(x+4)=0
⇔ [x2(x+4)-(x+4)] :(x+4)=0
⇔ (x+4)(x2-1):(x+4)=0
⇔(x-1)(x+1)=0
⇔ \(\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)
\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)
\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)
\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)
\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)
\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)
\(\Leftrightarrow4x^2+6x-51=0\)
\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)
Câu hỏi của Vogsi Tú Anh - Toán lớp 8 | Học trực tuyến