Câu hỏi của Pham Thi Minh Phuong

Question

tìm x biết

a.(x-2)2-(x-3)(x+3)=6

b.9x2-4-(3x-2)(4x-5)=0

c.x2(x+3)-x2-3x=0

katherina · Accepted Answer

a. $\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\Leftrightarrow x^2-4x+4-x^2+9-6=0\Leftrightarrow-4x+7=0\Leftrightarrow x=\dfrac{7}{4}$

b. $9x^2-4-\left(3x-2\right)\left(4x-5\right)=0\Leftrightarrow9x^2-4-12x^2+23x-10=0\Leftrightarrow-3x^2+23x-14=0\Leftrightarrow\left(x-7\right)\left(-3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-7=0\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\x=\dfrac{2}{3}\end{matrix}\right.$

c. $x^2\left(x+3\right)-x^2-3x=0\Leftrightarrow x^3+2x^2-3x=0\Leftrightarrow x\left(x^2+2x-3\right)=0\Leftrightarrow x\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\x-1=0\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\x=-3\end{matrix}\right.$Câu trả lời: a. $\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\Leftrightarrow x^2-4x+4-x^2+9-6=0\Leftrightarrow-4x+7=0\Leftrightarrow x=\dfrac{7}{4}$

b. $9x^2-4-\left(3x-2\right)\left(4x-5\right)=0\Leftrightarrow9x^2-4-12x^2+23x-10=0\Leftrightarrow-3x^2+23x-14=0\Leftrightarrow\left(x-7\right)\left(-3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-7=0\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\x=\dfrac{2}{3}\end{matrix}\right.$

c. $x^2\left(x+3\right)-x^2-3x=0\Leftrightarrow x^3+2x^2-3x=0\Leftrightarrow x\left(x^2+2x-3\right)=0\Leftrightarrow x\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\x-1=0\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\x=-3\end{matrix}\right.$

Đời về cơ bản là buồn...... · Answer

a) $\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6$

$\Rightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)=6$

$\Rightarrow x^2-4x+4-x^2+9=6$

$\Rightarrow-4x+13=6$

$\Rightarrow-4x=-7$

$\Rightarrow x=1,75$Câu trả lời: a) $\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6$

$\Rightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)=6$

$\Rightarrow x^2-4x+4-x^2+9=6$

$\Rightarrow-4x+13=6$

$\Rightarrow-4x=-7$

$\Rightarrow x=1,75$

Mysterious Person · Answer

a) $\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\Leftrightarrow x^2-4x+4-\left(x^2-9\right)=6$

$\Leftrightarrow x^2-4x+4-x^2+9=6\Leftrightarrow4x=4+9-6\Leftrightarrow4x=7$

$\Leftrightarrow x=\dfrac{7}{4}$ vậy $x=\dfrac{7}{4}$

b) $9x^2-4-\left(3x-2\right)\left(4x-5\right)=0$

$\Leftrightarrow9x^2-4-\left(12x^2-15x-8x+10\right)=0$

$\Leftrightarrow9x^2-4-12x^2+15x+8x-10=0\Leftrightarrow-3x^2+23x-14=0$

$\Leftrightarrow-3x^2+2x+21x-14=0\Leftrightarrow-x\left(3x-2\right)+7\left(3x-2\right)=0$

$\Leftrightarrow\left(7-x\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}7-x=0\3x-2=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=7\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\x=\dfrac{2}{3}\end{matrix}\right.$

vậy $x=7;x=\dfrac{2}{3}$

c) $x^2\left(x+3\right)-x^2-3x=0\Leftrightarrow x^3+3x^2-x^2-3x=0$

$\Leftrightarrow x^3+2x^2-3x=0\Leftrightarrow x\left(x^2+2x-3\right)=0$

$\Leftrightarrow x\left(x^2+3x-x-3\right)=0\Leftrightarrow x\left(x\left(x+3\right)-\left(x+3\right)\right)=0$

$\Leftrightarrow x\left(x-1\right)\left(x+3\right)=0$ $\Leftrightarrow\left[{}\begin{matrix}x=0\x-1=0\x+3=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\x=-3\end{matrix}\right.$

vậy $x=0;x=1;x=-3$Câu trả lời: a) $\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\Leftrightarrow x^2-4x+4-\left(x^2-9\right)=6$