解方程 $\displaystyle \frac{3-2x}{7} = \frac{1}{3}(2x-1).$ 答 x = ... 化简 $\displaystyle \frac{25a^{2}-9b^{2}}{15a-10ab+6b^{2}-9b}$ 答 ... $\displaystyle \frac{3c}{4ab}\div\frac{c^{3}}{12a^{2}b}$ 答 ... 用配方法解方程 $-x^{2}+9x-\frac{7}{2}=0$。将你的解保留到两位小数。 答 x = ... 或 ... 解释为什么 $-x^{2}+9x-\frac{7}{2}=18$ 没有实数解。

1. Clear the fractions by multiplying both sides by $21$.

$\frac{3-2x}{7}=\frac{1}{3}(2x-1)$
Multiply both sides by $21$:
$3(3-2x)=7(2x-1)$

2. Expand and solve for $x$.

$9-6x=14x-7$
$16=20x$
$x=\frac{16}{20}=\frac{4}{5}$

1. Factorise the numerator using the difference of two squares.

$25a^2-9b^2=(5a)^2-(3b)^2=(5a+3b)(5a-3b)$

2. Factorise the denominator by grouping.

$15a-10ab+6b^2-9b$

$=5a(3-2b)-3b(3-2b)$

$=(3-2b)(5a-3b)$

3. Cancel the common algebraic factor and write the simplified expression.

$\dfrac{25a^2-9b^2}{15a-10ab+6b^2-9b}=\dfrac{(5a+3b)(5a-3b)}{(3-2b)(5a-3b)}$

$=\dfrac{5a+3b}{3-2b}$

1. Rewrite the division as multiplication by the reciprocal.

$\dfrac{3c}{4ab}\div\dfrac{c^{3}}{12a^{2}b}=\dfrac{3c}{4ab}\times\dfrac{12a^{2}b}{c^{3}}$

2. Cancel common factors and simplify the product.

$\dfrac{3c}{4ab}\times\dfrac{12a^{2}b}{c^{3}}=\dfrac{3\times12\times c\times a^{2}\times b}{4\times a\times b\times c^{3}}=\dfrac{9a}{bc^{2}}$

1. Rearrange the quadratic equation so the coefficient of $x^2$ is positive.

$-x^{2}+9x-\dfrac{7}{2}=0$

Multiply throughout by $-1$:

$x^{2}-9x+\dfrac{7}{2}=0$

2. Complete the square for $x^2-9x$.

$x^{2}-9x+\dfrac{7}{2}=0$

$\left(x-\dfrac{9}{2}\right)^2-\left(\dfrac{9}{2}\right)^2+\dfrac{7}{2}=0$

$\left(x-\dfrac{9}{2}\right)^2-\dfrac{81}{4}+\dfrac{14}{4}=0$

$\left(x-\dfrac{9}{2}\right)^2-\dfrac{67}{4}=0$

3. Take square roots and solve for $x$.

$\left(x-\dfrac{9}{2}\right)^2=\dfrac{67}{4}$

$x-\dfrac{9}{2}=\pm \sqrt{\dfrac{67}{4}}$

$x=\dfrac{9}{2}\pm \dfrac{\sqrt{67}}{2}$

4. Evaluate and round both solutions to two decimal places.

$x=\dfrac{9+\sqrt{67}}{2}\quad \text{or}\quad x=\dfrac{9-\sqrt{67}}{2}$

Using a calculator,

$x=8.592\ldots \quad \text{or}\quad x=0.407\ldots$

Correct to two decimal places:

$x=8.59 \quad \text{or}\quad x=0.41$

1. Write the quadratic expression in completed-square form.

$-x^{2}+9x-\frac{7}{2}=-\left(x^{2}-9x\right)-\frac{7}{2}$
$=-\left[\left(x-\frac{9}{2}\right)^2-\frac{81}{4}\right]-\frac{7}{2}$
$=-\left(x-\frac{9}{2}\right)^2+\frac{81}{4}-\frac{14}{4}$
$=-\left(x-\frac{9}{2}\right)^2+\frac{67}{4}$

2. Explain why the expression cannot be equal to 18 for any real value of $x$.

For real $x$,
$\left(x-\frac{9}{2}\right)^2\geq 0.$
So
$-\left(x-\frac{9}{2}\right)^2\leq 0,$
and hence
$-\left(x-\frac{9}{2}\right)^2+\frac{67}{4}\leq \frac{67}{4}=16.75.$
Since $16.75<18$, the equation cannot be satisfied by any real $x$.