将 $\frac{12xy^2}{15} \div \frac{2xy^3}{(3x)^2}$ 化简为最简的单个分数。已知 $\dfrac{v}{3} = \sqrt[3]{\dfrac{2s+v}{s-3}}$ ，用 $v$ 表示 $s$ 。解方程 $\dfrac{3}{2x-1}=\dfrac{5}{x-2}$ 。化简 $(x^2+y^2)(x^2-y^2)$ 。证明对于所有正整数 $n$ ， $2(9^{n+1})+3^{2n+3}-9^n$ 都是 11 的倍数。

分式乘除公式变形可化为一元一次方程的分式方程乘法公式指数幂运算法则代数符号理解整式乘法展开一元一次方程一次式的加减数系与四则运算指数去括号与提公因式 English

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

1((a))

1. Rewrite the divisor by simplifying $(3x)^2$ and change division to multiplication by the reciprocal.

$\dfrac{12xy^2}{15} \div \dfrac{2xy^3}{(3x)^2} = \dfrac{12xy^2}{15} \div \dfrac{2xy^3}{9x^2} = \dfrac{12xy^2}{15} \times \dfrac{9x^2}{2xy^3}$

2. Cancel common numerical and algebraic factors to obtain the simplest single fraction.

$\dfrac{12xy^2}{15} \times \dfrac{9x^2}{2xy^3} = \dfrac{12 \times 9 \times x \times x^2 \times y^2}{15 \times 2 \times x \times y^3} = \dfrac{18x^2}{5y}$

答案

$\dfrac{18}{5y}$

1((b))

1. Cube both sides to eliminate the cube root.

$\frac{v}{3}=\sqrt[3]{\frac{2s+v}{s-3}}$
Cubing both sides,
$\left(\frac{v}{3}\right)^3=\frac{2s+v}{s-3}$
$\frac{v^3}{27}=\frac{2s+v}{s-3}$

2. Clear the denominator and collect the terms involving $s$.

$v^3(s-3)=27(2s+v)$
$v^3s-3v^3=54s+27v$
$v^3s-54s=3v^3+27v$

3. Factorise and divide to express $s$ in terms of $v$.

$s(v^3-54)=3v^3+27v$
$s=\frac{3v^3+27v}{v^3-54}$
$s=\frac{3v(v^2+9)}{v^3-54}$

Equivalently, leaving the numerator as $3v^3+27v$ is acceptable.

答案

$s=\dfrac{v^3(v+81)}{v^3-54}$

1((c))((i))

1. State the restrictions and cross-multiply to remove the fractions.

The denominators must not be zero, so $x\ne \dfrac12$ and $x\ne 2$ .

$\frac{3}{2x-1}=\frac{5}{x-2}$
$3(x-2)=5(2x-1)$

2. Expand and solve the resulting linear equation.

$3x-6=10x-5$
$-1=7x$
$x=-\frac17$

答案

$x=-1$

1((c))((ii))

1. Apply the difference of two squares identity.

$(x^2+y^2)(x^2-y^2)=(x^2)^2-(y^2)^2=x^4-y^4$

答案

$x^4-y^4$

1((d))

1. Rewrite the powers so that each term is expressed in terms of $9^n$.

Since $9^{n+1}=9\cdot 9^n$ and
$3^{2n+3}=3^{2n}\cdot 3^3=(3^2)^n\cdot 27=27\cdot 9^n,$
we have
$2(9^{n+1})+3^{2n+3}-9^n=2(9\cdot 9^n)+27\cdot 9^n-9^n.$

2. Collect like terms involving $9^n$.

$2(9\cdot 9^n)+27\cdot 9^n-9^n=18\cdot 9^n+27\cdot 9^n-9^n$
$=(18+27-1)9^n=44\cdot 9^n.$

3. Factor out 11 to prove the expression is a multiple of 11.

$44\cdot 9^n=11(4\cdot 9^n).$
Since $n$ is a positive integer, $4\cdot 9^n$ is an integer. Hence $11(4\cdot 9^n)$ is a multiple of $11$ .

答案

$2(9^{n+1})+3^{2n+3}-9^n=11(4\times 9^n),$
so it is a multiple of $11$ for all positive integer values of $n$ .

AI 分析，可能出错

用 AI 批改，圈出写错的那一步，并自动将你的薄弱点存入错题本，形成你的错因库。