Search
You can find the results of your search below.
Fulltext results:
- Exercise 6.1
- )$, $54(27x^4-x)$ **Solution:**\\ (i) $\begin{align} x^2+5x+6&=x^2+3x+2x+6,\\ &=x(x+3)+2(x+3)\\ &=(x+3)(x+2) \end{align}$ $\begin{align} x^2-4x-12&=x^2-6x+2x-12,\\ &=x(x-6)+2(x-6)\\ &=(x-6)(x+2) \end{align}$ H.C.F= $x+2$ (ii) $\begin{align} x^3-27 &=x
- Exercise 6.3
- $4x^2-12xy +9y^2$\\ **Solution:**\\ $\begin{align}4x^2-12xy +9y^2\\&=4x^2-6xy-6xy +9y^2\\&= 2x(2x-3... )-3y(2x-3y)\\&= (2x-3y)(2x-3y)\\&= (2x-3y)^2 \end{align}$\\ $\begin{align} \sqrt{4x^2-12xy +9y^2}&= \pm (2x-3y) \end{align}$ (ii) $x^2-1+\frac{1}{4x^2}, (x\neq 0)$\\ **Solution
- Review exercise
- 128$ , $12x^3-96$\\ **Solution:**\\ $\begin{align}8x^4-128 &= 8(x^4-16)\\&=8[(x^2)^2-(4)^2]\\&= 2 \... &=2 \times 2\times 2 \times (x^2+4)(x-2)(x+2)\end{align}$\\ $\begin{align}12 x^3-96&=12(x^3-8)\\&=2 \times 2\times 3 \times (x-2)(x^2+ 2 x+4)\end{align}$\\ $\begin{align} H.C.F. &= 2 \times 2 (x-2
- Exercise 6.2
- 2+2x-24}{x^2-x-12}$\\ **Solution:**\\ $\begin{align} \frac{x^2-x-6}{x^2-9}&+\frac{x^2+2x-24}{x^2-x-12... {x+3}\\&=\frac{2x+8}{x+3}\\&=2\frac{x+4}{x+3}\end{align}$ ====Question 2:==== $\left[\frac{x+1}{x-1}-\f... ]+\frac{4x}{x^4-1}$\\ **Solution:**\\ $\begin{align} \left[\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{... \frac{2+1}{x^4-1}\right)\\&=\frac{12x}{x^4-1}\end{align}$\\ ====Question 3:==== $\frac{1}{x^2-8x+15}+\f
- Exercise 4.1
- \ (i) $\frac{120 x^2y^3z^5}{30x^3yz^2}$\\ $\begin{align}\frac{30\times 4 y^(3-1)z^(5-2)}{30x^(3-2)}\\&=\frac{4 y^2 z^3}{x}\end{align}$\\ **Solution:**\\ (ii) $\frac{8 a(x+1)}{2(x^2-1)}$\\ $\begin{align}\frac{2\times 4a(x+1)}{2(x+1)(x-1)}\\&= \frac{4a}{x-1}\end{align}$\\ **Solution:**\\ (iii) $\frac{(x+y)^2-4xy}{(x-
- Exercise 11.1 (Solutions) @matric:9th_science:unit11
- gle B=m\angle D=130^\circ$ We know that\\ \begin{align} & m\angle A +\,\,m\angle B=180^\circ \\ & m\ang... {130}^{\circ }}\\ & m\angle A={{50}^{\circ }}\end{align} Also we have $m\angle A=m\angle C$ (Opposite an... ngle DAB=?$, $m\angle C=?$, $m\angle D=?$ \begin{align} & m\angle DAM+m\angle DAB=180^\circ \\ & 40^\cir... 0^\circ- 40^\circ \\ & m\angle DAB=140^\circ \end{align} Also \begin{align} & m\angle DAB+m\angle B=180^\