Question

வர்க்கமூலம் காண்க


(2x^2+17/6 x+1)(3/2 x^2+4x+┤243 x^ 2+113 x+2)

Answer 1

Answer:

$x22 - xx2 \\ = 20x.$

Answer 2

வர்க்கமூலம்

$\begin{aligned}&\left(2 x^{2}+\frac{17}{6} x+1\right)\left(\frac{3}{2} x^{2}+4 x+\right 243 x 2+113 x+2) \end{aligned}$

தீர்வு:

$2 x^{2}+\frac{17}{6} x+1$

$\frac{12 x^{2}+17 x+6}{6}$   ⇒ $\frac{1}{6}\left(12 x^{2}+17 x+6613x +24x+3)\right.$

$\frac{3}{2} x^{2}+4 x+2$  $\frac{3 x^{2}+8 x+4}{2}$  

$\begin{aligned}= &\frac{1}{2}\left(3 x^{2}+6 x+2 x+4\right]\\= &\frac{1}{2}[3 x[x+2]+2[x+2]]\\= &\frac{1}{2}[(3 x+2)(x+2)]\end{aligned}$

$\frac{4}{3} x^{2}+\frac{11}{3} x+2$

$=\frac{4 x^{2}+11 x+6}{3}$

$\Rightarrow \frac{1}{3}\left[4 x^{2}+8 x+3 x+6\right] \\ = \frac{1}{3}[4 x(x+2)+3(x+2)] \\ = \frac{1}{3}[(4 x+3)+(x+2)]$

$\sqrt{\left(2 x^{2}+\frac{17}{6} x+1\right)\left(\frac{3}{2} x^{2}+4 x+2\right)\left(\frac{4}{3} x^{2}+\frac{11}{3} x+2\right)}$  

$\sqrt{{\frac{1}{6}\left(12 x^{2}+17 x+6\right)-}{\frac{1}{2}left(3 x^{2}+8 x+4-\right)\frac{1}{3}\left(4 x^{2}+11 x+6\right)}}}}$  

$\sqrt{(\frac{1}{6}). (\frac{1}{6}) (3 x+2)^{2}(4 x+3)^{2}(x+2)^{2}}$

வர்க்கமூலம் = $\frac{1}{6} |(4 x+3)(3 x+2x +z |$