Question

6/5 ab×2b^2×0 solve this​

Answer 1

Answer:

answers is o

bcz any number multiplied by 0 and the answer is 0

Answer 2

x=3a+2b or x=32a+b .    

To find:

Solve by Factorization: 9 \mathrm{x}^{2}-9(\mathrm{a}+\mathrm{b}) \mathrm{x}+\left(2 \mathrm{a}^{2}+5 \mathrm{ab}+2 \mathrm{b}^{2}\right)=09x2−9(a+b)x+(2a2+5ab+2b2)=0

Solution:

\begin{gathered}\begin{array}{l}{9 \mathrm{x}^{2}-9(\mathrm{a}+\mathrm{b}) \mathrm{x}+(2 \mathrm{a} 2+4 \mathrm{ab}+\mathrm{ab}+2 \mathrm{b} 2)=0} \\ \\ {9 \mathrm{x}^{2}-(9 \mathrm{a}+9 \mathrm{b}) \mathrm{x}+[2 \mathrm{a}(\mathrm{a}+2 \mathrm{b})+\mathrm{b}(\mathrm{a}+2 \mathrm{b})]=0}\end{array}\end{gathered}9x2−9(a+b)x+(2a2+4ab+ab+2b2)=09x2−(9a+9b)x+[2a(a+2b)+b(a+2b)]=0

\begin{gathered}\begin{array}{l}{9 \mathrm{x}^{2}-(9 \mathrm{a}+9 \mathrm{b}) \mathrm{x}+[(\mathrm{a}+2 \mathrm{b})(2 \mathrm{a}+\mathrm{b})]=0} \\ \\ {9 \mathrm{x}^{2}-3[(\mathrm{a}+2 \mathrm{b})+(2 \mathrm{a}+\mathrm{b})] \mathrm{x}+\{(\mathrm{a}+2 \mathrm{b})(2 \mathrm{a}+\mathrm{b})\}=0}\end{array}\end{gathered}9x2−(9a+9b)x+[(a+2b)(2a+b)]=09x2−3[(a+2b)+(2a+b)]x+{(a+2b)(2a+b)}=0

\begin{gathered}\begin{array}{l}{9 \mathrm{x}^{2}-3(\mathrm{a}+2 \mathrm{b}) \mathrm{x}-3(2 \mathrm{a}+\mathrm{b}) \mathrm{x}+\{(\mathrm{a}+2 \mathrm{b})(2 \mathrm{a}+\mathrm{b})\}=0} \\ \\ {3 \mathrm{x}[3 \mathrm{x}-(\mathrm{a}+2 \mathrm{b})]-(2 \mathrm{a}+\mathrm{b})[3 \mathrm{x}+(\mathrm{a}-2 \mathrm{b})]=0}\end{array}\end{gathered}9x2−3(a+2b)x−3(2a+b)x+{(a+2b)(2a+b)}=03x[3x−(a+2b)]−(2a+b)[3x+(a−2b)]=0

[3 \mathrm{x}-(\mathrm{a}+2 \mathrm{b})][3 \mathrm{x}-(2 \mathrm{a}+\mathrm{b})]=0[3x−(a+2b)][3x−(2a+b)]=0

[3 x-(a+2 b)]=0[3x−(a+2b)]=0 or we can have

[3 x-(2 a+b)]=0[3x−(2a+b)]=0

3 \mathrm{x}=(\mathrm{a}+2 \mathrm{b}) \text { or } 3 \mathrm{x}=(2 \mathrm{a}+\mathrm{b})3x=(a+2b) or 3x=(2a+b)

Hence, x=\frac{a+2 b}{3} \text { or } x=\frac{2 a+b}{3}x=3a+2b or x=32a+b