Please refer to the above attachment and give me the solution of the both the questions.
this question is from the chapter Algebraic expressions and Indentities
please don't write any useless thing otherwise your answer will be reported and please don't delete this question also
it's a request please give me all the steps.
Answers
Answer:
5(i)
x2(3−2x+x2)
For x=1;x=−1;x=2/3 and x=−1/2
x2(3−2x+x2)=3x2−2x3+x4
(a) For x=1
3x2−2x3+x4=3(1)2−2(1)3+(1)4
= 3×1−2×1+1
= 3−2+1=2
(b) For x=−1
3x2−2x3+x4=3(4)2−2(4)3+(−1)4
= 3×1−2×(−1)+1
= 3+2+1=6
(c) For x=2/3
3x2−2x3+x4=3(2/3)2−2(2/3)3+(2/3)4
= 3×(4/9)−2×(8/27)+(16/81)
= (4/3)−(16/27)+(16/81)
= (108−48+16)/81
= (124−48)/81
= 76/81
(d) For x = -1/2
3x^2-2x^3+x^4 = 3(-1/2)^2-2(-1/2)^3+(-1/2)^4
= 3 × (1/4) – 2 × (-1/8) + (1/16)
= (3/4) + ¼ + (1/16)
= (12 + 4 + 1)/16
= 17/16
(ii) 5xy(3x+4y-7)-3y(xy-x^2+9)-8
= 15x^2y+20xy^2-35xy-3xy^2+3x^2y-21y-8
= 18x^2y+17xy^2-35xy-27y-8
When x = 2, y = -1, we have
= 18(2)^2\times(-1)+17(2)(-1)^2-35(2)(-1)-27(-1)-8
= 18 × 4 × (-1) + 17 × 2 × 1 – 35 × 2 × (-1) – 27 × (-1) – 8
= -74 + 34 + 70 + 27 – 8
= 131 – 80
= 51
Answer:
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