9-10 G math 4.1

 অনুশীলনী-৪.১

সরল কর (১-৮):

সমাধানঃ সমাধানসমুহ পর্যায়ক্রমে নিচে দেওয়া হলোঃ

১.	73✕7-3 -------- 3✕3-4
=	73-3 ----- 31-4
=	70 ----- 3-3
=	1 ---  1 ---- 27
=	27

২.	3√72.3√7 ----------- √7
=	72/3.71/3 ---------- 71/2
=	7(2/3+1/3) ----------- 71/2
=	73/3 ------ 71/2
=	71 ----- 71/2
=	71-1/2
=	71/2
=	√7

৩. (2^-1+5^-1)^-1

=(1/2+1/5)^-1

=(5/10+2/10)^-1

=(7/10)^-1

=10/7

৪.	(2a-1+3b-1)-1
=	(2✕1/a+3✕1/b)-1
=	(2/a+3/b)-1
=	{(2b+3a)/ab}-1
=	ab ------- 2b+3a

৫.	(	a2b-1 -------- a-1b	)2
=	(	a2.1/b --------- 1/a2.b	)2
=	(	a2/b ------- b/a2	)2
=	(	a2.a2 ------ b.b	)2
=	(	a4 ---- b2	)2
=		a8 ---- b4

৬.	√(x-1y)	.	√(y-1z	.	√z-1x
=	√(y/x)	.	√(z/y)	.	√(x/z)
=	(y/x)1/2	.	(z/y)1/2	.	(x/z)1/2
=	y1/2 ----- x1/2	.	z1/2 ----- y1/2	.	x1/2 ----- z1/2
=	y1/2 ----- y1/2	.	z1/2 ----- z1/2	.	x1/2 ----- x1/2
=	y1/2-1/2	.	z1/2-1/2	.	x1/2-1/2
=	y0	.	z0	.	x0
=	1	.	1	.	1
=	1

৭.	2n+4-4.2n+1 --------------   2n+2÷2
=	2n24-4.2n21 --------------  2n.22÷2
=	2n(24-4.2) ------------   2n.4÷2
=	2n(16-8) ----------    2n.2
=	8 -- 2
=	4

৮.	3m+1 --------- (3m)m-1	÷	9m+1 ------------ (3m-1)m+1
=	3m+1 --------  3m(m-1)	÷	32(m+1) --------------  3(m-1)(m+1)
=	3m+1-m(m-1) ÷ 32(m+1)-(m-1)(m+1)
	3m+1-m(m-1)-{2(m+1)-(m-1)(m+1)}
এখন,
=	m+1-m(m-1)-{2(m+1)-(m-1)(m+1)}
=	m+1-m2-m-{2m+2-(m2-12)}
=	m+1-m2+m-(2m+2-m2+1}
=	m+1-m2+m-2m-2+m2-1
=	-2
প্রদত্ত রাশি		=	3-2
		=	1 -- 32
		=	1 -- 9

প্রমাণ কর (৯-১৫):

৯.	4n-1 ----- 2n-1	=	2n+1
	LHS	=	4n-1 ----- 2n-1
		=	22n-1 ------ 2n-1
		=	(22n-1) (2n+1) ---------------- (2n-1) (2n+1)
		=	(22n-1) (2n+1) ----------------     (2n)2-12
		=	(22n-1) (2n+1) ---------------     22n-1
		=	2n+1
		=	RHS	(Proved)

১০.	22p+1.32p+q.5p+q.6p ---------------------- 3p-2.62p+2.10p.15q	=	1 -- 2
LHS=	22p+1.32p+q.5p+q.(3✕2)p ---------------------------------- 3p-2.(3✕2)2p+2.(5✕2)p.(5✕3)q
=	22p+1.32p+q.5p+q.3p.2p -------------------------------- 3p-2.32p+2.22p+2.5p.2p.5q.3q
=	22p+1+p.32p+q+p.5p+q ------------------------------ 3p-2+2p+2+q.22p+2+p.5p+q
=	23p+1.33p+q.5p+q ------------------- 33p+q.23p+2.5p+q
=	33p+q-3p-q.23p+1-3p-2.5p+q-p-q
=	30.2-1.50
=	1.2-1.1
=	2-1
=	1 2
=	RHS (Proved)

১১.	(al/am)n	.	(am/an)l	.	(an/al)m	=1
LHS=	(al/am)n	.	(am/an)l	.	(an/al)m
=	aln ---- amn	.	aml ---- anl	.	anm ---- aml
=	aln+ml+nm -------------  amn+nl+ml
=	1
=	RSH (Proved)

১২.	ap+q ---- a2r	✕	aq+r ---- a2p	✕	ar+p ----- a2q	=	1
LHS=	ap+q ---- a2r	✕	aq+r ---- a2p	✕	ar+p ----- a2q
=	ap+q+q+r+r+p ------------  a2r+2p+2q
=	a2r+2p+2q ----------- a2r+2p+2q
=	1
=	RHS (Proved)

১৩.	(xa/xb)1/ab	.	(xb/xc)1/bc	.	(xc/xa)1/ca	=	1
LHS=	(xa/xb)1/ab	.	(xb/xc)1/bc	.	(xc/xa)1/ca
=	(xa)1/ab -------- (xb)1/ab	.	(xb)1/bc -------- (xc)1/bc	.	(xc)1/ab --------- (xa)1/ab
=	xa/ab ------- xb/ab	.	xb/bc ------ xc/bc	.	xc/ca ------ xa/ca
=	x1/b ----- x1/a	.	x1/c ----- x1/b	.	x1/a ----- x1/c
=	1
=	RHS (Proved)

১৪.	(xa/xb)a+b	.	(xb/xc)b+c		.	(xc/xa)c+a	=1
LHS=	(xa/xb)a+b	.	(xb/xc)b+c		.	(xc/xa)c+a
=	x(a-b)(a+b)	.	x(b-c)(b+c)		.	x(c-a)(c+a)
=	x(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)
এখন,	(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)
=	a2-b2+b2-c2+c2-a2
=	0
প্রদত্ত রাশি= x0  =		1	= RHS (Proved)

১৫.	(xp/xq)p+q-r.(xq/xr)q+r-p.(xr/xp)r+p-q=1
LHS=	x(p-q)(p+q-r).x(q-r)(q+r-p).x(r-p)(r+p-q)
=	x(p-q)(p+q-r)+ (q-r)(q+r-p)+ (r-p)(r+p-q)
এখন,	(p-q)(p+q-r)+ (q-r)(q+r-p)+ (r-p)(r+p-q)
=	p2-pq+pq-q2-rp+qr+q2-q2-qr+qr-r2-pq+pr+ r2-pr+pr-qr+pq
=	0
প্রদত্ত রাশি= x0  = 1		=	RHS (Proved)

১৬. যদি a^x=b, b^y=c এবং c^z=a হয়, তবে দেখাও যে, xyz=1

সমাধানঃ

দেওয়া আছে,  a^x=b, b^y=c এবং c^z=a
রাশি তিনটি যোগ করে পাই,
a^x+b^y+c^z=b+c+a
বা,  a^x+b^y+c^z=a¹+b¹+c¹
এখানে, a^x=a¹
তাহলে। x=1
একইভাবে, y=1; z=1
LHS=xyz=1.1.1=1=RHS (Proved)

সমাধান কর (১৭-২০):

১৭. 4^x=8

সমাধানঃ

4^x=8
বা,  (2)^2x=2³
বা,  2x=3
বা,  x=3/2

১৮. 2^2x+1=128

সমাধানঃ

2^2x+1=128
বা,  2^2x+1=2⁷
বা,  2x+1=7
বা,  2x=7-1
বা,  2x=6
বা,  x=6/2
বা,  x=3

১৯. (√3)^x+1=(³√3)^2x-1

সমাধানঃ

বা,  (√3)^x+1=(³√3)^2x-1
বা,  3^(1/2)(x+1)=3^(1/3)(2x-1)
বা,  (x+1)/2=(2x-1)/3
বা,  3(x+1)=2(2x-1)
বা,  3x+3=4x-2
বা,  3x-4x=-2-3
বা,  -x=-5
বা,  x=5

২০. 2^x+2^1-x=3

সমাধানঃ

2^x+2^1-x=3
বা,  2^x+2¹/2^x=3
বা,  (2^2x+2)/2^x=3
বা,  2^2x+2=3.2^x
বা,  2^2x+2-3.2^x=0
বা,  (2^x)²-2.2^x.1+1²-2^x+1=0
বা,  (2^x-1)²-1(2^x-1)=0
বা,  (2^x-1)(2^x-1-1)=0
বা,  (2^x-1)(2^x-2)=0
বা,  2^x-1=0     অথবা, 2^x-2=0
বা,  2^x=1                বা,  2^x=2
বা,  2^x=2⁰      বা,  2^x=2¹
বা, x=0                  বা,  x=1
∴x=0,1

২১. P=x^a, Q=x^b এবং R=x^c

ক) P^bc.Q^-ca = এর মান নির্ণয় কর।

খ) (Q/P)^a+b✕(Q/R)^b+c÷2(RP)^a-c এর মান নির্ণয় কর।

গ) দেখাও যে, (P/Q)^a2+ab+b2✕(Q/R)^b2+bc+c2✕(R/P)^c2+ca+a2=1

সমাধানঃ

ক)
P^bc.Q^-ca
=(x^a)^bc.(x^b)^-ca
=x^abc.x^-abc
=x^abc-abc
=x⁰
=1

খ)

(Q/P)^a+b✕(Q/R)^b+c÷2(RP)^a-c
=(x^a/x^b)^a+b ✕ (x^b/x^c)^b+c ÷ 2(x^c.x^a)^a-c
= x^(a-b)(a+B) ✕ x^(b-c)(b+c) ÷ 2x^(a+c)(a-c)
=x^a2-b2✕x^b2-c2÷2x^a2-c2
= x^a2-b2✕x^b2-c2✕1/(2x^a2-c2)
=x^{a2-b2+b2-c2-a2+c2}✕1/2
=x⁰✕1/2
=1✕1/2
=1/2

গ)

LHS
=(P/Q)^a2+ab+b2✕(Q/R)^b2+bc+c2✕(R/P)^c2+ca+a2
=(x^a/x^b) ^a2+ab+b2✕(x^b/x^c)^b2+bc+c2✕(x^c/x^a)^c2+ca+a2
=(x^a-b) ^a2+ab+b2✕(x^b-c)^b2+bc+c2✕(x^c-a)^c2+ca+a2
=x^a3-b3✕x^b3-c3✕x^c3-a3
=x^{a3-b3+b3-c3+c3-a3}
=x⁰
=1
=LHS (Proved)

২২. x=(2a^-1+3b^-1)^-1, y=^pq√(x^p/x^q)✕ ^qr√(x^q/x^r)✕ ^rp√(x^r/x^p)

এবং,

z

=

5m+1 -------- (5m)m-1

÷

25m+1 --------- (5m-1)m+1

যেখানে x,p,q,r>0

ক) x এর মান নির্ণয় কর।

খ) দেখাও যে, y+⁴√81=5

গ) দেখাও যে, y÷z=25

সমাধানঃ
ক)
x	= (2a-1+3b-1)-1
=	(2✕1/a+3✕1/b)-1
=	(2/a+3/b)-1
=	{(2b+3a)/ab}-1
=	ab ------ 2b+3a

খ)

দেওয়া আছে, y=^pq√(x^p/x^q)✕ ^qr√(x^q/x^r)✕ ^rp√(x^r/x^p)
LHS
= y+⁴√81
= ^pq√(x^p/x^q)✕ ^qr√(x^q/x^r)✕ ^rp√(x^r/x^p)+ ⁴√81
= ^pq√(x^p-q)✕ ^qr√(x^q-r)✕ ^rp√(x^r-p)+ ⁴√81
= (x^p-q)^1/pq✕(x^q-r)^1/qr✕(x^r-p)^1/rp+ ⁴√81
= x^(p-q)/pq✕x^(q-r)/qr✕x^(r-p)rp+ 81^1/4
=x^{(pr-qr+pq-pr+qr-ppq)/pqr}+(3⁴)^1/4
=x^0/ppqr+3^4/4
=x⁰+3
=1+3
=4
=RHS [Note textbook RHS=5 is not correct]

গ)
দেওয়া আছে,
	z	=	5m+1 -------- (5m)m-1	÷	25m+1 --------- (5m-1)m+1
		=	5m+1 ------ 5m2-m	÷	(52)m+1 ------- 5m2-12
		=	5m+1 ------- 5m2-m	÷	52m+2 -------- 5m2-12
		=	5m+1 ------- 5m2-m	✕	5m2-12 --------- 52m+2
		=	5m+1+m2-1-m2+m-2m-2
		=	5-2
		=	1 52
খ হতে পাই, y=1
∴	y	÷	x = 1 ÷	1 52
			=1✕52
			=1✕25
			=25
÷ y ÷ z =25 (দেখানো হলো)