NCTB Class 7 Math Chapter Five Exercise 5.1 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 5.1 “Algebraic Formulaic and Applications” Exercise 5.1 Solution.

Board |
NCTB |

Class |
7 |

Subject |
Math |

Chapter |
5 |

Chapter Name |
“Algebraic Formulaic and Applications” |

Exercise |
5.1 Solution |

**Exercise:- 5.1**

**(1) Solution:-** (a+5)^{2}

= a^{2}+2x z x 5+5^{2}

= a^{2}+10a+25

**(2) Solution:-** (5x-7)^{2}

= (5x)^{2} – 2x5xX7+ (7)^{2}

= 25x^{2}-70x+49

**(3) Solution:-** (3a-11xy)^{2}

= (3a)2 – 2x3ax11xy+ (11xy)2

= 9a^{2}– 66axy+ 121x^{2} y^{2}

**(4) Solution:-** (5a^{2}+9m^{2})^{2}

= (5a^{2})^{2}+2x5a^{2}.9m^{2}+ (9m^{2})^{2}

= 25a^{4}+ 90a^{2} m^{2}+ 81 m^{4}

**(5) Solution:-** (55)^{2}

= (50+5)^{2}

= (50)2+ 2x50x5 +5^{2}

= 2500+500+25

= 3025

**(6) Solution:-** (990)^{2}

= (1000-10)^{2}

= (1000)^{2} – 2x1000x10+(10)^{2}

= 1000000-20000+100

= 980100

**(7) Solution:-** (xy-6y)^{2}

= (x y)^{2}– 2.xy.6y+ (6y)^{2}

= x^{2} y^{2}– 12xy^{2} +30y^{2}

**(8) Solution:-** (a x-by)^{2}

= (a x)^{2}– 2.ax.by + (by)^{2}

= a^{2} x^{2}– 2axby + b^{2} y^{2}

**(9) Solution:-** (97)2

= (100-3)^{2}

= (100)^{2}– 2.100.3+ 3^{2}

= 10000- 600+9

= 9409

**(10) Solution:-** (2x+y-z)^{2}

= {(2x+y)-z}^{2}

= (2x+y)^{2} – 2. (2x+y).z+ z^{2}

= (2x)2 + 2X2xXy+y^{2}– 4zx-2yz+z^{2}

= 4x^{2}+4xy+y^{2}-4zx-2yz+x^{2}

**(11) Solution:-** (2a-b+3c)^{2}

= {(2a-b)+ 3c}^{2}

= (2a-b)^{2}+2. (2a-b).3c+(3c)^{2}

= (2a)^{2}– 2.2a.b.+b^{2}+12ac-6bc+9c^{2}

= 4a^{2}-4ab+b^{2}+12ac-6bc+9c^{2}

= 4a^{2}+b^{2}+9c^{2}-4ab+12ac-6bc

**(13) Solution:-** (x^{2}+y^{2}-z^{2})^{2}

= {(x^{2}+y^{2})^{2} – (z^{2})^{2}}

= (x2+y^{2})^{2}– 2. (x^{2}+y^{2}). Z^{2} + (z^{2})^{2}

= (x^{2})^{2}+ 2x^{2}y^{2}+ (y^{2})^{2}– 2x^{2} z^{2} – 2y^{2} z^{2}+ z^{4}

= x^{4}+y^{4}+z^{4}+2x^{2} y^{2}+ 2y^{2} z^{2}– 2z^{2} x^{2}

**(14) Solution:-** (3x-2y+z)^{2}

= {(3x-2y) + z}^{2}

= (3x-2y)^{2}+ 2. (3x-2y) . z +z^{2}}

= 9x^{2}-12xy+4y^{2}+6xz-4yz+z^{2}

= 9x^{2}+4y^{2}+z^{2}+12xy+6xz-4yz

**(15) Solution:-** (b c +ca +a b)^{2}

= {(b c +c a) + a b}^{2}

= (b c+ c a)^{2} + 2. (b c +c a). a b+ (a b)^{2}}

= (b c)2+2.bc.ca+ (ca)2 + 2ab2c +2a2bc+ a2 b2

=b^{2}c^{2}+ c^{2}a^{2}+a^{2}b^{2}+ 2ac^{2}+2ab^{2}c+2a^{2}bc

**(17) Solution:-** (2a+1)^{2} – 4a(2a+1)+ 4a^{2}

= (2a1)^{2}– 2. (2a+1). 2a + (2a)^{2}

= {(2a+1) – 2a}^{2}

= (2a+1-2a)^{2}

= (1)^{2} = 1

**(18) Solution:-** Let, 5a+3b= x ; 4a-3b= y

Given, x^{2}+2xy+y^{2}

= (x +y)^{2}

= (5a+3b+4a-3b)^{2}

= (9a)^{2}

= 81a^{2}

**(19) Solution:-** Let, 7a+b= x ; 7a-b= y

Therefore, given, x^{2}-2xy+y^{2}

= (x-y)^{2}

= (7a+b-7a+b)^{2}

= (2b)^{2}

= 4b^{2}

**(20) Solution:-** Let, (2x+3y)= a ; 2x-3y= b

Given, a^{2}+ 2ab+b^{2}

= (a +b)^{2}

= (2x+3y+2x-3y)^{2}

= (4x)^{2}

= 16x^{2}

**(21) Solution:-** (5x-2) = a ; 5x+7= b

Given, a2+b2- 2ab

= (a-b)2

= (5x-2-5x-7)^{2}

= (-9)^{2}

= 81

**(22) Solution:-** (3ab-cd)^{2} + 9 (c d-a b) + 6 (3ab-cd) (c d-a b)

= (3ab-cd)^{2} + 2. (3ab-cd) x 3 (c d-a b) + [3 (c d- a b)]^{2}

Let, 3ab-cd= x ; 3 (c d-a b) = y

Therefore, x^{2}+ 2xy +y^{2}

= (x +y)^{2}

= {(3ab-cd) +3 (c d-a b)}^{2}

= (3ab-cd+3cd-3ab)^{2}

= (2cd)^{2}

= 4c^{2}d^{2}

**(23) Solution:-** (2x+5y+3z)2 + (5y+3z-x)2- 2 (5y+3z-x) (2x+5y+3z)

Let, (2x+5y+3z) = a ; 5y+3z-x = b

Therefore, a^{2}+b^{2}-2a b

= (a-b)^{2}

= {2x+5y+3z-5y-3z+x}^{2}

= (3x)^{2}

= 9x^{2}

**(24) Solution:-** (2a-3b+4c)2 + (2a+3b-4c)2 + 2 (2a-3b+4c) (2a+3b-4c)

Therefore, Let, 2a-3b+4c= x and, (2a+3b-4c) = y

Therefore, x^{2}+ y^{2}+2xy

= (x +y)^{2}

= (2a-3b+4c+2a+3b-4c)^{2}

= (4a)^{2}

= 16a^{2}

**(25) Solution:-** 25x^{2}+36y^{2}– 60xy ; when x= -4, y= -5

Therefore, 25x^{2} + 36y^{2}– 60xy

= (5x)^{2} + 2.5x.6y+ (6y)^{2}

= (5x-6y)^{2}

= {5(-4) -6 (-5)}^{2}

= (-20+30)^{2}

= (10)^{2} = 100

**(26) Solution:-** 16a^{2}– 24ab- 9b^{2}

= (4a)^{2} – 2.4a.3b+ (3b)^{2}

= (4a-3b)^{2}

= (4×7-3×6)^{2}

= (28-18)^{2}

= (10)^{2}

= 100

**(27) Solution:-** (9x^{2}+30x+25)

= (3x)^{2}+2.3x.5+5^{2}

= (3x+5)^{2}

= {3x (-2) + 5}^{2}

= (-6+5)^{2}

= (-1)^{2}

=1

**(28) Solution:-** 81a^{3}+18ac+c^{2}

= (9a)^{2}+ 2.9a.1 +c^{2}

= (9a+c)^{2}

= ({9×7-67}^{2}

= (63-67)^{2}

= (-4)^{2}

= 16

**(29) Solution:-** a^{2}+ b^{2}= (a +b)^{2}– 2ab

= (5)^{2}– 2x 12

= 25-24

= 1

**(31) Solution:-** x+ 1/x = 5 (given)

=> (x+1/x)^{2}= (5)^{2}

=> x^{2}+ 1/x^{2} = 25-2

= x^{2}+ 1/x^{2} = 23

Both side square,

(x^{2}+1/x^{2})^{2} = (23)^{2}

=> (x^{2}-1/x^{2})^{2} + 4.x^{2}. 1/x^{2}= 529

=> (x^{2}– 1/x^{2})

= 529-4

= 525

**(32) Solution:-** a +b= 8: a-b- 4

We know that,

4ab= (a +b)^{2} – (a -b)^{2}

= 8^{2}– 4^{2}

= 64-16

= 48

Therefore, a b= 12

Where is 33 and 34 ??

Wait available soon

Sir 22 Number is it correct? Because,, I’m little bit confused.

Reason is,, In Text book the question is different. Kindly reply me.In this pandemic situation our only trust and options is Online.And when I solve this question.I face problem just only this question.In the Text book The question like this,,,

(3ab-cd)^2 + 9(cd-ab)^2+6(3ab-cd)(cb-ab)❓❓

⭕️⭕️ ^2 It means (Hole Square)

Please reply as soon as possible.

Thank you ????

Joty please tell Question number. We will solve your doubt

5.1 Question no.(22)

33,34?

Pls send 33 and 34 fast.

There is no solution of 33 and 34.

Where is namber33.34????

5.1 Question no.(22)

Where r the solutions of 33 & 34

Tahmid will available soon

Where is 30 no solution sir

13Number not correct

From 29 to 31 incomplete

Please apply formula

can anyone pls solve this problem for me ?

if, a-b=7 & ab=12, now show that, (a+b)²=61