Algebraic Expressions and Arithmetic - Class VII

$Multiplying\quad (3y+4z)(3y-4z),\quad (2y+7z)(y+z),\ =-16z^{ 2 }+12yz-12yz+9y^{ 2 }\quad +\quad 7z^{ 2 }+7yz+2yz+2y^{ 2 }\ =-9z^{ 2 }+9yz+11y^{ 2 }$

$(a+b)(c-d)+(a-b)(c+d)+2(ac+bd)$
$=a(c-d)+b(c-d)+a(c+d)-b(c+d)+2(ac+bd)$
$=ac-ad+bc-bd+ac+ad-bc-db+2ac+2bd$
$=4ac$

$\sqrt {13+\sqrt {44+10^2}}\$

We apply BODMAS rule to find the value of the given expression. According to BODMAS rule, if an expression contains brackets we have to first solve or simplify the bracket followed by of (powers and roots etc.), then division, multiplication, addition and subtraction from left to right.

Given exp.$\displaystyle \frac{2}{1+\frac{1}{\frac{1}{2}}}\times\frac{3}{\frac{15}{12}\div \frac{5}{4}}=\frac{2}{1+2}\times\frac{3}{\frac{15}{12}\times\frac{4}{5}}=\frac{2}{3}\times\frac{3}{1}=2$

$((x^3-2)\div2^2)\times 4+16$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $((x^3-2)\div2^2)\times 4+16$
$=$ $\dfrac{x^3-2}{4}\times 4+16$
$=$ $x^3+14$

$3x[x^2+1]-[2x(x^2+x-1)+1]-x^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$3x[x^2+1]-[2x(x^2+x-1)+1]-x^2$
$=$ $3x.x^2+3x.1-[2x.x^2+2x.x-2x.1+1]-x^2$
$=$ $3x^3+3x-2x^3-2x^2+2x-1-x^2$
$=$ $x^3-3x^2+5x-1$

$4-x^2\div x +(4\times-(\dfrac{2x^3}{x^2}))-3^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $4-x^2\div x +(4\times-(\dfrac{2x^3}{x^2}))-3^2$
$=$ $4-x+(4\times -2x)-9$
$=$ $4-x-8x-9$
$=$ $-9x-5$

$[((100+x)x^4)\div x^2]\times 2 - (x+x^2-1)$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $[((100+x)x^4)\div x^2]\times 2 - (x+x^2-1)$
$=$ $\dfrac{(100+x)(x^4)}{x^2}\times 2-x-x^2+1$
$=$ $200x^2+2x^3-x-x^2+1$
$=$ $2x^3+199x^2-x+1$

$x-1[(x^2+x-2)(x^2-1^2)\div (x-1)^2]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $\frac{x-1(x^2+x-2)(x+1)(x-1)}{(x-1(x-1)}$
$=$ $x^3+x^2-2x+x^2+x-2$
$=$ $x^3+2x^2-x-2$

$12-[5y+2x(y^2-2x+2)+6y-(y^2-1)]\times 0.1$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $12-[5y+2xy^2-4x^2+4x+6y-y^2+1]\times 2$
$=$ $12- [2xy^2-4x^2+4x+11y-y^2+1]\times 2$
$=$ $12-[4xy^2-8x^2+8x+22y-2y^2+2]$
$=$ $12-4xy^2+8x^2-8x-22y+2y^2-2$
$=$ $8x^2+2y^2-4xy^2-8x-22y+10$

$x^2-x[(-x)(-2+x)]\div x+x^3-3x^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $-2x^2-x[\dfrac{(-x)(-2+x)}{x}]+x^3$
$=$ $-2x^2-x[2-x]+x^3$
$=$ $x^3-2x^2-2x+x^2$
$=$ $x^3-x^2-2x$

$2y-1(y-y^2)+5y[(-2y)(y^2-1)]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $2y-y+y^2+5y[-2y^3+2y]$
$=$ $y+y^2-10y^4+10y^2$
$=$ $-10y^4+11y^2+y$

$4x^3[(3x-x^2)-1]+(x^2)[x+1]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $4x^3[(3x-x^2)-1]+(x^2)[x+1]$
$=$ $4x^3[3x-x^2-1]+x^3+x^2$
$=$ $12x^4-4x^5-4x^3+x^3+x^2$
$=$ $-4x^5+12x^4-3x^3+x^2$

$5x[2x(x^2+x^3)-x^3]-4x^2\div x^2-12x$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $5x[2x(x^2+x^3)-x^3]-4x^2\div x^2-12x$
$=$ $5x[2x^3+2x^4-x^3]-\dfrac{4x^2}{x^2}-12x$
$=$ $5x[x^3+2x^4]-4-12x$
$=$ $5x^4+10x^4-12x-4$
$=$ $15x^4-12x-4$

$xy^2+x^3-x[x^2-x][2x]+(x-1)x^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $xy^2+x^3+[-x^3+x^2]2x+x^3-x^2$
$=$ $xy^2+x^3-x^4+2x^3+x^3-x^2$
$=$ $xy^2+4x^3-x^4-x^2$
$=$ $-x^4+4x^3-x^2+xy^2$

$x^2\times(x-1)+[(2x+2)\times 4x]-1$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $x^2\times(x-1)+[(2x+2)\times 4x]-1$
$=$ $x^3-x^2+8x^2+8x-1$
$=$ $x^3+7x^2+8x-1$

$24[x+1]-[x^2-24+x]-[2x^2]\div [x^2]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $24[x+1]-[x^2-24+x]-[2x^2]\div [x^2]$
$=$ $24x+24-x^2+24-x-2$
$=$ $-x^2+46+23x$
$=$ $-x^2+23x+46$

$3x(x-2)+x(x^2\times 2x)-12x$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $3x(x-2)+x(x^2\times 2x)-12x$
$=$ $3x^2-6x+x^3\times2x^2-12x$
$=$ $3x^2-6x+2x^5-12x$
$=$ $2x^5+3x^2-18x$

$\frac{1}{3}(-2p+6q-9r)-\frac{1}{6}(-4p -18q +24r) = \frac{1}{3}(-2p+6q-9r)-\frac{1}{3}(-2p -9q +12r) $
$=\frac{1}{3} [\left (-2p+6q-9r)-(-2p -9q +12r) \right]$
$=\frac{1}{3} (-2p + 6q - 9r + 2p + 9q -12r)$
$=\frac{1}{3} (15q - 21r)$
$=(5q - 7r)$

$ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left ( y + a -\frac{1}{4}(a+y) \right )\right ]$
= $ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left (\frac{3}{4}(a+y) \right) \right ]$
= $ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{4}(a+y)\right ]$
= $ \frac{3}{4}(a+y) \left [ \frac{3}{4}(a+y)\right ]$
= $ \frac{9}{16}(a+y)^2$

Given Exp. $=-84\times (30-1)+365$
$=-(84\times 30)+84+365$
$=-2520+449$
$=-2071$

Given Exp.$=35+15\times \cfrac{3}{2}=35+\cfrac{45}{2}=35+22.5=57.5$

Given Exp. $=\cfrac { 800 }{ 64 } \times \cfrac { 1296 }{ 36 } =450$

Given the sum $=9+\cfrac { 3 }{ 4 } +7+\cfrac { 2 }{ 17 } -\left( 9+\cfrac { 1 }{ 15 }  \right) $
$=(9+7-9)+\left( \cfrac { 3 }{ 4 } +\cfrac { 2 }{ 17 } -\cfrac { 1 }{ 15 }  \right) $
$=7+\cfrac { 765+120-68 }{ 1020 } $
$=7+\cfrac { 817 }{ 1020 } $