Math, asked by Dɪʏᴀ4Rᴀᴋʜɪ, 8 months ago

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1)Simplify by using identity-:

A)(856+167)^2+(856-167)^2/856×856+167×167

2)If \:a^2+b^2+c^2=>79\: and \:ab +bc+ca=>45 .Find\: the\: value\: of\: a+b+c

3)If x\:+y+z=>14\: and \:x^2 +y^2+z^2=>50 .Find\: the\: value\: of\: xy+yz+zx

Answers

Answered by Mora22
9

Answer:

A) \frac{(856+167)^2+(856-167)^2}{856×856+167×167}

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

so \:  \frac{2( {856}^{2} +  {167}^{2} ) }{( {856}^{2}  +  {167}^{2} )}  = 2

1st question answer is 2

2)\:a^2+b^2+c^2=>79

and \: ab +bc+ca=>45

 {(a + b + c)}^{2}  =  {a}^{2}  +  {b}^{2}  +  {c}^{2}  + 2(ab + bc + ca)

so \:  {(a + b + c)}^{2}  = 79 + 2(45) = 79 + 90 = 169

 {(a + b + c)} = 13 \: or \:  - 13

2nd question answer is 13 or -13

3)x + y + z = 14

 {x}^{2}  +  {y}^{2}  +  {z}^{2}  = 50

 {(x + y + z)}^{2}  =  {x}^{2}  +  {y}^{2}  +  {z}^{2}  + 2(xy + yz + xz)

 {14}^{2}  = 50 + 2(xy + yz + xz)

196 - 50 = 2(xy + yz + zx)

146 = 2(xy + yz + zx)

(xy + yz + zx) = 73

3rd answer is 73.

Answered by Anonymous
0

Answer:

)856×856+167×167(856+167)2+(856−167)2

{(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )(a+b)2+(a−b)2=2(a2+b2)

so \: \frac{2( {856}^{2} + {167}^{2} ) }{( {856}^{2} + {167}^{2} )} = 2so(8562+1672)2(8562+1672)=2

1st question answer is 2

2)\:a^2+b^2+c^2= > 792)a2+b2+c2=>79

and \: ab +bc+ca= > 45andab+bc+ca=>45

{(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)(a+b+c)2=a2+b2+c2+2(ab+bc+ca)

so \: {(a + b + c)}^{2} = 79 + 2(45) = 79 + 90 = 169so(a+b+c)2=79+2(45)=79+90=169

{(a + b + c)} = 13 \: or \: - 13(a+b+c)=13or−13

2nd question answer is 13 or -13

3)x + y + z = 143)x+y+z=14

{x}^{2} + {y}^{2} + {z}^{2} = 50x2+y2+z2=50

{(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + xz)(x+y+z)2=x2+y2+z2+2(xy+yz+xz)

{14}^{2} = 50 + 2(xy + yz + xz)142=50+2(xy+yz+xz)

196 - 50 = 2(xy + yz + zx)196−50=2(xy+yz+zx)

146 = 2(xy + yz + zx)146=2(xy+yz+zx)

(xy + yz + zx) = 73(xy+yz+zx)=73

3rd answer is 73.

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