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Lời giải:
Từ \(10x^2=10y^2+z^2\Rightarrow 10x^2-10y^2=z^2\)
Theo hằng đẳng thức đáng nhớ ta có:
\((7x-3y+2z)(7x-3y-2z)=(7x-3y)^2-(2z)^2\)
\(=(7x-3y)^2-4z^2=(49x^2-42xy+9y^2)-4(10x^2-10y^2)\)
\(=9x^2-42xy+49y^2=(3x)^2-2.(3x).(7y)+(7y)^2=(3x-7y)^2\)
Ta có đpcm.
\(5x^2\left(3y-1\right)-\left[3x^2\left(5y+2\right)-2x\left(3x^2-7x\right)\right]=15x^2y-5x^2-\left(15x^2y+6x^2-6x^3+14x^2\right)=15x^2y-5x^2-15x^2y-6x^2+6x^3-14x^2=6x^3-25x^2\)
\(VT=\left(7x-3y+2z\right)\left(7x-3y-2z\right)\)
\(=\left(7x-3y\right)^2-4z^2\)
\(=49x^2-42xy+9y^2-4z^2\)
\(=4\cdot10x^2+9x^2-42xy+9y^2-4z^2\)
mà 10x2 = 10y2 + z2
\(\Rightarrow VT=4\left(10y^2+z^2\right)+9x^2-42xy+9y^2-4z^2\)
\(=40y^2+4x^2+9x^2-42xy+9y^2-4z^2\)
\(=9x^2-42xy+49y^2\)
\(=\left(3x-7y\right)^2=VP\)
Ta có :
10x2=10y2+z2
=>40x2=40y2+4z2
=>49x2-9x2-49y2+9y2-4z2=0
=>49x2+9y2-4z2=9x2+49y2
=>49x2-2.7x.3y+9y2-4z2=9x2-2.3x.7y+49y2
=>(7x-3y)2-4z2=(3x-7y)2
=>(7x-3y+2z)(7x-3y-2z)=(3x-7y)2
kbnha
`a)7x^3y^2+14x^2y^3+7xy^4`
`=7xy^2(x^2+2xy+y^2)`
`=7xy^2(x+y)^2`
______________________________________________
`b)x^2-xy+5x-5y`
`=x(x-y)+5(x-y)`
`=(x-y)(x+5)`
______________________________________________
`c)3x^2-6xy-12+3y^2`
`=3(x^2-2xy-4+y^2)`
`=3[(x-y)^2-4]`
`=3(x-y-2)(x-y+2)`
a)7x3y2+14x2y3+7xy4
=7xy2(x2+2xy+y2)
=7xy2(x+y)2
b)x2-xy + 5x - 5y
=x(x-y) + 5(x-y)
=(x-y) (x+5)
\(a,A+B-C=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)
\(=\left(16x^4-15x^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)+\left(-9y^4-6y^4-17y^4\right)-1\)
\(=x^4-10x^3y-x^2y^2-32y^4-1\)
\(b,A-C+B=A+B-C\) ( giống câu a )
\(a,\)
\(A+B+C\)
\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-\left(5x^3y+3x^2y^2+17y^4+1\right)\)
\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)
\(=\left(16x^4-15x^4\right)+\left(-9y^4-6y^4-17y^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)-1\)
\(=x^4-32y^4-10x^3y-x^2y^2-1\)
\(b,\)
\(A-C+B=A+B-C=x^4-32y^4-10x^3y-x^2y^2-1\)