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.

\(VT=\left(7x-3y+2z\right)\left(7x-3y-2z\right)\)

\(=\left(7x-3y\right)^2-4z^2\)

\(=49x^2-42xy+9y^2-4z^2\)

\(=4\cdot10x^2+9x^2-42xy+9y^2-4z^2\)

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

1,Áp dụng hằng đẳng thức ( hình như bn viết sai)

\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

2, I am stupid so I don't know.

a) \(5x\left(x-1\right)-3y\left(x-1\right)=\left(x-1\right)\left(5x-3y\right)\)

b) \(7x^2-7y^2=7\left(x^2-y^2\right)=7\left(x-y\right)\left(x+y\right)\)

c) \(x^2y^2z+xy^2z^2+x^2yz^2=xyz\left(xy+yz+xz\right)\)