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.

\(5x^2\left(3y-1\right)-\left[3x^2\left(5y+2\right)-2x\left(3x^2-7x\right)\right]=15x^2y-5x^2-\left(15x^2y+6x^2-6x^3+14x^2\right)=15x^2y-5x^2-15x^2y-6x^2+6x^3-14x^2=6x^3-25x^2\)

Ta có: 6x - 2y = 7y - 3x

=> 6x + 3x = 7y + 2y

=> 9x = 9y => x = y

=> x - y = 0

mà x - y = 10 (đb)

=> ko có x; t tm

7x - 2y = 5x - 3y

=> 7x - 5x = -3y + 2y

=> 2x = -y

=> \(\frac{x}{-1}=\frac{y}{2}\) => \(\frac{2x}{-2}=\frac{3y}{6}\)

áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{2x}{-2}=\frac{3y}{6}=\frac{2x+3y}{-2+6}=\frac{20}{4}=5\)

=> \(\hept{\begin{cases}\frac{x}{-1}=5\\\frac{y}{2}=5\end{cases}}\) => \(\hept{\begin{cases}x=5.\left(-1\right)=-5\\y=5.2=10\end{cases}}\)

ta có 6x-2y=7y-3x chuyển vế sang

=>9x=9y

do x-y=10 nên x=10+y

=>9(10+y)=9y

=>90+9y=9y 

=>90=0y

=>y=0=>x=10

mik ko biét

Các bạn giúp mk với ạ

giúp mk đi mn ơi please

ban hoi tung cau mot thoi

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

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

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

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

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) Ta có: \(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\cdot...\cdot\left(a^{32}+b^{32}\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\cdot...\cdot\left(a^{32}+b^{32}\right)\)

\(=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\cdot...\cdot\left(a^{32}+b^{32}\right)\)

\(=\left(a^4-b^4\right)\left(a^4+b^4\right)\cdot...\cdot\left(a^{32}+b^{32}\right)\)

\(=\left(a^8-b^8\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\left(a^{32}+b^{32}\right)\)

\(=\left(a^{16}-b^{16}\right)\left(a^{16}+b^{16}\right)\left(a^{32}+b^{32}\right)\)

\(=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)\)

\(=a^{64}-b^{64}\)

`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)

p