a.

$7x-2y=5x-3y$

$\Leftrightarrow 2x=-y$. Thay vào điều kiện số 2 ta có:

$-y+3y=20$

$2y=20$

$\Rightarrow y=10$. 

$x=\frac{-y}{2}=\frac{-10}{2}=-5$

b.

$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$

$3y=4z-2y\Rightarrow 5y=4z\Rightarrow \frac{y}{4}=\frac{z}{5}$

$\Rightarrow \frac{x}{6}=\frac{y}{4}=\frac{z}{5}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{6}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{6+4+5}=\frac{45}{15}=3$

$\Rightarrow x=6.3=18; y=4.3=12; z=5.3=15$ 

c.

$3x=4y-2x$

$\Rightarrow 5x=4y\Rightarrow x=\frac{4}{5}y$

$3x=7z-4y$

$\Leftrightarrow \frac{12}{5}y=7z-4y$

$\Leftrightarrow \frac{32}{5}y=7z\Rightarrow z=\frac{32}{35}y$

Khi đó:

$x+y-2z=10$

$\frac{4}{5}y+y-2.\frac{32}{35}y=10$

$y.\frac{-1}{35}=10$

$y=-350$

$x=\frac{4}{5}y=\frac{4}{5}.(-350)=-280$

$z=\frac{32}{35}y=\frac{32}{35}.(-350)=-320$

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