\(\left(x-1\right)^2+\left(y+2\right)^2=0\)

Vì \(\left(x-1\right)^2\ge0\forall x\)

       \(\left(y+2\right)^2\ge0\forall y\)

Nên \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\) \(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy x = 1 và y = -2

4x+2x+68-2³

^{Suy ra 6x+68-8}

           ^6x+60

           ^6*(x+10)

\(4x+3.\left(1-x\right)=2.\left(x-2\right)\)

\(4x+3-3x=2x-4\)

\(\left(4x-3x\right)+3=2x-4\)

\(x+3=2x-4\)

\(x-2x=-4-3\)

\(-x=-7\)

\(x=7\)

Các bạn ơi giúp mình đi , minh đang cần gấp

\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)

\(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)

\(=-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\)

\(=-2x^3-13x^2-x-12\)

\(3x-\left(15-x\right)=17+x\)

\(\Rightarrow3x-15+x=17+x\)

\(\Rightarrow3x+x-x=17+15\)

\(\Rightarrow3x=32\)

\(\Rightarrow x=\frac{32}{3}\)

trả lời:

3x-(15-x)=17+x

(=)3x-15+x=17+x

(=)3x+x-x=17+15

(=)3x=32

(=)x=\(\frac{32}{3}\)

Ta có : 3x + ( 15 - x ) = 17 + x .

=>        3x - 15 + x     = 17 + x .

=>        3x + x - x       = 17 + 15 .

=>               4x           = 32 .

=>                 x           = 32 : 4 .

=>                 x           = 8 .

Vậy x = 8 .

3^x+2.5^y=45^x

=>3^x+2.5^y=(3²)^x.5^x

=>\(\hept{\begin{cases}x+2=2x\\y=x\end{cases}}\Rightarrow x=y=2\)