x.(5-2x)+2x.(x-1)=15

<=>5x-2x²+2x²-2x=15

<=>3x=15

<=>x=5

cũng vậy @@

Đề @@

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Ta có: *x²+y²=(x+y)²-2xy=20²-2.(-44)=488

*x³+y³=(x+y)(x²-xy+y²)=20.[488-(-44)]=10640

Suy ra: (x²+y²)(x³+y³)=x⁵+x²y³+x³y²+y⁵=x⁵+y⁵+x²y².(x+y)

=>488.10640=x⁵+y⁵+(-44)².20

=>5192320=x⁵+y⁵+38720

=>x⁵+y⁵=5153600

thiếu đề a phải thuộc Z thì phải

\(a,\frac{0,25}{2}=\frac{0,5}{4}\)

b)\(\frac{4}{6}=\frac{10}{15}\)

a) ĐKXĐ: \(\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\begin{cases}x\ne0\\x\ne-5\end{cases}\)

b)\(A=\frac{x^2+2x}{2x+10}+\frac{x+5}{x}-\frac{50-5x}{2x\left(x+5\right)}=\frac{x^2+2x}{2.\left(x+5\right)}+\frac{x+5}{x}-\frac{50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^2+2x}{2x.\left(x+5\right)}+\frac{2\left(x+5\right)^2}{2x\left(x+5\right)}-\frac{50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^2+2x+2x^2+20x+50-50+5x}{2x\left(x+5\right)}=\frac{3x^2+27x}{2x\left(x+5\right)}=\frac{3x.\left(x+9\right)}{2x\left(x+5\right)}=\frac{3x+27}{2x+10}\)

c)Để A=1 thì: \(\frac{3x+27}{2x+10}=1\Rightarrow3x+27=2x+10\Leftrightarrow x=-17\)(nhận)

Vậy x=-17 thì A=1

\(\frac{8^2.4^3}{16^3}=\frac{\left(2^3\right)^2.\left(2^2\right)^3}{\left(2^4\right)^3}=\frac{2^6.2^6}{2^{12}}=1\)

\(\frac{25^2.125}{5^3.5^4}=\frac{\left(5^2\right)^2.5^3}{5^3.5^4}=\frac{5^4.5^3}{5^3.5^4}=1\)