x^2+1>=1

=>(x^2+1)^2>=1

y^2+2>=2

=>(y^2+2)^4>=16

=>(x^2+1)^2+(y^2+2)^4>=17

=>(x^2+1)^2+(y^2+2)^4-2>=15

Dấu = xảy ra khi x=y=0

Đề bằng 1 thì (x-2)(x+3)=0 suy ra x=2 hoặc x=-3.

thanks bạn

\(5^{\left(x-2\right)\left(x+3\right)}=1\)

đặt tổng của (x-2)(x+3) là a

\(\Leftrightarrow5^a=1\)

\(\Leftrightarrow x=0\)

thay vào \(\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

vậy x=2 hoặc x=-3

\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)

\(=>11-3x+1=\frac{9}{2}-5+3,5x\)

\(=>-3x+12=3,5x-\frac{1}{2}\)

\(=>-3x-3,5x=-\frac{1}{2}-12\)

\(=>-6,5x=-12,5\)

\(=>x=\frac{-12,5}{-6,5}=\frac{25}{13}\)

Ủng hộ nha

\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)

\(11-3x+1=\frac{9}{2}-5+3,5x\)

\(12-3x=-\left(0,5\right)+3,5x\)

\(12,5-3x=3,5x\)

\(12,5=6,5x\)

\(x=12,5:6,5=\frac{25}{13}\)

\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)

\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)

\(x=\)-\(\dfrac{27}{10}\)

\(x:\dfrac{2}{9}=\dfrac{9}{2}\)

\(x.\dfrac{9}{2}=\dfrac{9}{2}\)

\(x=\dfrac{9}{2}:\dfrac{9}{2}\)

\(x=1\)

\(\dfrac{x}{9}=\dfrac{5}{3}\)

\(x.3=5.9\)

\(x.3=45\)

\(x=45:3=15\)

\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)

\(x:\dfrac{8}{125}=\dfrac{125}{8}\)

\(x.\dfrac{125}{8}=\dfrac{125}{8}\)

\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)

\(\Rightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)-6=0\)

\(\Rightarrow x^2-4x+4-x^2+9-6=0\)

\(\Rightarrow-4x=-7\Rightarrow x=\frac{7}{4}\)

bạn Nguyễn Gia Triệu ơi :

Cho mik hỏi là làm sao bạn ra được -7 vậy

\(x^2-4x+4-x^2+9-6=-4x+7=0\Rightarrow-4x=-7\)

e sẽ cố gắng !!! 

\(3x-15=2x\left(x-5\right)\)

\(3x-15=2x^2-10x\)

\(3x-15-2x^2+10x=0\)

\(13x-15-2x^2=0\)

\(x\left(13-2x\right)-15=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

\(f,x\left(2x-7\right)-4x+14=0\)

\(2x^2-7x-4x+14=0\)

\(2x^2-11x+14=0\)

\(x\left(2x-11\right)=-14\)

\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)

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

\(3x+36-8x=x^2+4x+4\)

\(-5x+36-x^2-4x-4=0\)

\(-9x+32-x^2=0\)

\(x\left(-9-x\right)+32=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\23-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=23\end{cases}}}\)

chúc cj hay a hc tốt 

a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)

\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)

\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)

\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)

b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)

c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)

\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)