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Xét ΔABC có MN//BC
nên AM/MB=AN/NC
=>4/x=3/2
=>x=8/3(cm)
=>AB=8/3+4=20/3(cm)
Xét ΔBAC có BN là phân giác
nên BC/BA=NC/AN
=>y:20/3=2/3
=>y=2/3*20/3=40/9(cm)
\(\left(x+\frac{2}{x}\right)^2+\left(y+\frac{2}{y}\right)^2=x^2+y^2+\frac{4}{x^2}+\frac{4}{y^2}+4+4\)
\(=\left(x^2+\frac{1}{x^2}\right)+\left(y^2+\frac{1}{y^2}\right)+\left(\frac{3}{x^2}+3x+3x\right)+\left(\frac{3}{y^2}+3y+3y\right)-6\left(x+y\right)+8\)
\(\ge2+2+9+9-6.2+8=18\)
a: \(\left(x+10\right)\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x+10=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=5\end{matrix}\right.\)
b: \(\left(2x+10\right)\left(4+x\right)=0\)
=>\(\left[{}\begin{matrix}2x+10=0\\4+x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-4\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)
c: \(\left(4x+20\right)\left(12x-24\right)=0\)
=>\(\left[{}\begin{matrix}4x+20=0\\12x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-20\\12x=24\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d: \(\left(x-2024\right)\left(4x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-2024=0\\4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2024\\4x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\\x=2024\end{matrix}\right.\)
e: \(\left(2x-6\right)\left(7+x\right)=0\)
=>\(\left[{}\begin{matrix}2x-6=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\x=-7\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
g: (4x+8)(6-x)=0
=>\(\left[{}\begin{matrix}4x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x=6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)
h: (2x+2)(4x-8)=0
=>2(x+1)*4*(x-2)=0
=>(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
i: (2x-2024)(8x-16)=0
=>\(2\left(x-1012\right)\cdot8\cdot\left(x-2\right)=0\)
=>\(\left(x-1012\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-1012=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1012\\x=2\end{matrix}\right.\)
`x xx 2 +x/2=10`
`=> x xx 2+ x xx 1/2 =10`
`=> x xx (2+1/2) =10`
`=> x xx ( 4/2 +1/2) =10`
`=> x xx 5/2 =10`
`=> x=10:5/2`
`=> x= 10 xx 2/5`
`=> x= 20/5`
`=>x=4`
Vậy `x=4`
Nhóc NgốcDiệp ViTú QuyênLizzy-Marie ??_Trang_??我爱你。♪aomike Nguyễn Minh Huyền