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a: \(\left(9-4\sqrt{5}\right)\left(4\sqrt{5}+9\right)\)
\(=9^2-\left(4\sqrt{5}\right)^2\)
=81-80
=1
b: \(A=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{x\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}}\)
\(=\dfrac{x+3\sqrt{x}+2-x+3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{x-4}{\sqrt{x}}\)
\(=\dfrac{6\sqrt{x}}{\sqrt{x}}=6\)
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a: \(-3x^2-8x+3=0\)
\(\Leftrightarrow3x^2+8x-3=0\)
\(\Leftrightarrow3x^2+9x-x-3=0\)
=>(x+3)(3x-1)=0
=>x=1/3 hoặc x=-3
b: \(\left\{{}\begin{matrix}2x+3y=4\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=4\\2x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-6\\x=11\end{matrix}\right.\)
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\(\frac{1}{a}-1=\frac{a+b+c}{a}-\frac{a}{a}=\frac{b+c}{a}\)
Tương tự : \(\frac{1}{b}-1=\frac{c+a}{b};\frac{1}{c}-1=\frac{a+b}{c}\)
Nhân theo vế ta đc :
\(VT=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Áp dụng bđt Cauchy :
\(VT\ge\frac{8abc}{abc}=8\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
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`d)D=1/2sqrt{a^10b^8}`
`=1/2sqrt{(a^5b^4)^2}`
`=1/2|a^5b^4|`
`=1/2b^4|a^5|`
`a>=0=>D=1/2b^4a^5`
`a<=0=>D=-1/2b^4a^5`
`e)sqrt{(2-sqrt3)^2}+sqrt2`
`=2-sqrt3+sqrt2``
`f)3sqrt5-sqrt{(1-sqrt5)^2}`
`=3sqrt5-(sqrt5-1)`
`=2sqrt5+1`
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a) Ta có: \(\angle LDI+\angle LBI=90+90=180\Rightarrow LDIB\) nội tiếp
\(\Rightarrow\angle DIL=\angle DBL=\angle DBC=45\) mà \(\Delta IDL\) vuông tại D
\(\Rightarrow\Delta DIL\) vuông cân tại D
b) Ta có: \(\dfrac{1}{DL^2}+\dfrac{1}{DK^2}=\dfrac{1}{DC^2}\) (hệ thức lượng trong tam giác LDK)
mà \(\Delta DIL\) vuông cân tại D \(\Rightarrow DL=DI\Rightarrow\dfrac{1}{DI^2}+\dfrac{1}{DK^2}=\dfrac{1}{DC^2}\)
mà CD là cố định \(\Rightarrow\) đpcm
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`\sqrt((2a)/3) . \sqrt((3a)/8)`
`= \sqrt((2a)/3 . (3a)/8)`
`= \sqrt((6a^2)/24)`
`=(a\sqrt6)/(2\sqrt6)`
`=1/2 a`
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Đặt x+y=a; x-y=b
Hệ \(\Leftrightarrow\left\{{}\begin{matrix}3a-2b=9\\2a+b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a-2b=9\\4a+2b=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a-2b=9\\7a=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-2b=9\\a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-3\\a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-3\\x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=-2\\x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\-1+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+5y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+5y=9\\15x+5y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x=-14\\y=\dfrac{9-x}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)