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\(2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow-5\sqrt{x}+2=0-2x\)
\(\Leftrightarrow-5\sqrt{x}+2=-2x\)
\(\Leftrightarrow-5\sqrt{x}=-2x-2\)
\(\Leftrightarrow\left(-5\sqrt{x}\right)^2=\left(-2x-2\right)^2\)
\(\Leftrightarrow25x=4x^2+8x+4\)
\(\Leftrightarrow4x^2+8x+4=25x\) (chuyển vế)
\(\Leftrightarrow4x^2+8x+4-25x=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(\Leftrightarrow x\left(4x-1\right)-\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=1\end{cases}}\)
Vậy nghiệm phương trình là: \(\left\{\frac{1}{4};1\right\}\)
ĐK:\(x\ge0\)
pt\(\Leftrightarrow\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}-2\right)=0\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\Rightarrow\orbr{\begin{cases}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{4}\\x=4\end{cases}}}\left(TMĐK\right)\)
Vậy....
\(a,\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)=\left(1+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1^2-\sqrt{a}^2=1-a\)
\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
\(c,\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)=\left(3+\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\right)\left(3-\frac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)
\(=\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)=3^2-\sqrt{a}^2=3-a\)
\(d,\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
\(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}.6\sqrt{3}=36\)
\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)
\(=11-4\sqrt{6}\)
\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
\(\sqrt{x^2+3x+3}=1\)
\(\Leftrightarrow x^2+3x+3=1\)
\(\Leftrightarrow x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
\(2\sqrt{x+2+2\sqrt{x+1}}-\sqrt{x+1}=4\)
\(\Leftrightarrow2\sqrt{x+1+2\sqrt{x+1}+1}-\sqrt{x+1}=4\)
\(\Leftrightarrow2\sqrt{\left(\sqrt{x+1}+1\right)^2}-\sqrt{x+1}=4\)
\(\Leftrightarrow2\left(\sqrt{x+1}+1\right)-\sqrt{x+1}=4\)
\(\Leftrightarrow2\sqrt{x+1}+2-\sqrt{x+1}=4\)
\(\Leftrightarrow\sqrt{x+1}=2\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
VÌ AM LÀ ĐƯỜNG TRUNG TUYẾN ỨNG VỚI CẠNH HUYỀN
SUY RA AM=1/2*BC=1/2*10=5 CM
XÉT TAM GIÁC AHM VUÔNG TẠI H[VÌ AH LÀ ĐƯỜNG CAO]
SUY RA MH^2=AM^2-AH^2[PI TA GO]
MH^2=5^2-4,8^2
MH^2=1,96
MH=1,4
LẠI CÓ
BH=BM+MH=1/2*BC+1,4=5+1,4=6,4[CM]
TA CÓ:
CH=CM-MH=1/2BC-MH=5-1,4=3,6
TAM GIÁC ABH
AB^2=BH^2+AH^2
SUY RA AB^2=6,4^2+4,8^2=64 AB=8[CM]
TAM GIÁC ABC
AC^2=BC^2-AB^2
AC^2=10^2-8^2=36 AC=6[CM]
Ta có : \(\frac{DE}{DF}=\frac{3}{7}\Rightarrow DE=\frac{3}{7}DF\)
Xét tam giác DEF vuông tại D, đường cao DI
* Áp dụng hệ thức : \(\frac{1}{DI^2}=\frac{1}{DE^2}+\frac{1}{DF^2}=\frac{1}{\left(\frac{3}{7}DF\right)^2}+\frac{1}{DF^2}\Rightarrow\frac{1}{1764}=\frac{1}{\left(\frac{3}{7}DF\right)^2}+\frac{1}{DF^2}\)
\(\Rightarrow DF=14\sqrt{58}\)cm
\(\Rightarrow DE=\frac{3}{7}DF=\frac{3}{7}.14\sqrt{58}=6\sqrt{58}\)cm
Áp dụng định lí Pytago tam giác DIE vuông tại I, đường cao DI
\(ED^2=EI^2+DI^2\Rightarrow EI=\sqrt{ED^2-DI^2}=18\)cm
* Áp dụng hệ thức : \(DI^2=EI.FI\Rightarrow FI=\frac{DI^2}{EI}=98\)cm
