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1) Thay x=16 vào A ta có:
A=\(\frac{16+\sqrt{16}+1}{\sqrt{16}+2}\)
A=\(\frac{16+4+1}{4+2}\)
A=\(\frac{21}{6}=\frac{7}{2}\)
\(2,\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x-\sqrt{x}+2}{x-\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2x-x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}\)\(\left(đpcm\right)\)
\(3,P=A.B=\frac{x+\sqrt{x}+1}{\sqrt{x}+2}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
Ta thấy \(\left(\sqrt{x}-1\right)^2>0\Rightarrow x-2\sqrt{x}+1>0\)
\(\Rightarrow x+\sqrt{x}+1>3\sqrt{x}\)
\(\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>\frac{3\sqrt{x}}{\sqrt{x}}\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>3\left(đpcm\right)\)
binh rồi căn thì cứ chuyển bỏ dấu âm đi nó tương tự dấu giá trị tuyệt đối thôi
a) \(sin\left(x\right)=\frac{2}{3}\)
\(x=\arcsin \left(\frac{2}{3}\right)+2\pi n,\:x=\pi -\arcsin \left(\frac{2}{3}\right)+2\pi n\)
b) \(sin\left(x\right)=\frac{\sqrt{3}}{2}\)
\(x=\frac{\pi }{3}+2\pi n,\:x=\frac{2\pi }{3}+2\pi n\)
Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\) ( ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\)
+ Với \(x=1\)\(\Rightarrow\)\(\frac{x+3}{2}=0\)\(\Leftrightarrow\)\(x=-3\)( loại )
+ Với \(x>1\)\(\Rightarrow\)\(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)
\(\Leftrightarrow\)\(2\sqrt{x-1}=\frac{x+3}{2}\)
\(\Leftrightarrow\)\(4\sqrt{x-1}=x+3\)
\(\Leftrightarrow\)\(16\left(x-1\right)=\left(x+3\right)^2\)
\(\Leftrightarrow\)\(16x-16=x^2+6x+9\)
\(\Leftrightarrow\)\(x^2-10x+25=0\)
\(\Leftrightarrow\)\(\left(x-5\right)^2=0\)
\(\Leftrightarrow\)\(x=5\)\(\left(TM\right)\)
Vậy \(S=\left\{5\right\}\)
ĐKXĐ: x≥1x≥1
Ta có:
√x+2√x−1+√x−2√x−1=x+32⇔√(√x−1+1)2+√(√x−1−1)2=x+32⇔√x−1+1+∣∣√x−1−1∣∣=x+32⇔√x−1+∣∣√x−1−1∣∣=x+12(1)x+2x−1+x−2x−1=x+32⇔(x−1+1)2+(x−1−1)2=x+32⇔x−1+1+|x−1−1|=x+32⇔x−1+|x−1−1|=x+12(1)
Ta xét 2 trường hợp sau:
TH1: x≥2x≥2
Khi đó:
(1)⇔2√x−1−1=x+12⇔2√x−1=x+32⇔16(x−1)=x2+6x+9⇔x2−10x+25=0⇔(x−5)2=0⇔x=5(TMĐK)(1)⇔2x−1−1=x+12⇔2x−1=x+32⇔16(x−1)=x2+6x+9⇔x2−10x+25=0⇔(x−5)2=0⇔x=5(TMĐK)
TH2: 1≤x<21≤x<2
Khi đó:
(1)⇔1=x+12⇔x=1(TMĐK)(1)⇔1=x+12⇔x=1(TMĐK)
Vậy x=1 hoặc x=5