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3:
a: \(\Leftrightarrow x+1-6\sqrt{x+1}-9=0\)
=>\(\left(\sqrt{x+1}-3\right)=0\)
=>x+1=9
=>x=8
b: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\left(\sqrt{\dfrac{1}{2}x+1}+3\right)}}=10\)
=>\(\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}-\dfrac{21}{4}}=10\)
=>\(\dfrac{1}{2}x-\dfrac{21}{4}-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}=100\)
=>\(\dfrac{7}{4}\cdot\sqrt{\dfrac{1}{2}x+1}=\dfrac{1}{2}x-\dfrac{21}{4}-100=\dfrac{1}{2}x-\dfrac{421}{4}\)
=>\(\sqrt{\dfrac{1}{2}x+1}=\dfrac{2}{7}x-\dfrac{421}{7}\)
=>1/2x+1=(2/7x-421/7)^2
=>1/2x+1=4/49x^2-1684/49x+177241/49
=>\(x\simeq249,77;x\simeq177,36\)
\(\Delta'=16-\left(3m+1\right)\ge0\Rightarrow m\le5\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-8\\x_1x_2=3m+1\end{matrix}\right.\)
Kết hợp điều kiện đề bài ta được: \(\left\{{}\begin{matrix}x_1+x_2=-8\\5x_1-x_2=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=-8\\6x_1=-6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=-7\end{matrix}\right.\)
Thế vào \(x_1x_2=3m+1\)
\(\Rightarrow\left(-1\right).\left(-7\right)=3m+1\)
\(\Rightarrow m=2\) (thỏa mãn)
\(x+2\sqrt{x}-3\\ =x-\sqrt{x}+3\sqrt{x}-3\\ =\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)\\ =\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)\)
\(A=\left(\dfrac{2\sqrt{x}+x+1}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
(ĐKXĐ: x\(\ge\) 0 ; x \(\ne\) 1 )
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
\(=\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right):\left(1-\sqrt{x}\right)\)
\(=\sqrt{x}+1\)
\(A=\left(\dfrac{2\sqrt{x}+x+1}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right)\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)=\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\left(1-\sqrt{x}\right)=\left(1-x\right)\left(1-\sqrt{x}\right)=1-\sqrt{x}-x+x\sqrt{x}=x\sqrt{x}-x-\sqrt{x}+1\)
a: \(Q=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{x-4}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)
\(=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{x-4}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{-x+2\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\cdot\left(-1\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b: Khi x=4-2căn 3 thì \(Q=\dfrac{\sqrt{3}-1+2}{\sqrt{3}-1-3}=\dfrac{\sqrt{3}+1}{\sqrt{3}-4}=\dfrac{-7-5\sqrt{3}}{13}\)
c: Q>1/6
=>Q-1/6>0
=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{1}{6}>0\)
=>\(\dfrac{6\sqrt{x}+12-\sqrt{x}+3}{6\left(\sqrt{x}-3\right)}>0\)
=>\(\dfrac{5\sqrt{x}+9}{6\left(\sqrt{x}-3\right)}>0\)
=>căn x-3>0
=>x>9
m2 -8m -16 =0
m2 -2.4m -4\(^2\) =0
(m - 4)\(^2\) = 0
=> m -4 = 0
=> m = 4
HT
m2 - 8m - 16 = 0 <=> m2 - 8m + 16 - 32 = 0
<=> ( m - 4 )2 - ( 4√2 )2 = 0 <=> ( m - 4 - 4√2 )( m - 4 + 4√2 ) = 0
<=> m = 4 ± 4√2