Mấy bạn tách pt này giúp mik với ạ cảm ơn mấy bn rất nhiều
m2 -8m -16 =0
Mik sẽ tích cho bn nào giải
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\(A=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)(đk: \(x>0,x\ne1\))
\(=\frac{\sqrt{x}\left[\left(\sqrt{x}\right)^3-1\right]}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
ĐK : x > 0 , x khác 1
\(A=\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
Trả lời:
a, \(P=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\frac{1}{\sqrt{x}+1}-\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{2}{x-1}\right)\) \(\left(ĐK:x\ge0;x\ne1\right)\)
\(=\left[\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left(\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\right)\)
\(=\left[\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{2}{x-1}\right]\)
\(=\frac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{x-1}:\frac{\sqrt{x}-1-\sqrt{x}\left(\sqrt{x}+1\right)+2}{x-1}\)
\(=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}:\frac{\sqrt{x}-1-x-\sqrt{x}+2}{x-1}\)
\(=\frac{4\sqrt{x}}{x-1}:\frac{1-x}{x-1}=\frac{4\sqrt{x}}{x-1}\cdot\frac{x-1}{1-x}=\frac{4\sqrt{x}}{1-x}\)
a, \(\left(\frac{1}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}+2}\right):\frac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)ĐK : x >= 0 ; \(x\ne1\)
\(=\left(\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b, \(F=\frac{5}{2}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\)
ĐK : x > 0 , x khác 1
\(bthuc=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
Để bthuc = 5/2 thì \(\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\left(tm\right)\)
HPT<=>\(\hept{\begin{cases}2\left(u+v\right)+v^2+2uv+u^2=15\\u^2+v^2=5\end{cases}}\)
\(< =>\hept{\begin{cases}\left(u+v+1\right)^2=16\\u^2+v^2=5\end{cases}}\)
\(< =>\hept{\begin{cases}u+v=3\\u^2+v^2=5\end{cases}or\hept{\begin{cases}u+v=-5\\u^2+v^2=5\end{cases}}}\)
đến đến thì dễ r haaaa
\(\hept{\begin{cases}uv+u+v=5\\u^2+v^2=5\end{cases}}\)
\(u^2+v^2=\left(u+v\right)^2-2uv=\left(u+v\right)^2-2\left[5-\left(u+v\right)\right]\)
\(=\left(u+v\right)^2+2\left(u+v\right)-10=5\)
\(\Leftrightarrow\left(u+v\right)^2+2\left(u+v\right)-15=0\)
\(\Leftrightarrow\left(u+v+5\right)\left(u+v-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}u+v=-5\\u+v=3\end{cases}}\)
- \(u+v=-5\Rightarrow uv=10\)
\(u,v\)là hai nghiệm của phương trình: \(x^2+5x+10=0\)(1)
mà \(x^2+5x+10=x^2+2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2+\frac{15}{4}=\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\)
nên phương trình (1) vô nghiệm.
- \(u+v=3\Rightarrow uv=2\)
\(u,v\)là hai nghiệm của phương trình \(x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy \(\left(u,v\right)\in\left\{\left(1,2\right),\left(2,1\right)\right\}\).
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