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Bài 1 :
\(a,2\sqrt{50}-3\sqrt{72}+\sqrt{98}=2\sqrt{2.25}-3\sqrt{2.36}+\sqrt{2.49}=10\sqrt{2}-18\sqrt{2}+7\sqrt{2}\) = \(-\sqrt{2}\)
\(b,\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}+\sqrt{28}\) = \(\left|3-\sqrt{5}\right|-\left|\sqrt{5}-\sqrt{7}\right|+\sqrt{7.4}=3-\sqrt{5}-\sqrt{5}+\sqrt{7}+2\sqrt{7}=3-2\sqrt{5}+3\sqrt{7}\)
\(c,\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{3+2.2\sqrt{3}+4}=\)\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}+2\right)^2}=\left|-\left(2-\sqrt{3}\right)\right|+\left|\sqrt{3}+2\right|=2-\sqrt{3}+\sqrt{3}+2=4\)
Bài 1:
a: ĐKXĐ: x>0; x<>1
b: \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2}{\sqrt{x}-1}\)
c: Thay \(x=6+2\sqrt{5}\) vào A, ta được:
\(A=\dfrac{2}{\sqrt{5}+1-1}=\dfrac{2\sqrt{5}}{5}\)
d: Để |A|>A thì A>0
=>\(\sqrt{x}-1>0\)
hay x>1
a, không nhìn rõ
b, \(\dfrac{a+2\sqrt{a}+1}{a-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)
Đề 1: TỰ LUẬN
Câu 1: sin 60o31' = cos 29o29'
cos 75o12' = sin 14o48'
cot 80o = tan 10o
tan 57o30' = cot 32o30'
sin 69o21' = cos 20o39'
cot 72o25' = 17o35'
- Chiều về mình làm cho nha nha Giờ mình đi học rồi Bạn có gấp lắm hông
Bài 1:
a)
\(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\) ĐKXĐ: x >1
\(=\left(\dfrac{2\sqrt{x}.\sqrt{x}}{2.2\sqrt{x}}-\dfrac{2}{2.2\sqrt{x}}\right)\left(\dfrac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)^2}-\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)^2}\right)\\ =\left(\dfrac{2x-2}{4\sqrt{x}}\right)\left(\dfrac{x\sqrt{x}-x-x+\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{\left(x-1\right)^2}\right)\\ =\left(\dfrac{x-1}{2\sqrt{x}}\right)\left(\dfrac{-4x}{\left(x-1\right)^2}\right)\\ =\dfrac{\left(x-1\right).\left(-4x\right)}{2\sqrt{x}.\left(x-1\right)^2}=\dfrac{-2\sqrt{x}}{x-1}\)
b)
Với x >1, ta có:
A > -6 \(\Leftrightarrow\dfrac{-2\sqrt{x}}{x-1}>-6\Rightarrow-2\sqrt{x}>-6\left(x-1\right)\)
\(\Leftrightarrow-2\sqrt{x}+6x-6>0\\ \Leftrightarrow x-\dfrac{2}{6}\sqrt{x}-1>0\\ \Leftrightarrow x-2.\dfrac{1}{6}\sqrt{x}+\left(\dfrac{1}{6}\right)^2>1+\dfrac{1}{36}\\ \Leftrightarrow\left(\sqrt{x}-\dfrac{1}{6}\right)^2>\dfrac{37}{36}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{6}-\sqrt{x}>\dfrac{\sqrt{37}}{6}\\\sqrt{x}-\dfrac{1}{6}>\dfrac{\sqrt{37}}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-\sqrt{x}>\dfrac{\sqrt{37}-1}{6}\\\sqrt{x}>\dfrac{\sqrt{37}+1}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-x>\dfrac{19-\sqrt{37}}{18}\\x>\dfrac{19+\sqrt{37}}{18}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{\sqrt{37}-19}{18}\\x>\dfrac{19+\sqrt{37}}{18}\end{matrix}\right.\)
Vậy không có x để A >-6
\(A=\left(1-\frac{\sqrt{x}}{1+\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)(ĐK: \(x\ge0,x\ne4,x\ne9\))
\(=\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\div\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{1}{1+\sqrt{x}}\times\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(A< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{2}\)(\(x\ge0,x\ne4,x\ne9\))
\(\Leftrightarrow\frac{2\sqrt{x}-4-\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-5< 0\)
\(\Leftrightarrow x< 25\).
Vậy \(0\le x< 25,x\ne4,x\ne9\)thì \(A< \frac{1}{2}\).
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\inℤ\Leftrightarrow\frac{3}{\sqrt{x}+1}\inℤ\)
suy ra \(\sqrt{x}+1\inƯ\left(3\right)=\left\{1,3\right\}\)vì \(x\inℤ\)
\(\Rightarrow x\in\left\{0,4\right\}\)
Đối chiếu điều kiện ta chỉ có \(x=0\)thỏa mãn.