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ĐK:x>0
a) \(I=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\left(2\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}+1\right)+1-2\sqrt{x}-1=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)b)
Ta có \(I=2\Leftrightarrow x-\sqrt{x}=2\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow x+\sqrt{x}-2\sqrt{x}-2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)=0\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=-1\left(ktm\right)\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
Vậy x=4 thì I=2
c)
Ta có x>1\(\Leftrightarrow x>\sqrt{x}\Leftrightarrow x-\sqrt{x}>0\)
Vậy \(I-\left|I\right|=x-\sqrt{x}-\left|x-\sqrt{x}\right|=x-\sqrt{x}-\left(x-\sqrt{x}\right)=0\)
d)\(I=x-\sqrt{x}=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{1}{4}\Leftrightarrow I\ge\dfrac{1}{4}\)
Dấu bằng xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy GTNN của I là \(\dfrac{1}{4}\) và xảy ra khi \(x=\dfrac{1}{4}\)
\(a.\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\left(1-\dfrac{1}{x-1}\right)=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}.\dfrac{x-2}{x-1}=\dfrac{\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|}{\left|x-2\right|}.\dfrac{x-2}{x-1}\left(x>1\right)\)
Tới đây dễ r , bạn tự chia TH ra làm nhé :D
\(b.\dfrac{1}{\sqrt{x}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{\sqrt{x^3}-x}{1-\sqrt{x}}=\dfrac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}+\dfrac{x\sqrt{x}-x}{\sqrt{x}-1}=-2\sqrt{x-1}+x\left(x\ge1\right)\)
Bạn ơi câu a có vẻ có vấn đề ý. Nếu bạn áp dụng HĐT thì phải là√(x-2)2 chứ nhỉ. Mong bạn giải đáp
Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
bài 2 :
ta có : \(2+\sqrt{3}\) = \(2+\sqrt{3}\) (1)
\(\sqrt{3}+2=2+\sqrt{3}\) (2)
từ 1 và 2 => \(2+\sqrt{3}=\sqrt{3}+2\) ( tính chất bắc cầu )
Bạn nào làm được bài này thì giúp mình với ạ ! mình đang cần gấp
Bài 4:
\(AH=\sqrt{9\cdot16}=12\left(cm\right)\)
\(AB=\sqrt{9\cdot25}=15\left(cm\right)\)
AC=căn(25^2-15^2)=20(cm)
Xét ΔABC vuông tại A có sin ABC=AC/BC=4/5
nên góc ABC=53 độ
ĐKXĐ: x≥0; x≠9
Đặt P= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
P=\(\dfrac{2\sqrt{x}\cdot\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}-3\right)}-\dfrac{3-11\sqrt{x}}{x-9}\)
\(P=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3-11\sqrt{x}}{x-9}\)
P=\(\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
P=\(\dfrac{3x+9\sqrt{x}}{x-9}\)
P=\(\dfrac{3\sqrt{x}\cdot\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}\)
P=\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
Vậy P=\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\) với x≥0: x≠9
\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
\(a.Để:\dfrac{1}{\sqrt{1-x}}\) xác định thì :
\(1-x\) ≥ 0 ⇔ x ≤ 1
\(b.Để:\sqrt{x^2+\text{ |}x\text{ |}+1}\) xác định thì :
\(x^2+\text{ |}x\text{ |}+1\) ≥ 0
+) Với : x ≥ 0 thì :
\(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
+) Với : x < 0 , thì :
\(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
KL.....
Câu a : x < 1 thui nhé ::v ( nhầm )