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Bài I:
1: Thay x=4 vào A, ta được:
\(A=\dfrac{4}{2+1}=\dfrac{4}{3}\)
2: \(B=\dfrac{3}{\sqrt{x}+1}+\dfrac{x+5}{x-1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3}{\sqrt{x}+1}+\dfrac{\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)+x+5-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\sqrt{x}-3+x-\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
3: P=A*B
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\)
P<=4
=>P-4<=0
=>\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-1}< =0\)
=>\(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}< =0\)
=>\(\sqrt{x}-1< 0\)
=>\(\sqrt{x}< 1\)
=>0<=x<1
Kết hợp ĐKXĐ, ta được: 0<=x<1
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Bài 1:
Phần a bạn tự làm nha! (Đ/S: 0,5)
b, B = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\) với \(x\ge0;x\ne4;x\ne9\)
B = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
B = \(\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
B = \(\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
B = \(\dfrac{1}{\sqrt{x}-2}=\dfrac{\sqrt{x}+2}{x-4}\)
Vậy ...
c, Ta có: A = \(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\)= \(\dfrac{1}{\sqrt{x}+1}\)
T = \(\dfrac{A}{B}\)= \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)= 1 - \(\dfrac{3}{\sqrt{x}+1}\)
Ta có: x \(\ge\) 0 \(\Leftrightarrow\) \(\sqrt{x}\ge0\) \(\Leftrightarrow\) \(\sqrt{x}+1\ge1\) \(\Leftrightarrow\) \(\dfrac{3}{\sqrt{x}+1}\le3\) \(\Leftrightarrow\) \(-\dfrac{3}{\sqrt{x}+1}\ge-3\) \(\Leftrightarrow\) T \(\ge\) -2
Vậy ...
Bài 2: ĐK: x \(\ge\) 0
Giả sử: \(P\) < \(\sqrt{P}\)
\(\Leftrightarrow\) \(\dfrac{\sqrt{x}+2}{\sqrt{x}+5}< \dfrac{\sqrt{\sqrt{x}+2}}{\sqrt{\sqrt{x}+5}}\)
\(\Leftrightarrow\) \(\dfrac{\sqrt{\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)}-\left(\sqrt{x}+2\right)}{\sqrt{x}+5}>0\)
\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)}-\left(\sqrt{x}+2\right)>0\) (\(\sqrt{x}+5>0\) với mọi x \(\ge\) 0)
\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x}+2\right)}\sqrt{\sqrt{x}+5-\sqrt{x}-2}>0\)
\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x}+2\right)}\sqrt{3}>0\)
\(\Leftrightarrow\) \(\sqrt{\sqrt{x}+2}>0\)
Vì x \(\ge\) 0 \(\Leftrightarrow\) \(\sqrt{x}+2\ge2\) \(\Leftrightarrow\) \(\sqrt{\sqrt{x}+2}\ge\sqrt{2}>0\) (Đpcm)
Vậy \(P\) < \(\sqrt{P}\)
Chúc bn học tốt!
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DKXD : \(x\ge-1;y\ne-1\)
Dat : \(\left\{{}\begin{matrix}\sqrt{x+1}=a\left(a\ge0\right)\\y+1=b\left(b\ne0\right)\end{matrix}\right.\)
hpt<=>\(\left\{{}\begin{matrix}a+2-\dfrac{2}{y+1}=2\\2a-\dfrac{1}{y+1}=\dfrac{3}{2}\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}a+2-\dfrac{2}{b}=2\\2a-\dfrac{1}{b}=\dfrac{3}{2}\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}a-\dfrac{2}{b}=0\\4a-\dfrac{2}{b}=3\end{matrix}\right.< =>\left\{{}\begin{matrix}3a=3\\a=\dfrac{2}{b}\end{matrix}\right.< =>\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)(tmdk)
\(=>\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)(tmdk)
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c: Xét ΔBAC vuông tại A có AH là đường cao
nên \(BH\cdot BC=AB^2\left(1\right)\)
Xét ΔABM vuông tại A có AK là đường cao
nên \(BK\cdot BM=AB^2\left(2\right)\)
Từ (1) và (2) suy ra \(BH\cdot BC=BK\cdot BM\)
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a: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
c: Ta có: \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-3\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\)
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\(2,\\ a,=6+12-15=3\\ b,=\left(2\sqrt{2}-3\sqrt{2}\right)\sqrt{2}=-\sqrt{2}\cdot\sqrt{2}=-2\\ c,=\dfrac{2\left(\sqrt{3}-1\right)}{2}-\dfrac{\sqrt{3}+2}{-1}+\dfrac{6\left(3-\sqrt{3}\right)}{6}\\ =\sqrt{3}-1+\sqrt{3}+2+3-\sqrt{3}=4+\sqrt{3}\)
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