Cho đường thẳng (d): y = -2x + 3. Xác định tọa độ giao điểm A;B của đường thẳng (d) với hai trục Ox, Oy. Tính khoảng cách từ điểm O(0;0) đến đường thẳng (d)
Chương I - Căn bậc hai. Căn bậc ba
* Giao điểm với trục Ox:
Ta có: -2x + 3 = 0
⇔ 2x = 3
⇔ x = 3/2
⇒ A(3/2; 0) là giao điểm với trục Ox
* Giao điểm với trục Oy:
x = 0 ⇔ y = 3
⇒ B(0; 3) là giao điểm với trục Oy
* Khoảng cách từ O(0; 0) tới (d):
Xét đồ thị:
Ta có:
AB² = OA² + OB² (Pytago)
= (3/2)² + 3²
= 45/4
⇒ AB = 3√5/2
Khoảng cách từ O đến (d) là đoạn thẳng OH
Ta có:
OH.AB = OA.OB
⇒ OH = OA.OB : AB
= 3/2 . 3 : (3√5/2)
= 3/√5
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khoảng cách là \(\dfrac{3}{\sqrt{5}}\)
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a) Thay x=36 vào B ta có:
\(B=\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)
b) \(B< \dfrac{1}{2}\) khi:
\(\dfrac{\sqrt{x}}{\sqrt{x}-3}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{2\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}-3\right)}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}< 0\)
Mà: \(\sqrt{x}+3\ge3>0\forall\left(x\ge0\right)\)
\(\Leftrightarrow2\left(\sqrt{x}-3\right)< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow x< 9\)
Kết hợp với đk:
\(0< x< 9\)
c) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\left[\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
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a: Khi x=36 thì \(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
b: B<1/2
=>B-1/2<0
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{1}{2}< 0\)
=>\(\dfrac{2\sqrt{x}-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}< 0\)
=>\(\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}< 0\)
=>căn x-3<0
=>0<=x<9
c: \(A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{x-1}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
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a: =>4-3x=64
=>3x=-60
=>x=-20
b: \(\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}=-1\)
=>\(-2\sqrt{x-2}=-1\)
=>căn x-2=1/2
=>x-2=1/4
=>x=9/4
c: \(\Leftrightarrow2x-4\sqrt{x}+\sqrt{x}-2-7=0\)
=>\(2x-3\sqrt{x}-9=0\)
=>\(2x-6\sqrt{x}+3\sqrt{x}-9=0\)
=>căn x-3=0
=>x=9
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a: \(=2\cdot3\sqrt{5}+\sqrt{5}-3\cdot4\sqrt{5}\)
\(=-5\sqrt{5}\)
b: \(=2-\sqrt{3}+\sqrt{3}-1-6\cdot\dfrac{4}{\sqrt{3}}\)
\(=1-4\sqrt{3}\)
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Bài 1:
a) \(\sqrt{4\cdot\dfrac{1}{2}}+\sqrt{32}-\sqrt{72}+\sqrt{162}\)
\(=\sqrt{\dfrac{4}{2}}+\sqrt{4^2\cdot2}-\sqrt{6^2\cdot2}+\sqrt{9^2\cdot2}\)
\(=\sqrt{2}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\)
\(=8\sqrt{2}\)
b) \(\dfrac{7+2\sqrt{12}}{\sqrt{3}+\sqrt{4}}-\dfrac{7-2\sqrt{12}}{\sqrt{4}-\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}}{\sqrt{3}+\sqrt{4}}-\dfrac{7-4\sqrt{12}}{\sqrt{4}-\sqrt{3}}\)
\(=\dfrac{2^2+2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}{\sqrt{3}+2}-\dfrac{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}{2-\sqrt{3}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)^2}{2+\sqrt{3}}-\dfrac{\left(2-\sqrt{3}\right)^2}{2-\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c) \(\left(\dfrac{3}{\sqrt{2}+1}+\dfrac{14}{2\sqrt{2}-1}-\dfrac{4}{2-\sqrt{2}}\right)\cdot\left(\sqrt{8}+2\right)\)
\(=\left[\dfrac{3\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\dfrac{14\left(2\sqrt{2}+1\right)}{\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)}-\dfrac{4\left(2+\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\right]\cdot\left(2\sqrt{2}+2\right)\)
\(=\left[\dfrac{3\left(\sqrt{2}-1\right)}{2-1}+\dfrac{14\left(2\sqrt{2}+1\right)}{8-1}-\dfrac{4\left(2+\sqrt{2}\right)}{4-2}\right]\cdot\left(2\sqrt{2}+2\right)\)
\(=\left[3\left(\sqrt{2}-1\right)+2\left(2\sqrt{2}+1\right)-2\left(2+\sqrt{2}\right)\right]\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
\(=10\cdot\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)
\(=10\cdot\left(2-1\right)\)
\(=10\)
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1:
a: \(\sqrt{4\dfrac{1}{2}}+\sqrt{32}-\sqrt{72}+\sqrt{162}\)
\(=\sqrt{\dfrac{9}{2}}+4\sqrt{2}-6\sqrt{2}+9\sqrt{2}\)
\(=\dfrac{3}{2}\sqrt{2}+7\sqrt{2}=\dfrac{17}{2}\sqrt{2}\)
2:
a: \(\Leftrightarrow x+7-\sqrt{x+7}=0\)
=>\(\sqrt{x+7}\left(\sqrt{x+7}-1\right)=0\)
=>x+7=0 hoặc x+7=1
=>x=-6 hoặc x=-7
b: \(\Leftrightarrow2x-1-\sqrt{2x+19}=0\)
=>\(2x+19-\sqrt{2x+19}-20=0\)
=>\(\left(\sqrt{2x+19}-5\right)\left(\sqrt{2x+19}+4\right)=0\)
=>2x+19=25
=>2x=6
=>x=3
c: \(\Leftrightarrow3\sqrt{x+1}\cdot\dfrac{1}{3}-2\sqrt{x+1}+8\cdot\dfrac{2}{5}\sqrt{x+1}=11\)
=>\(\dfrac{11}{5}\sqrt{x+1}=11\)
=>\(\sqrt{x+1}=5\)
=>x+1=25
=>x=24
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Cho A = 6/(x - 3sqrt(x)) B= (2sqrt(x))/(x - 9) - 2 sqrt x +3 (x>0,x ne9) a) Tính giá trị của A khi x = 16 b) Rút gọn biểu thức P = A/B c) So sánh P với 1. d) Tính x biết P * sqrt(x) >= x/4 + 4
a: Khi x=16 thì \(A=\dfrac{6}{16-3\cdot4}=\dfrac{6}{4}=\dfrac{3}{2}\)
b: P=A:B
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{6}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
c: \(P-1=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}}=\dfrac{3}{\sqrt{x}}>0\)
=>P>1
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1: \(P=\left(\dfrac{-1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\dfrac{-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\dfrac{-\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{x-1}+\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\dfrac{-1}{x-1}+\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\right]\)
\(=\dfrac{-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\dfrac{-x+\sqrt{x}-1+x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\right]\)
\(=\dfrac{-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\dfrac{-\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{-2\sqrt{x}+1}=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
P=7/3
=>\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{7}{3}\)
=>\(3x-3\sqrt{x}+3-7\sqrt{x}=0\)
=>\(3x-10\sqrt{x}+3=0\)
=>\(\left(\sqrt{x}-3\right)\left(3\sqrt{x}-1\right)=0\)
=>x=9 hoặc x=1/9
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bài 1 côsi ngc dấu cho a,b,c >0 và a+b+c=3 cmr a^3/b^3+ab
26 tháng 8 2023 lúc 20:40