Các bạn giúp mình làm vs ạ
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Bài 3:
1: =>2x=-5/3-1/2=-10/6-3/6=-13/6
hay x=-13/12
2: =>3/5x=1/7+3/5=5/35+21/35=26/35
hay x=26/3
3: =>-3x=5/6+3/4=10/12+9/12=19/12
hay x=-19/36
4: =>1/2x=3/7-5/4=12/28-35/28=-23/28
hay x=-23/14
5: =>1/4x=-3/5-7/5=-2
hay x=-8
6: =>3x=1/42+1/7=1/42+6/42=1/7
hay x=1/21
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!
a) 2112 . 3163
= ...1 . ...6
= ...6
b) 415
= (42)7 . 4
= (...6). ...4
= ...4
c) 7123
= (74)30 . 73
= (...1)30 . ...3
= ...1 . ...3
= ...3
d) 8567
= (84)141 . 83
= (...6)141 . ...2
= ...6 . ...2
= ...2
~Study well~
#Thạc_Trân
have played
Were - was
enjoys collecting - has had
goes - has been
is said to be
Bài 5:
a: ĐKXĐ: a>0; a<>1; a<>4
b: \(B=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1+a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{2a-5}\)
\(=\dfrac{\sqrt{a}-2}{\sqrt{a}\left(2a-5\right)}\)