giúp mình vs ạ mình cần gấp lắm
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
3 careless
4 bleed
V
1 Hung has colllected stamps since 2000
2 The front yard isn't large enough to play soccer in
3 Milk is delivered twice a week by the milkman
4 When did you bought this computer?
5 His car doesn't run as fasst as a race car
6 Two languages can be spoken well by Lan
VI
1 T
2 F
3 T
4 T
VII
1 Mr. Nam is arriving in Hue tonight
2 They are going back to Englang in two months
3 It is not difficult to tránlate this sentence into English
4 Would you mind lending me your dictionary?
3 careless
4 bleed
V
1 Hung has colllected stamps since 2000
2 The front yard isn't large enough to play soccer in
3 Milk is delivered twice a week by the milkman
4 When did you bought this computer?
5 His car doesn't run as fasst as a race car
6 Two languages can be spoken well by Lan
VI
1 T
2 F
3 T
4 T
VII
1 Mr. Nam is arriving in Hue tonight
2 They are going back to Englang in two months
3 It is not difficult to tránlate this sentence into English
4 Would you mind lending me your dictionary?
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
a: Xét tứ giác ABDC có
M là trung điểm của BC
M là trung điểm của AD
Do đó: ABDC là hình bình hành
mà \(\widehat{BAC}=90^0\)
nên ABDC là hình chữ nhật
Lời giải:
Đặt \(\sqrt[3]{5\sqrt{2}+7}=m; \sqrt[3]{5\sqrt{2}-7}=n\)
\(m^3-n^3=14\)
\(mn=1\)
\((a+b+c)^3=(m-n)^3=m^3-3mn(m-n)-n^3=14-3(m-n)\)
\(\Leftrightarrow (a+b+c)^3=14-3(a+b+c)\)
\(\Leftrightarrow (a+b+c)^3+3(a+b+c)-14=0\)
\(\Leftrightarrow (a+b+c)^2[(a+b+c)-2]+2(a+b+c)(a+b+c-2)+7(a+b+c-2)=0\)
\(\Leftrightarrow (a+b+c-2)[(a+b+c)^2+2(a+b+c)+7]=0\)
Dễ thấy biểu thức trong ngoặc vuông $>0$ nên $a+b+c-2=0$
$\Leftrightarrow a+b+c=2$
$ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{2^2-1}{2}=\frac{3}{2}$