CẢM ƠN TRƯỚC ^^ (lm bnhiêu cũng đc == lm hết càng tốt ạ •cảm ơn• ^^ [Mong ai đó giúp🙏]
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![](https://rs.olm.vn/images/avt/0.png?1311)
11)11) 3x(x-5)2-(x+2)3+2(x-1)3-(2x+1)(4x2-2x+1)=3x(x2-10x+25)-(x3+6x2+12x+8)+2(x3-3x2+3x-1)-(8x3+1)=3x3-30x2+75x-x3-6x2-12x-8+2x3-6x2+6x-2-8x3-1=-4x3-42x2+63x-11
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a. $\frac{x}{2}=\frac{3,6}{1,2}=3$
$x=3.2=6$
b.
$\frac{8}{2x+1}=\frac{4}{3}$
$2x+1=\frac{8.3}{4}=6$
$2x=6-1=5$
$x=\frac{5}{2}$
c. $\frac{x}{4}=\frac{9}{x}$
$x^2=9.4=36=6^2=(-6)^2$
$\Rightarrow x=\pm 6$
d.
$\frac{x+1}{2}=\frac{32}{x+1}$
$(x+1)^2=32.2=64=8^2=(-8)^2$
$\Rightarrow x+1=8$ hoặc $x+1=-8$
$\Rightarrow x=7$ hoặc $x=-9$
![](https://rs.olm.vn/images/avt/0.png?1311)
9. There are four floors in my house.
10. Your bag is smaller than my bag.
9, There are four floors in my house.
10, Your bag is smaller than mine
![](https://rs.olm.vn/images/avt/0.png?1311)
11 c)
\(a^2+2\ge2\sqrt{a^2+1}\Leftrightarrow a^2+1-2\sqrt{a^2+1}+1\ge0\Leftrightarrow\left(\sqrt{a^2+1}-1\right)^2\ge0\) (luôn đúng)
12 a) Có a+b+c=1\(\Rightarrow\) (1-a)(1-b)(1-c)= (b+c)(a+c)(a+b) (*)
áp dụng BĐT cô-si: \(\left(b+c\right)\left(a+c\right)\left(a+b\right)\ge2\sqrt{bc}2\sqrt{ac}2\sqrt{ab}=8\sqrt{\left(abc\right)2}=8abc\) ( luôn đúng với mọi a,b,c ko âm )
b) áp dụng BĐT cô-si: \(c\left(a+b\right)\le\dfrac{\left(a+b+c\right)^2}{4}=\dfrac{1}{4}\)
Tương tự: \(a\left(b+c\right)\le\dfrac{1}{4};b\left(c+a\right)\le\dfrac{1}{4}\)
\(\Rightarrow abc\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{4}\dfrac{1}{4}\dfrac{1}{4}=\dfrac{1}{64}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có Cu ko tác dụng được vs H2SO4
-> 2,24 lít khí H2 đc tạo ra là do phản ứng HH của Fe va H2SO4
\(PTHH:\) \(Fe+H_2SO_4->FeSO_4+H2\)
n H2 = 2,24 :22,4=0,1 mol
n H2 = n Fe =0,1 mol
m Fe = 0,1.56=5,6 g
m Cu = 10-5,6=4,4 g
vậy giá trị của x là 4,4 g
26
PTHH : \(Zn+2HCl->ZnCl_2+H_2\)
n Zn = 6,5:65=0,1 mol
n ZnCl2 = n Zn =0,1 mol
m ZnCl2 = 0,1.(65+35,5.2)=13,6 g
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(x-3.5\right)^2\ge0\forall x\)
\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)
Do đó: \(\left(x-3.5\right)^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(\dfrac{7}{2};\dfrac{1}{10}\right)\)
do
\(\left(x-3.5\right)^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
mà ta có \(\left(x-3.5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)
nên \(\left(x-3.5\right)^2+\left(y-\dfrac{1}{10}\right)^4=0\)
suy ra \(\left\{{}\begin{matrix}x-3,5=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}\end{matrix}\right.\)
tick mik nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Ex 1:
1. listen
2. rains
3. will recognise
4. don’t want
5. does/ begin
6. will not
7. does/ watches
8. plant
9. play
10. leaves
Ex 2:
1. watching
2. listening
3. to buy
4. to speak
5. making
6. to eat
7. working
8. to call
9. to build
10. doing
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: - \(\dfrac{5}{7}\) x \(\dfrac{31}{33}\) + \(\dfrac{-5}{7}\) x \(\dfrac{2}{33}\) + 2\(\dfrac{5}{7}\)
= - \(\dfrac{5}{7}\) \(\times\) ( \(\dfrac{31}{33}\) + \(\dfrac{2}{33}\)) + 2 + \(\dfrac{5}{7}\)
= - \(\dfrac{5}{7}\) + 2 + \(\dfrac{5}{7}\)
= 2
2, \(\dfrac{3}{14}\): \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\): \(\dfrac{1}{28}\) - 8
= (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) : \(\dfrac{1}{28}\) - 8
= \(\dfrac{2}{7}\) x 28 - 8
= 8 - 8
= 0
câu 63: A
câu 64: B
câu 65: C
mấy câu còn lại phải từ từ đã!
Câu 63: A
Câu 64: B
Câu 65: C
Câu 71: D
Câu 72: A
Câu 73: C
Câu 74: B
Câu 75: C
Câu 76: B
Câu 77: A
Câu 78: C
Câu 79: B
Câu 80: C