1 staying

2 her

3 was

4 There

5 him

6 He

7 gave

8 of

9 you

10 at

40.B         41.A        42.A       43.D          44.B        45.A        46.D

Câu 3 : 

\(n_{SO_3}=\dfrac{8}{80}=0.1\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(0.1....................0.1\)

b) Cho quỳ tím vào => quỳ tím hóa đỏ

\(m_{H_2SO_4}=0.1\cdot98=9.8\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)

Câu 2 : 

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe_2O_3+3H_2\underrightarrow{^{t^0}}2Fe+3H_2O\)

\(0.1.........0.3.........0.2\)

\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

Câu 2,3,5 í ạ

Bài 5:

3xy+6x=1-y

=>\(3x\left(y+2\right)-1+y=0\)

=>\(3x\left(y+2\right)+y+2-3=0\)

=>\(3x\left(y+2\right)+\left(y+2\right)=3\)

=>(y+2)(3x+1)=3

=>\(\left(3x+1\right)\cdot\left(y+2\right)=1\cdot3=3\cdot1=\left(-1\right)\cdot\left(-3\right)=\left(-3\right)\cdot\left(-1\right)\)

=>\(\left(3x+1;y+2\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(3x,y\right)\in\left\{\left(0;1\right);\left(2;-1\right);\left(-2;-5\right);\left(-4;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;1\right);\left(\dfrac{2}{3};-1\right);\left(-\dfrac{2}{3};-5\right);\left(-\dfrac{4}{3};-3\right)\right\}\)

mà x,y nguyên

nên \(\left(x,y\right)\in\left(0;1\right)\)

Bài 2:

a: \(3\left(2x+1\right)-6=27\)

=>\(3\left(2x+1\right)=33\)

=>\(2x+1=\dfrac{33}{3}=11\)

=>2x=11-1=10

=>\(x=\dfrac{10}{2}=5\)

b: \(5+3^{x+1}+2\cdot3^{x+2}=194\)

=>\(5+3^x\cdot3+2\cdot3^x\cdot9=194\)

=>\(21\cdot3^x=189\)

=>\(3^x=9\)

=>x=2

c: \(\left(x^3+8\right)\left(x^2-4\right)=0\)

=>\(\left[{}\begin{matrix}x^3+8=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^3=-8\\x^2=4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-2\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow x\in\left\{2;-2\right\}\)

d: \(3x-2⋮x-3\)

=>\(3x-9+7⋮x-3\)

=>\(7⋮x-3\)

=>\(x-3\in\left\{1;-1;7;-7\right\}\)

=>\(x\in\left\{4;2;10;-4\right\}\)

Bài 3:

Gọi số học sinh khối 6 là x(bạn)

(Điều kiện: \(x\in Z^+\))

\(15=3\cdot5;20=2^2\cdot5;25=5^2\)

=>\(BCNN\left(15;20;25\right)=5^2\cdot3\cdot2^2=300\)

Vì số học sinh khi xếp hàng 15;20;25 đều thiếu 2 người

nên ta có: \(x+2\in BC\left(15;20;25\right)\)

=>\(x+2\in B\left(300\right)\)

=>\(x+2\in\left\{300;600;...\right\}\)

=>\(x\in\left\{298;598;...\right\}\)

mà x<400

nên x=298(nhận)

Vậy: Khối 6 có 298 bạn

Bài 1

a) \(-452-\left(-67+75-452\right)\)

\(=-452+67-75+452\)

\(=\left(-452+452\right)+\left(67-75\right)\)

\(=0-8\)

\(=-8\)

b) \(61.64+32.\left(-7\right)+15.\left(-32\right)\)

\(=61.32.2-32.7-32.15\)

\(=32.\left(61.2-7-15\right)\)

\(=32.\left(122-22\right)\)

\(=32.100\)

\(=3200\)

c) \(\left(-3\right)^2.125.11.\left(-2\right)^3\)

\(=9.125.11.\left(-8\right)\)

\(=\left(9.11\right).\left[125.\left(-8\right)\right]\)

\(=99.\left(-1000\right)\)

\(=-99000\)

d) \(2353-\left(473+2353\right)+\left(-55+373\right)\)

\(=2353-473-2353-55+373\)

\(=\left(2353-2353\right)-\left(473-373\right)-55\)

\(=0-100-55\)

\(=-155\)

Bài 2

a) \(3\left(2x+1\right)-6=27\)

\(3\left(2x+1\right)=27+6\)

\(3\left(2x+1\right)=33\)

\(2x+1=33:3\)

\(2x+1=11\)

\(2x=11-1\)

\(2x=10\)

\(x=\dfrac{10}{2}\)

\(x=5\)

b) \(5+3^{x+1}+2.3^{x+2}=194\)

\(3^{x+1}.\left(1+2.3\right)=194-5\)

\(3^{x+1}.7=189\)

\(3^{x+1}=\dfrac{189}{7}\)

\(3^{x+1}=27\)

\(3^{x+1}=3^3\)

\(x+1=3\)

\(x=3-1\)

\(x=2\)

c) \(\left(x^3+8\right)\left(x^2-4\right)=0\)

\(\Rightarrow x^3+8=0;x^2-4=0\)

*) \(x^3+8=0\)

\(x^3=-8\)

\(x^3=\left(-2\right)^3\)

\(x=-2\)

*) \(x^2-4=0\)

\(x^2=4\)

\(x=-2;x=2\)

Vậy \(x=-2;x=2\)

d) \(\left(3x-2\right)⋮\left(x-3\right)\)

\(\Rightarrow\left(3x-9+7\right)⋮\left(x-3\right)\)

\(\Rightarrow\left[3\left(x-3\right)+7\right]⋮\left(x-3\right)\)

\(\Rightarrow7⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow x\in\left\{-4;2;4;10\right\}\)

Bài 1:

a: \(-452-\left(-67+75-452\right)\)

\(=-452+67-75+452\)

\(=\left(-452+452\right)+\left(67-75\right)\)

=-8+0

=-8

b: \(61\cdot64+32\left(-7\right)+15\left(-32\right)\)

\(=61\cdot64+32\left(-15-7\right)\)

\(=32\left(2\cdot61-22\right)=32\cdot100=3200\)

c: \(\left(-3\right)^2\cdot125\cdot11\cdot\left(-2\right)^3\)

\(=9\cdot125\cdot11\cdot\left(-8\right)\)

\(=\left(-8\cdot125\right)\cdot\left(9\cdot11\right)\)

\(=-99\cdot1000=-99000\)

d: \(2353-\left(473+2153\right)+\left(-55+373\right)\)

\(=2353-473-2153-55+373\)

\(=\left(2353-2153\right)+\left(373-473\right)-55\)

\(=200-100-55=45\)

Bài 2: 

a: \(x^2-6x-y^2-4y+5\)

\(=x^2-6x+9-\left(y^2+4y+4\right)\)

\(=\left(x-3\right)^2-\left(y+2\right)^2\)

b: \(4a^2-12a-b^2+2b+8\)

\(=4a^2-12a+9-\left(b^2-2b+1\right)\)

\(=\left(2a-3\right)^2-\left(b-1\right)^2\)

c: \(\left(x+y-3\right)\left(x+y+3\right)\)

\(=\left(x+y\right)^2-3^2\)

d: \(\left(3z+x+2y\right)\left(2y-x+3z\right)\)

\(=\left(2y+3z\right)^2-x^2\)

Viết 2 bức xong trả lời câu hỏi giúp mình luôn nha <3  

Giúp mình với mình đang cần gấp ạ :< 

Giúp mình với mình đang cần gấp ạ :<