\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

a

PTHH:

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,2--->0,6----->0,2------->0,3

b

\(C\%_{dd.HCl.đã.dùng}=\dfrac{0,6.36,5.100\%}{200}=10,95\%\)

c

\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(\left(12x-5\right)\left(4x-1\right)-\left(3x-7\right)\left(1+16x\right)=81\\ 48x^2-12x-20x+5-\left(3x+48x^2-7-112x\right)=81\\48x^2-12x-20x+5-3x-48x^2+7+112x=8\\ \left(48x^2-48x^2\right)+\left(-12x-20x-3x+112x\right)+\left(5+7\right)=81\\ 77x+12=81\\ 77x=81-12=69\\ x=\dfrac{69}{77} \)

Vậy \(x=\dfrac{69}{77}\)

Phản ứng trên không xảy ra ở điều kiện thường (không có điều kiện phản ứng)

PTHH:

\(H_2+Cl_2\underrightarrow{á.s}2HCl\) (xảy ra ở điều kiện có chiếu sáng)

ĐK: \(x\ge0\)

\(5\sqrt{x}< 15\\ \Leftrightarrow\sqrt{x}< \dfrac{15}{5}\\ \Leftrightarrow\sqrt{x}< 3\\ \Leftrightarrow x< 3^2\\ \Leftrightarrow x< 9\)

Mà \(x\in Z;x\ge0\) nên \(x\in\left\{0;1;2;3;4;5;6;7;8\right\}\)

Vậy phương trình có nghiệm \(S=\left\{0;1;2;3;4;5;6;7;8\right\}\)

8

a

\(\sqrt{\left(5-\sqrt{3}\right)^2}\\ =\left|5-\sqrt{3}\right|\)

\(=5-\sqrt{3}\) (\(5>\sqrt{3}\))

b

\(\sqrt{\left(1-\sqrt{2}\right)^2}\\ =\left|1-\sqrt{2}\right|\)

\(=-\left(1-\sqrt{2}\right)\) (\(1< \sqrt{2}\))

\(=\sqrt{2}-1\)

7

a

\(x^2+6x+9\\ =x^2+2.3.x+3^2\\ =\left(x+3\right)^2\)

b

\(9x^2-6x+1\\ =\left(3x\right)^2-2.3x.1+1^2\\ =\left(3x-1\right)^2\)

c

\(x^2y^2+xy+\dfrac{1}{4}\\ =\left(xy\right)^2+2.\dfrac{1}{2}.xy+\left(\dfrac{1}{2}\right)^2\\ =\left(xy+\dfrac{1}{2}\right)^2\)

d

\(\left(x-y\right)^2+6\left(x-y\right)+9\\ =\left(x-y\right)^2+2.3\left(x-y\right)+3^2\\ =\left(x-y+3\right)^2\)

8

a

\(x^2+6x+9=\left(x+3\right)^2\)

b

\(4x^2-4x+1=\left(2x-1\right)^2\)

c

\(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)

d

\(\left(x-\dfrac{y}{3}\right)\left(x+\dfrac{y}{3}\right)=x^2-\dfrac{y^2}{9}\)

1

Với \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\)

\(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\left(\dfrac{x^2+2x+1}{4x^4-4x^2+1}\right)\\ =\left(\dfrac{\left(x-1\right)\left(x+1\right)}{\left(2-x\right)\left(x+1\right)}+\dfrac{x^2}{\left(x+1\right)\left(2-x\right)}\right)\left(\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\right)\\ =\dfrac{x^2-1+x^2}{\left(x+1\right)\left(2-x\right)}\left(\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\right)\\ =\dfrac{\left(2x^2-1\right)\left(x+1\right)^2}{\left(x+1\right)\left(2-x\right)\left(2x^2-1\right)^2}\\ =\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}\)

2

Để M = 0 thì \(\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}=0\Rightarrow x+1=0\Rightarrow x=-1\) (loại)

Vậy không có giá trị x thỏa mãn M = 0

Bài 1

1

\(\sqrt{28a^4}=\sqrt{4.7.\left(a^2\right)^2}=\sqrt{\left(2a^2\right)^2.7}=\sqrt{\left(2a^2\right)^2}.\sqrt{7}=2a^2\sqrt{7}\)

2

\(A=\left(\dfrac{\sqrt{21}-\sqrt{7}}{\sqrt{3}-1}+\dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{2}-1}\right):\left(\dfrac{1}{\sqrt{7}-\sqrt{5}}\right)\\ =\left(\dfrac{\sqrt{7}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\right):\left(\dfrac{1}{\sqrt{7}-\sqrt{5}}\right)\\ =\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\\ =7-5\\ =2\)

Bài 2

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)

Hệ phương trình khi đó trở thành:

\(\left\{{}\begin{matrix}\dfrac{3}{2}a-y^2b=6\\a+2y^2b=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-2y^2b=12\\a+2y^2b=-4\end{matrix}\right.\\ \Leftrightarrow\left(3a+a\right)+\left(-2y^2b+2y^2b\right)=12+\left(-4\right)\\ \Leftrightarrow4a=8\\ \Leftrightarrow a=2\)

\(\Rightarrow x=\dfrac{1}{a}=\dfrac{1}{2}\\ \Rightarrow y=\dfrac{3}{2.\dfrac{1}{2}}-6=-3\)

Vậy hệ phương trình có tập nghiệm \(S=\left\{\dfrac{1}{2};-3\right\}\)

Biết a//b theo đề có: \(\widehat{A_1}-\widehat{C_1}=40^o\) \(\Rightarrow\widehat{A_1}=40^o+\widehat{C_1}\left(1\right)\)

Mà \(\widehat{A_1}\) và \(\widehat{C_1}\) bù nhau (do ở vị trí trong cùng phía)

\(\Rightarrow\widehat{A_1}+\widehat{C_1}=180^o\) (2)

Thế (1) vào (2) có: \(40^o+\widehat{C_1}+\widehat{C_1}=180^o\)

\(2\widehat{C_1}=180^o-40^o=140^o\)

\(\Rightarrow\widehat{C_1}=\dfrac{140^o}{2}=70^o\\ \Rightarrow\widehat{A_1}=70^o+40^o=110^o\)

Do có a//b nên:

\(\widehat{A_2}=\widehat{C_1}=70^o\\ \widehat{C_2}=\widehat{A_1}=110^o\) (so le trong)