b: \(AC^2-HC^2=AH^2\)

\(AB^2-BH^2=AH^2\)

Do đó: \(AC^2-HC^2=AB^2-HB^2\)

hay \(AC^2+HB^2=AB^2+HC^2\)

b, Ta có \(AC^2-HC^2=AH^2;AB^2-HB^2=AH^2\left(pytago\right)\)

Do đó \(AC^2-CH^2=AB^2-HB^2\)

\(\Rightarrow AC^2+HB^2=AB^2+HC^2\)

a: Thay m=3 vào hệ, ta được:

\(\begin{cases}\left(3+1\right)x-y=\left(3+1\right)\\ x+\left(3-1\right)y=2\end{cases}\Rightarrow\begin{cases}4x-y=4\\ x+2y=2\end{cases}\)

=>\(\begin{cases}8x-2y=8\\ x+2y=2\end{cases}\Rightarrow\begin{cases}8x-2y+x+2y=8+2\\ x+2y=2\end{cases}\)

=>\(\begin{cases}9x=10\\ x+2y=2\end{cases}\Rightarrow\begin{cases}x=\frac{10}{9}\\ 2y=2-x=2-\frac{10}{9}=\frac89\end{cases}\Rightarrow\begin{cases}x=\frac{10}{9}\\ y=\frac89:2=\frac49\end{cases}\)

b: Để hệ có nghiệm duy nhất thì \(\frac{m+1}{1}<>\frac{-1}{m-1}\)

=>\(m^2-1<>-1\)

=>\(m^2<>0\)

=>m<>0

\(\begin{cases}\left.\left(m+1\right)x-y=m+1\right.\\ x+\left(m-1\right)y=2\end{cases}\Rightarrow\begin{cases}\left(m+1\right)x-y=\left(m+1\right)\\ \left(m+1\right)x+\left(m^2-1\right)y=2\left(m+1\right)\end{cases}\)

=>\(\begin{cases}x\left(m+1\right)-y-x\left(m+1\right)-\left(m^2-1\right)y=\left(m+1\right)-2\left(m+1\right)\\ x+\left(m-1\right)y=2\end{cases}\)

=>\(\begin{cases}-y\left(1+m^2-1\right)=-\left(m+1\right)\\ x=2-\left(m-1\right)y\end{cases}\Rightarrow\begin{cases}y=\frac{m+1}{m^2}\\ x=2-\frac{\left(m-1\right)\left(m+1\right)}{m^2}=2-\frac{m^2-1}{m^2}=2-1+\frac{1}{m^2}=1+\frac{1}{m^2}\end{cases}\)

Để x,y nguyên thì 1⋮\(m^2\) và m+1⋮\(m^2\)

=>\(\begin{cases}1\vdots m^2\\ m^2-1\vdots m^2\end{cases}\Rightarrow\begin{cases}1\vdots m^2\\ -1\vdots m^2\end{cases}\Rightarrow1\vdots m^2\)

=>\(m^2=1\)

=>m=1 hoặc m=-1

3.\(A=\left(\dfrac{x-4}{\sqrt{x}-2}+\dfrac{x-2\sqrt{x}}{\sqrt{x}}\right):\sqrt{x}\)

\(=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}}\right).\dfrac{1}{\sqrt{x}}\)

\(=\left(\sqrt{x}+2+\sqrt{x}-2\right).\dfrac{1}{\sqrt{x}}=2\sqrt{x}.\dfrac{1}{\sqrt{x}}=2\)

Tham Khảo!!!

\(3.A=\left(\dfrac{x-4}{\sqrt{x-2}}+\dfrac{x-2\sqrt{x}}{\sqrt{x}}\right):\sqrt{x}\)

\(=\left(\dfrac{\sqrt{x-2}.\sqrt{x+2}}{\sqrt{x-2}}+\dfrac{\sqrt{x}\left(\sqrt{x-2}\right)}{\sqrt{ }}x\right).\dfrac{1}{\sqrt{x}}\)

\(=\left(\sqrt{x}+2+\sqrt{x}-2\right).\dfrac{1}{\sqrt{x}}\)

\(=2\sqrt{x}.\dfrac{1}{\sqrt{x}}\)

\(=2\)

Câu 3: 

Trong 1 giờ vòi 1 chảy được 1/3(bể)

TRong 1 giờ vòi 2 chảy được 1/5(bể)

Vì 1/3>1/15 nên trong 1 giờ, vòi 1 chảy được nhiều hơn và nhiều hơn 2/15(bể)

Câu 3:

a) Gọi số mol Al, Fe là a, b (mol)

=> 27a + 56b = 1,1 (1)

\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

             a----->1,5a---------------->1,5a

             Fe + H₂SO₄ --> FeSO₄ + H₂

               b---->b---------------->b

=> 1,5a + b = 0,04 (2)

(1)(2) => a = 0,02 (mol); b = 0,01 (mol)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02.27}{1,1}.100\%=49,1\%\\\%m_{Fe}=\dfrac{0,01.56}{1,1}.100\%=50,9\%\end{matrix}\right.\)

b) \(m_{H_2SO_4\left(lý.thuyết\right)}=\left(1,5a+b\right).98=3,92\left(g\right)\)

=> \(m_{H_2SO_4\left(tt\right)}=\dfrac{3,92.120}{100}=4,704\left(g\right)\)

=> \(C\%=\dfrac{4,704}{49}.100\%=9,6\%\)

c) \(\left\{{}\begin{matrix}n_{Al}=\dfrac{0,66.49,1\%}{27}=0,012\left(mol\right)\\n_{Fe}=\dfrac{0,66.50,9\%}{56}=0,006\left(mol\right)\end{matrix}\right.\)

PTHH: 2Al + 6H₂SO₄ --> Al₂(SO₄)₃ + 3SO₂ + 6H₂O

          0,012------------------------->0,018

            2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

           0,006------------------------->0,009

=> V_SO2 = (0,018 + 0,009).22,4 = 0,6048 (l)

Question3:

1- Her father works in a factory.  (work==>works)

2- My mother can cook well.   (OK)

3- He has learnt English for 3 years. (since==>for)