Mình cân bằng pt rồi bạn tự chọn tỉ lệ theo cặp chất nhé! :))

1. H₂SO₄ + K₂CO₃ \(\rightarrow\) K₂SO₄  + H₂O + CO₂\(\uparrow\)

2. Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

3, 2Al + 3PbCl₂ \(\rightarrow\) 2AlCl₃ + 3Pb

1: \(=\dfrac{4\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{3\left(x+1\right)}{-20\left(x-1\right)}=\dfrac{-12}{20}\cdot\dfrac{1}{x+1}=\dfrac{-3}{5x+5}\)

2: \(=\dfrac{x^2-xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)

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

3: \(=\dfrac{1-4x^2-1}{1-2x}:\dfrac{4x^2-2x-4x^2}{2x-1}\)

\(=\dfrac{4x^2}{2x-1}\cdot\dfrac{2x-1}{-2x}\)

=-2x

1. I'd rather you didn't ask me that question

2. I haven't seen Bob seen I was in HCM City

3. He would rather read books than watch TV

4. It took Peter three hours to repaint his house

5. He asked me if I knew to speak English

6. We haven't met each other for ten years

7. The film was so boring that she fell asleep

8. The furniture was too expensive for me to buy

9. The weather is so good that they are going for a picnic

10. The coffee is too hot for me to drink

1: AD=8-2=6cm

AD/AB=6/8=3/4

AE/AC=9/12=3/4

=>AD/AB=AE/AC

2: Xét ΔADE và ΔABC có

AD/AB=AE/AC
góc A chung

=>ΔADE đồng dạng với ΔABC

3: AI là phân giác

=>IB/IC=AB/AC

=>IB/IC=AD/AE

=>IB*AE=AD*IC

c3: R2 nt( R1//Rv)

a,\(\Rightarrow U1=Uv=60V\Rightarrow I1=\dfrac{60}{R1}=\dfrac{60}{2000}=\dfrac{3}{100}A\)

\(\Rightarrow U2=U-U1=180-60=120V\Leftrightarrow I2=\dfrac{U2}{R2}=\dfrac{120}{3000}=0,04A\)

b, khi mach von ke song song vs R2

do von ke mang dien tro Rv(khong ly tuong)ne coi von ke la 1 dien tro

=>(R2//Rv)nt R1

\(\Rightarrow Iv=I2-I1=0,04-\dfrac{3}{100}=0,01A\)

\(\Rightarrow Rv=\dfrac{U1}{Iv}=\dfrac{60}{0,01}=6000\Omega\)

\(\Rightarrow Uv=Im.\left(\dfrac{R2.Rv}{R2+Rv}\right)=\left(\dfrac{180}{R1+\dfrac{R2Rv}{R2+Rv}}\right)\left(\dfrac{R2.Rv}{R2+Rv}\right)=90V\)

c4:

a, von ke chi 0V=>(R1 nt R2)//(R3 nt R4)

\(Uv=0\Rightarrow-U1+U3=0\Leftrightarrow U3=U1\)

\(\Rightarrow I34.R3=I12.R1\)

\(\Rightarrow\dfrac{U34}{R3+R4}.R3=\dfrac{U12}{R1+R2}.R1\Leftrightarrow\dfrac{20R3}{20+R3}=\dfrac{40}{10}\Rightarrow R3=5\Omega\)

b, von ke chi 4V \(\Rightarrow4=-U1+U3\)

\(\Rightarrow U3=4+U1\Rightarrow\dfrac{20R3}{20+R3}=4+\dfrac{40}{10}\Rightarrow R3=13,333\Omega\)

a: Xét tứ giác BFCE có

D là trung điểm của BC

D là trung điểm của FE

Do dó: BFCE là hình bình hành

b: Xét tứ giác ABFE có 

AB//FE

AB=FE

Do đó: ABFE là hình bình hành

mà \(\widehat{FAB}=90^0\)

nên ABFE là hình chữ nhật

vcl