Khi gặp mặt nước vỏ trai hé mở.mặt mở là:

A.mặt bụng

B.mặt lưng

C.bản lề vỏ

C.cả a,b,c đều đúng

Khi gặp mặt nước vỏ trai hé mở.mặt mở là:

A.mặt bụng

B.mặt lưng

C.bản lề vỏ

C.cả a,b,c đều đúng

\(1.Ca+2HCl->CaCl2+H2\)

\(2.H2+Cl2->2HCl\)

\(3.2Mg+O2->2MgO\)

\(4.FeO+C0->Fe+CO2\)

Ca + 2HCl -> CaCl₂ + H₂

4Na + O₂ -> 2Na₂O

H₂ + Cl₂ -> 2HCl

2Mg + O₂ -> 2MgO

FeO + CO -> Fe + CO₂

Bài 1:

a) Nhẹ hơn

b) Nhẹ hơn

c) Nặng hơn

d) Nhẹ hơn

Bài 2:

a) Ta có: \(M_X=16\cdot2=32\left(đvC\right)\)

  Vậy X là Lưu huỳnh  (S)

b) Nhẹ hơn

c) 16 hạt

d) 17 hạt

ừm:)

hí lô:> ko giúp âu:>

\(\dfrac{x}{-3}=\dfrac{y}{7}\Rightarrow\dfrac{x}{6}=\dfrac{y}{-14};\dfrac{y}{-2}=\dfrac{z}{5}\Rightarrow\dfrac{y}{-14}=\dfrac{z}{35}\\ \Rightarrow\dfrac{x}{6}=\dfrac{y}{-14}=\dfrac{z}{35}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{6}=\dfrac{y}{-14}=\dfrac{z}{35}=\dfrac{2x}{12}=\dfrac{4y}{-56}=\dfrac{5z}{175}=\dfrac{-2x-4y+5z}{-12+56+175}=\dfrac{146}{219}=\dfrac{2}{3}\\ \Rightarrow\left\{{}\begin{matrix}x=6\cdot\dfrac{2}{3}=4\\y=-14\cdot\dfrac{2}{3}=-\dfrac{28}{3}\\z=35\cdot\dfrac{2}{3}=\dfrac{70}{3}\end{matrix}\right.\)

x/-3=y/7;y/-2=z/5 và -2x-4y+5z=146
BCNN(7,2)=14
=>x/-3=y/7;y/-2=z/5
=>x/-3=y/7=>x/6=y/14(1)
=>y/-2=z/5=>y/-14=z/35(2)
từ(1) và (2) =>x/6=y/-14=z/35 và -2x-4y+5z=146
Sử dụng tính chất dãy tỉ số bằng nhau:
=>x/6=y/-14=z/35=>-2x-4y+5z/(-2).6-4.(-14)+5.35=146/219=2/3
=>x/6=2/3=>x=2.6/3=4
=>y/-14=2/3=>y=-14.2/3=-28/3
=>z/35=2/3=>z=35.2/3=70/3

Bài 4:

\(28x^3+6x^2+12x+8=0\)

\(\Leftrightarrow28x^3+14x^2-8x^2-4x+16x+8=0\)

\(\Leftrightarrow14x^2\left(2x+1\right)-4x\left(2x+1\right)+8\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(14x^2-4x+8\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x^2-\dfrac{2}{7}x+\dfrac{4}{7}\right)=0\)

\(\Leftrightarrow2x+1=0\) hay \(\left(x^2-\dfrac{2}{7}x+\dfrac{4}{7}\right)=0\)

\(\Leftrightarrow x=\dfrac{-1}{2}\) hay \(x^2-2.\dfrac{1}{7}x+\dfrac{1}{49}+\dfrac{27}{49}=0\)

\(\Leftrightarrow x=\dfrac{-1}{2}\) hay  \(\left(x-\dfrac{1}{7}\right)^2+\dfrac{27}{49}=0\) (vô nghiệm vì \(\left(x-\dfrac{1}{7}\right)^2+\dfrac{27}{49}\ge\dfrac{27}{49}\))

-Vậy \(S=\left\{\dfrac{-1}{2}\right\}\)

Bài 3: 

a) AB//CD \(\Rightarrow\widehat{BAM}=\widehat{ACD}\) (so le trong)

\(\widehat{AMB}=\widehat{ADC}=90^0\)

\(\Rightarrow\)△ABM∼△CAD (g-g).

b) △ADC vuông tại D \(\Rightarrow AD^2+DC^2=AC^2\Rightarrow AD^2+AB^2=AC^2\Rightarrow AC=\sqrt{AD^2+AB^2}=\sqrt{9^2+12^2}=15\left(cm\right)\)△ADC có DN phân giác \(\Rightarrow\dfrac{NA}{NC}=\dfrac{DA}{DC}\)

\(\Rightarrow\dfrac{NA}{DA}=\dfrac{NC}{DC}=\dfrac{NA+NC}{DA+DC}=\dfrac{AC}{DA+DC}\)

\(\Rightarrow NC=\dfrac{AC.DC}{DA+DC}=\dfrac{15.12}{9+12}=\dfrac{60}{7}\left(cm\right)\)

△ADC có NK//AD (cùng vuông góc với DC) \(\Rightarrow\dfrac{NK}{AD}=\dfrac{NC}{AC}\)

\(\Rightarrow NK=\dfrac{NC}{AC}.AD=\dfrac{\dfrac{60}{7}}{15}.9=\dfrac{36}{7}\left(cm\right)\)

c) △ABM∼△CAD \(\Rightarrow\dfrac{BM}{AD}=\dfrac{AM}{CD}\Rightarrow\dfrac{BM}{AM}=\dfrac{AD}{CD}\Rightarrow\dfrac{BM}{AM}=\dfrac{AN}{CN}\)

\(\Rightarrow BM.CN=AM.AN\)

△BMC∼△ABC (g-g)\(\Rightarrow\dfrac{BM}{AB}=\dfrac{BC}{AC}\Rightarrow BM=\dfrac{AB.BC}{AC}\Rightarrow\dfrac{1}{BM}=\dfrac{AC}{AB.BC}\Rightarrow\dfrac{1}{BM^2}=\dfrac{AC^2}{AB^2.BC^2}=\dfrac{AB^2+BC^2}{AB^2.BC^2}=\dfrac{1}{AB^2}+\dfrac{1}{BC^2}\)

       Christmas is a magical and enchanting time of year. It is a joyous and festive occasion filled with love, happiness, and excitement. The air is filled with a warm and cozy atmosphere, as families come together to celebrate. The beautifully decorated Christmas tree sparkles with twinkling lights and shimmering ornaments. Delicious and indulgent feasts are prepared, filling the air with mouth-watering aromas. Generosity and kindness abound as people exchange thoughtful and heartfelt gifts. The joyful laughter of children fills the air, accompanied by the melodic sounds of Christmas carols. The winter scenery is breathtaking, with glistening snowflakes and frosty landscapes. The spirit of Christmas is truly magical, bringing warmth, joy, and togetherness to all.

    Christmas is a magical and enchanting time of year. It is a joyous and festive occasion filled with love, happiness, and excitement. The air is filled with a warm and cozy atmosphere, as families come together to celebrate. The beautifully decorated Christmas tree sparkles with twinkling lights and shimmering ornaments. Delicious and indulgent feasts are prepared, filling the air with mouth-watering aromas. Generosity and kindness abound as people exchange thoughtful and heartfelt gifts. The joyful laughter of children fills the air, accompanied by the melodic sounds of Christmas carols. The winter scenery is breathtaking, with glistening snowflakes and frosty landscapes. The spirit of Christmas is truly magical, bringing warmth, joy, and togetherness to all.

Sửa đề: I là trung điểm của BC

AB=AM+MB=3+2=5(cm)

AC=AN+NC=7,5+5=12,5(cm)

Xét ΔABC có \(\frac{AM}{MB}=\frac{AN}{NC}\)

nên MN//BC

Xét ΔABC có MN//BC

nên \(\frac{MN}{BC}=\frac{AM}{AB}=\frac35\)

Xét ΔOMN và ΔOCB có

\(\hat{OMN}=\hat{OCB}\) (hai góc so le trong, MN//CB)

\(\hat{MON}=\hat{COB}\) (hai góc đối đỉnh)

Do đó: ΔOMN~ΔOCB

=>\(\frac{OM}{OC}=\frac{ON}{OB}=\frac{MN}{CB}=\frac{2\cdot MK}{2\cdot CI}=\frac{MK}{CI}\)

Xét ΔOMK và ΔOCI có

\(\frac{OM}{OC}=\frac{MK}{CI}\)

\(\hat{OMK}=\hat{OCI}\) (hai góc so le trong, MK//CI)

Do đó: ΔOMK~ΔOCI

=>\(\hat{MOK}=\hat{COI}\)

mà \(\hat{MOK}+\hat{KOC}=180^0\) (hai góc kề bù)

nên \(\hat{COI}+\hat{KOC}=180^0\)

=>K,O,I thẳng hàng

mà A,K,I thẳng hàng

nên A,K,O,I thẳng hàng

=>A,O,I thẳng hàng

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