Our team played well (too/ that/ so/ enough) to win the gold medal

\(\rightarrow\)Our team played well enough to win the gold medal

1.the party will start (at seven o'clock in the morning).

\(\rightarrow\)When will the party start?

2.i am talking to (Mrs.Hoa).

\(\rightarrow\)Who are you talking to?

3.it is (five kilometers) from our house to the mountain.

\(\rightarrow\)How far is it from your house to the mountain?

4.there are (two pictures) in my room.

\(\rightarrow\)How many picture are there in you room?

\(x=\dfrac{2a-1}{3}\)

1)

Ta có:

\(x>0\)

\(\Leftrightarrow\dfrac{2a-1}{3}>0\)

suy ra 2a - 1và 3 cùng dấu

mà 3>0

\(\Rightarrow2a-1>0\)

\(\Rightarrow a>\dfrac{1}{2}\)

2)

Ta có :

\(x< 0\)

\(\Leftrightarrow\dfrac{2a-1}{3}< 0\)

suy ra 2x-1 và 3 trái dấu

mà 3>0

\(\Rightarrow2a-1< 0\)

\(\Rightarrow a< \dfrac{1}{2}\)

3)

Ta có : x không là số dương không là số âm

\(\Rightarrow x=0\)

\(\Rightarrow\dfrac{2a-1}{3}=0\)

mà \(3\ne0\)

\(\Rightarrow2x-1=0\)

\(\Rightarrow x=\dfrac{1}{2}\)

a > 0

Áp dụng BĐT côsi dạng \(a+b\ge2\sqrt{a.b}\) ta có:

\(a+\dfrac{1}{a}\ge2\sqrt{a.\dfrac{1}{a}}=2\)

Dấu "=" xảy ra khi a=1(đpcm)

It rained yesterday after it _____________ dry for months.

A. was

B. have been

C. had been

D. was being

\(a.\left(2x+1\right)^2=25\)

\(\Leftrightarrow\left(2x+1\right)^2-25=0\)

\(\Leftrightarrow\left(2x+1+5\right)\left(2x+1-5\right)=0\)

\(\Leftrightarrow\left(2x+6\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+6=0\\2x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-3;2\right\}\)

Ta có :\(A^{\Lambda}+B^{\Lambda}+C^{\Lambda}+D^{\Lambda}=360^o\)(\(\Lambda\)là kí hiệu góc nhé)

mà \(A^{\Lambda}+B^{\Lambda}=235^o\)

\(\Rightarrow\)\(C^{\Lambda}+D^{\Lambda}=360^o-235^o=125^o\)

mặt khác:\(C^{\Lambda}-D^{\Lambda}=25^o\Rightarrow D^{\Lambda}=C^{\Lambda}+25^o\)

Gọi x là số đo góc C(\(x\in^{\cdot}\)N^*)(độ)

Khi đó : số đo góc D là x +25(độ)

Theo đề ra ta có phương trình:

\(x+\left(x+5\right)=125\)

\(\Rightarrow2x+5=125\)

\(\Rightarrow x=\dfrac{125-5}{2}=60^o\)

Vậy góc C=60^o

góc D= 65^o

\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)

\(\Leftrightarrow8x^2-24x-\left(8x-8\right)\left(x+1\right)=20\)

\(\Leftrightarrow8x^2-24x-8x^2-8x+8x+8=20\)

\(\Leftrightarrow-24x=12\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

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

a.

\(nNa_2O=\dfrac{m}{M}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

\(P.tr:Na_2O+H_2O\rightarrow2NaOH\)

\(..........1......1.......2\)

\(........0,25.....0,25.....0,5\)

Nồng độ mol của NaOH:

\(C_MNaOH=\dfrac{n}{V}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b)

\(Ptr:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(........2..........1............1.........2\)

\(.......0,5..........0,25.............0,25.......0,5\)

\(mH_2SO_4=n.M=0,25.98=24,5\left(g\right)\)

\(\Rightarrow mddH_2SO_4=\dfrac{mH_2SO_4.100\%}{C\%}=\dfrac{24,5.100\%}{20\%}=122,5\left(g\right)\)

\(\Rightarrow VddH_2SO_4=\dfrac{m}{D}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)

Vậy thể tích dung dịch H₂SO₄ cần dùng là 107,46l

c.Đổi:\(107,46ml=0,10746l\)

Nồng độ mol của Na₂SO₄ thu được sau phản ứng là:

\(C_MNa_2SO_4=\dfrac{n}{V}=\dfrac{0,25}{0,10746}\approx2,33\left(M\right)\)