a, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)    \(\left(1\right)\)

b, Đổi \(20ml=0,02l\)

\(n_{H_2SO_4}=0,02.1=0,02\left(mol\right)\)

Thep phương trình \(\left(1\right)\) ta được:

\(n_{NaOH}=2n_{H_2SO_4}=2.0,02=0,04\left(mol\right)\\ \Rightarrow m_{NaOH}=0,04.40=1,6\left(g\right)\\ \Rightarrow m_{dd NaOH}=\dfrac{1,6}{20}.100=8\left(g\right)\)

c, \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)    \(\left(2\right)\)

Theo phương trình \(\left(2\right)\):

\(n_{KOH}=2n_{H_2SO_4}=2.0,02=0,04\left(mol\right)\\ \Rightarrow m_{KOH}=0,04.56=2,24\left(g\right)\\ \Rightarrow m_{dd KOH}=\dfrac{2,24}{5,6}.100=40\left(g\right)\\ \Rightarrow V_{dd KOH}=\dfrac{40}{1.045}=38,3\left(ml\right)\)

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5,

a. \(\left(x+y+z\right)^2\)

\(=\left[\left(x+y\right)+z\right]^2\\ =\left(x+y\right)^2+2\left(x+y\right)z+z^2\\ =x^2+2xy+y2+\left(2x+2y\right)z+z^2\\ =x^2+y^2+z^2+2xy+2yz+2zx\)

3

a,\(\left(x-5\right)\left(x+5\right)-x\left(x-4\right)=0\)

\(\Leftrightarrow x^2-5^2-x^2+4x=0\\ \Leftrightarrow\left(x^2-x^2\right)+4x-5^2\\ \Leftrightarrow4x=25\\ \Leftrightarrow x=\dfrac{25}{4}\)

Vậy....

b,\(\left(2x-1\right)\left(4x^2+2x+1\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow8x^3-1-\left(x^2-1^2\right)=0\\ \Leftrightarrow8x^3-x^2-\left(1-1\right)=0\\ \Leftrightarrow8x^3-x^2=0\\ \Leftrightarrow x^2\left(8x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\8x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{8}\end{matrix}\right.\)

Vậy...

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

\(\Leftrightarrow x^2-6x+9+2x^2-3x=3x^2-6x+9\\ \Leftrightarrow x^2+2x^2-3x^2-6x+6x-3x=9-9\\ \Leftrightarrow-3x=0\\ \Leftrightarrow x=0\)

Vậy...

Ta có: \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)+4\left(x-7\right)\)

\(=x^2-4x+4-x^2+3^2+4x-28\\ =\left(x^2-x^2\right)-\left(4x-4x\right)+\left(4+3^2-28\right)\\=-15 \)

⇒ Gía trị của biểu thức \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)+4\left(x-7\right)\) không phụ thuộc vào biến \(x\) \(\left(đcpcm\right)\)

@Kudo Shinichi AKIRA^_^ 

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