Giúp câu 14 , 15 nha
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X x 14 + X x 15 = 292929
Xx(14+15)=292929
X x 29=292929
X=292929:29
X=10101
Học tốt!!!!
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\(\dfrac{3}{1-\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{3\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)^2}{-1}=-\left(3\sqrt{2}+3-3+2\sqrt{2}\right)=-5\sqrt{2}\)
\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}=\dfrac{\left(\sqrt{5}-1\right).\left(1-\sqrt{5}\right)+6.\left(\sqrt{5}+1\right)}{-4}=\dfrac{6-2\sqrt{5}-6\sqrt{5}-6}{4}=\dfrac{-8\sqrt{5}}{4}=-2\sqrt{5}\)
\(\dfrac{\sqrt{2}-\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}=\dfrac{\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{6}+2\right)+\left(\sqrt{3}-\sqrt{2}\right).\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\left(\sqrt{12}-\sqrt{18}\right)}{-2}=\sqrt{18}-\sqrt{12}\)
\(\dfrac{-31+8\sqrt{x}-x}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
\(=\dfrac{-31+8\sqrt{x}-x}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}+\dfrac{3\sqrt{x}-1}{\sqrt{x}-5}\)
\(=\dfrac{-31+8\sqrt{x}-x-x+25+3x-9\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
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Câu 14:
\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\Rightarrow n_H=0,2.2=0,4\left(mol\right)\)
Có: mC + mH = 5,2 (g) ⇒ mC = 5,2 - 0,4.1 = 4,8 (g)
\(\Rightarrow n_{CO_2}=n_C=\dfrac{4,8}{12}=0,4\left(mol\right)\)
A: CnH2n-2
⇒ nA = nCO2 - nH2O = 0,4 - 0,2 = 0,2 (mol)
\(\Rightarrow n=\dfrac{n_{CO_2}}{n_A}=2\)
→ A là C2H2.
→ Đáp án: B
Câu 15:
X: CnH2n-2
\(\Rightarrow\dfrac{2n-2}{12n+2n-2}=0,11111\Rightarrow n=4\)
→ X là C4H6.
CTCT: \(CH\equiv C-CH_2-CH_3\)
\(CH_3-C\equiv C-CH_3\)
→ Đáp án: B
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 14:
Ta có: \(n_{C_2H_4}+n_A=\dfrac{3,36}{22,4}=0,15\left(1\right)\)
m bình tăng = mC2H4 = 1,4 (g) \(\Rightarrow n_{C_2H_4}=\dfrac{1,4}{28}=0,05\left(mol\right)\) (2)
Từ (1) và (2) \(\Rightarrow n_A=0,1\left(mol\right)\)
Gọi CTPT của A là CnH2n+2.
Có: \(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\)
\(\Rightarrow n=\dfrac{n_{CO_2}}{n_A}=2\)
Vậy: A là C2H6.
→ Đáp án: A.
Câu 15:
Ta có: \(n_{CO_2}=n_{H_2O}=x\left(mol\right)\)
\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
BTNT O, có: \(2n_{CO_2}+n_{H_2O}=2n_{O_2}\)
\(\Rightarrow2x+x=2.0,6\Leftrightarrow x=0,4\left(mol\right)\)
BTNT C, có: \(n_{CaCO_3}=n_{CO_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,4.100=40\left(g\right)\)
→ Đáp án: C
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 13:
1:
a: \(2x^2+2x=2x\cdot x+2x\cdot1=2x\left(x+1\right)\)
b: \(9x^2-4y^2\)
\(=\left(3x\right)^2-\left(2y\right)^2\)
=(3x-2y)(3x+2y)
2:
\(\dfrac{xy+2x+1}{xy+x+y+1}+\dfrac{yz+2y+1}{yz+y+z+1}+\dfrac{zx+2z+1}{zx+z+x+1}\)
\(=\dfrac{xy+2x+1}{\left(y+1\right)\left(x+1\right)}+\dfrac{yz+2y+1}{\left(z+1\right)\left(y+1\right)}+\dfrac{z\left(x+2\right)+1}{\left(z+1\right)\left(x+1\right)}\)
\(=\dfrac{\left(xy+2x+1\right)\left(z+1\right)+\left(yz+2y+1\right)\left(x+1\right)+\left(xz+2z+1\right)\left(y+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\dfrac{xyz+xy+2xz+2x+z+1+xyz+yz+2xy+2y+x+1+\left(xz+2z+1\right)\left(y+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\dfrac{2xyz+3xy+2xz+3x+z+2+yz+2y+x+xyz+xz+2zy+2z+y+1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\dfrac{3xyz+3xy+3xz+3yz+3x+3z+3y+3}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\dfrac{3\left(xyz+xy+xz+yz+x+z+y+1\right)}{\left(xy+x+y+1\right)\left(z+1\right)}\)
=3
Câu 14:
1:
f(0)=0+5=5
2:
Vì hệ số góc của y=ax+b là -1 nên a=-1
=>y=-x+b
Thay x=1 và y=2 vào y=-x+b, ta được:
b-1=2
=>b=3
14: B
15: A
Câu 14: B
Câu 15: A