GIẢI GẤP GIÚP MÌNH Ạ, MÌNH CẦN GẤP
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PTHH : 2Al + 6HCl --> 2AlCl3 + 3H2 ↑ (1)
nAlCl3 = \(\dfrac{m}{M}=\dfrac{13,35}{27+35,5.3}=0.1\left(mol\right)\)
Từ (1) => nHCl = 2nH2 = 0.2 (mol)
=> mHCl = n.M = 0.2 x 36.5 = 7.3 (g)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=\dfrac{m}{M}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\\ Theo.PTHH:n_{HCl}=3.n_{AlCl_3}=3.0,1=0,3\left(mol\right)\\ m_{HCl}=n.M=0,3.36,5=10,95\left(g\right)\)
Lời giải:
a. ĐKXĐ: $x>0; x\neq 4$
b.
\(M=\sqrt{x}.\left[\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right].\frac{x-4}{2\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{x-4}{2}=\frac{2\sqrt{x}}{x-4}.\frac{x-4}{2}=\sqrt{x}\)
c. Để $M>3\Leftrightarrow \sqrt{x}>3\Leftrightarrow x>9$
Kết hợp đkxđ suy ra $x>9$ thì $M>3$
5:
Th1: m=0
=>6x-27=0
=>x=27/6(loại)
TH2: m<>0
Δ=(6m-6)^2-4m(9m-27)
=36m^2-72m+36-36m^2+108m=36m+36
Để phương trình có hai nghiệm pb thì 36m+36>0
=>m>-1
x1+x2=x1x2
=>6(m-1)=9(m-3)
=>9m-27=6m-6
=>3m=21
=>m=7
Câu 1:
\(\sqrt{16}=4\)
\(\sqrt{36}=6\)
\(\sqrt{81}=9\)
\(\sqrt{144}=12\)
\(\sqrt{625}=25\)
\(\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}\)
\(\sqrt{\dfrac{36}{25}}=\dfrac{6}{5}\)
\(\sqrt{\dfrac{64}{49}}=\dfrac{8}{7}\)
\(\sqrt{\dfrac{169}{400}}=\dfrac{13}{20}\)
\(\sqrt{11\dfrac{1}{9}}=\sqrt{\dfrac{100}{9}}=\dfrac{10}{3}\)
\(\sqrt{1\dfrac{11}{25}}=\sqrt{\dfrac{36}{25}}=\dfrac{6}{5}\)
\(\sqrt{1\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}=\dfrac{7}{6}\)
Câu 2:
a) \(3.\sqrt{16}-4\sqrt{\dfrac{1}{4}}\)
\(=3.4-4.\dfrac{1}{2}\)
\(=4.\left(3-\dfrac{1}{2}\right)\)
\(=4.\dfrac{5}{2}\)
\(=10\)
b) \(-5\sqrt{\dfrac{9}{16}}+4\sqrt{0,36}-6\sqrt{0,09}\)
\(=-5.\dfrac{3}{4}+4.0,6-6.0,3\)
\(=\dfrac{-15}{4}+\dfrac{12}{5}-\dfrac{9}{5}\)
\(=\dfrac{-75+48-36}{20}=\dfrac{-63}{20}\)
c) \(2.\sqrt{9}-10.\sqrt{\dfrac{1}{25}}\)
\(=2.3-10.\dfrac{1}{5}\)
\(=6-2\)
\(=4\)
d) \(-3\sqrt{\dfrac{25}{16}}+5\sqrt{0,16}-7\sqrt{0,64}\)
\(=-3.\dfrac{5}{4}+5.0,4-7.0,8\)
\(=\dfrac{-15}{4}+2-\dfrac{28}{5}\)
\(=\dfrac{-75+40-28}{20}=\dfrac{-63}{20}\)
e) \(3\sqrt{25}-27\sqrt{\dfrac{4}{81}}\)
\(=3.5-27.\dfrac{2}{9}\)
\(=15-6\)
\(=9\)
f) \(-21\sqrt{\dfrac{100}{49}}+3\sqrt{0,04}-5\sqrt{0,25}\)
\(=-21.\dfrac{10}{7}+3.0,2-5.0,5\)
\(=-30+\dfrac{3}{5}-\dfrac{5}{2}\)
\(=\dfrac{-300+6-25}{10}=\dfrac{-319}{10}\)
h) \(5\sqrt{9}-4\sqrt{\dfrac{1}{16}}+6\sqrt{25}\)
\(=5.3-4.\dfrac{1}{4}+6.5\)
\(=15-1+30\)
\(=14+30\)
\(=44\)
g) \(10\sqrt{\dfrac{9}{25}}-14\sqrt{\dfrac{36}{49}}+24\sqrt{\dfrac{81}{64}}\)
\(=10.\dfrac{3}{5}-14.\dfrac{6}{7}+24.\dfrac{9}{8}\)
\(=6-12+27\)
\(=\left(-6\right)+27=21\)
Câu 3:
a) \(\sqrt{x}=7\)
\(=>x=49\)
b) \(\sqrt{x}=12\)
\(=>x=144\)
c) \(\sqrt{x}=15\)
\(=>x=225\)
d) \(\sqrt{x}=20\)
\(=>x=400\)
e) \(4\sqrt{x}=8\)
\(\sqrt{x}=8:4\)
\(\sqrt{x}=2\)
\(=>x=4\)
f) \(6\sqrt{x}=3\)
\(\sqrt{x}=\dfrac{3}{6}=\dfrac{1}{2}\)
\(=>x=\dfrac{1}{4}\)
g) \(\sqrt{x-1}=1\)
\(x-1=1\)
\(x=1+1\)
\(=>x=2\)
h) \(\sqrt{x+1}=2\)
\(x+1=4\)
\(x=4-1\)
\(=>x=3\)
i) \(\sqrt{x}-2=7\)
\(\sqrt{x}=7+2\)
\(\sqrt{x}=9\)
\(=>x=81\)
j) \(14-\sqrt{x}=12\)
\(\sqrt{x}=14-12\)
\(\sqrt{x}=2\)
\(=>x=4\)
k) \(12-\sqrt{x-1}=2\)
\(\sqrt{x-1}=12-2\)
\(\sqrt{x-1}=10\)
\(x-1=100\)
\(x=100+1\)
\(=>x=101\)
l) \(\sqrt{x+5}+10=20\)
\(\sqrt{x+5}=20-10\)
\(\sqrt{x+5}=10\)
\(x+5=100\)
\(x=100-5\)
\(=>x=95\)
# Wendy Dang
3:
a: ĐKXĐ: x>=0
\(\sqrt{x}=7\)
=>x=7^2=49
b: ĐKXĐ: x>=0
\(\sqrt{x}=12\)
=>x=12^2=144
c: ĐKXĐ: x>=0
\(\sqrt{x}=15\)
=>x=15^2=225
d: ĐKXĐ: x>=0
\(\sqrt{x}=20\)
=>x=20^2=400
e: ĐKXĐ: x>=0
\(4\sqrt{x}=8\)
=>\(\sqrt{x}=2\)
=>x=4
f: ĐKXĐ: x>=0
\(6\cdot\sqrt{x}=3\)
=>\(\sqrt{x}=\dfrac{3}{6}=\dfrac{1}{2}\)
=>x=1/4
g: ĐKXĐ: x>=1
\(\sqrt{x-1}=1\)
=>x-1=1
=>x=2
h: ĐKXĐ: x>=-1
\(\sqrt{x+1}=2\)
=>x+1=4
=>x=3
i: ĐKXĐ: x>=0
\(\sqrt{x}-2=7\)
=>\(\sqrt{x}=9\)
=>x=81
j: ĐKXĐ: x>=0
\(14-\sqrt{x}=12\)
=>\(\sqrt{x}=14-12=2\)
=>x=4
k: ĐKXĐ: x>=1
\(12-\sqrt{x-1}=2\)
=>\(\sqrt{x-1}=10\)
=>x-1=100
=>x=101
i: ĐKXĐ: x>=-5
\(\sqrt{x+5}+10=20\)
=>\(\sqrt{x+5}=10\)
=>x+5=100
=>x=95