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1 tháng 1 2022
1: \(x=\dfrac{3}{4}-\dfrac{2}{4}=\dfrac{1}{4}\)
2: \(x=\dfrac{2}{11}+\dfrac{1}{5}=\dfrac{10+11}{55}=\dfrac{21}{55}\)
27 tháng 8 2021
Bài 4:
d: Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
e: Ta có: \(x^3-y^3-3x+3y\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2-3\right)\)
DA
4
20 tháng 8 2021
2.
\(a,x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(b,x^2-3y^2=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
\(c,\left(3x-2y\right)^2-\left(2x-3y\right)^2\\ =\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\\ =\left(x+y\right)\left(5x-5y\right)=5\left(x-y\right)\left(x+y\right)\)
\(d,9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)
\(e,\left(4x^2-4x+1\right)-\left(x+1\right)^2\\ =\left(2x-1\right)^2-\left(x+1\right)^2\\ =\left(2x-1-x-1\right)\left(2x-1+x+1\right)\\ =3x\left(x-2\right)\)
\(f,x^3+27=\left(x+3\right)\left(x^2+3x+9\right)\)
\(g,27x^3-0,001=\left(3x-0,1\right)\left(9x^2+0,027x+0,01\right)\)
\(h,125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
20 tháng 8 2021
Bài 3 :
a) \(x^4+2x^2+1=\left(x^2+1\right)^2\)
b) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
c) \(-x^2-2xy-y^2=-\left(x+y\right)^2\)
e) \(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
f) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
g) \(x^3+6x^2+12x+8=\left(x+2\right)^3\)
h) \(x^3+1-x^2-x=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=\left(x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)\left(x-1\right)^2\)
l) \(\left(x+y\right)^2-x^3-y^3=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)=3xy\left(x+y\right)\)
1 tháng 3 2022
a: =1/3x7/5=7/15
b: =11/9x1/2=11/18
c: =1/3x3/4=1/4
d: =1/5x4/5=4/25
e: =2/3x1/4=2/12=1/6
f: =1/2x1/3=1/6
g: =1/3x1/2=1/6
NT
0
NT
0
HP
0
LH
1
30 tháng 1 2022
Bài 2:
\(a.\left(y+3,02\right):1,5=6,9\times0,3.\\ \Leftrightarrow\left(y+3,02\right):1,5=2,07.\\ \Leftrightarrow y+3,02=3,105.\\ \Leftrightarrow y=0,085.\)
\(b.16,15:\left(y\times19\right)=17.\\ \Leftrightarrow y\times19=0,95.\\ \Leftrightarrow y=0,05.\)
\(c.0,5\times y:5=10\times0,2.\\ \Leftrightarrow0,5\times y:5=2.\\ \Leftrightarrow0,5\times y=10.\\ \Leftrightarrow y=20.\)
1 tháng 3 2022
uses crt;
var st:array[1..100]of string;
a,b,c:array[1..100]of real;
i,n:integer;
max:real;
begin
clrscr;
readln(n);
for i:=1 to n do readln(st[i],a[i],b[i],c[i]);
max=(a[1]+b[1]+c[1])/3;
for i:=1 to n do
if (max<(a[i]+b[i]+c[i])/3) then max:=(a[i]+b[i]+c[i])/3;
writeln(max:4:2);
readln;
end.
Bài 1 :
\(1) Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O\\ (2) FeCl_3 + 3AgNO_3 \to 3AgCl + Fe(NO_3)_3\\ (3)Fe(NO_3)_3 + 3KOH \to Fe(OH)_3 + 3KNO_3\\ (4)2Fe(OH)_3 +3 H_2SO_4 \to Fe_2(SO_4)_3 + 6H_2O\\ (5)Fe_2(SO_4)_3 + 6NaOH \to 2Fe(OH)_3 + 3Na_2SO_4\\ (6)2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ b) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O\\ Al_2(SO_4)_3 + 3BaCl_2 \to 3BaSO_4 + 2AlCl_3\\ AlCl_3 + 3AgNO_3 \to 3AgCl + Al(NO_3)_3\\ Al(NO_3)_3 + 3KOH \to Al(OH)_3 + 3KNO_3\\ \)
\(2Al(OH)_3 \xrightarrow{t^o} Al_2O_3 + 3H_2O\\ c) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ Fe_2(SO_4)_3 + 3BaCl_2 \to 3BaSO_4 + 2FeCl_3\\ FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl\\ 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ Fe + 2HCl \to FeCl_2 + H_2\\ 2FeCl_2 + Cl_2 \to 2FeCl_3\\ d) CuCl_2 + 2AgNO_3 \to 2AgCl + Cu(NO_3)_2\\ Cu(NO_3)_2 + 2KOH \to Cu(OH)_2 +2KNO_3\\ Cu(OH)_2 + H_2SO_4 \to CuSO_4 + 2H_2O\\ CuSO_4 + Fe \to FeSO_4 + Cu\\ \)
\(2Cu + O_2 \to 2CuO\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\)
Bài 2 :
$(1) 2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3$
$(2) SO_3 + H_2O \to H_2SO_4$
$(3) BaCl_2 + H_2SO_4 \to BaSO_4 + 2HCl$
$(4) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
$(5) 2NaOH + SO_2 \to Na_2SO_3 + H_2O$
$(6) Na_2SO_3 + 2HCl \to 2NaCl + SO_2 + H_2O$