cho a/k=x/9;b/k=y/b
CMR:a^2/b^2 = x/y cho a/k=x/9;b/k=y/b
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
uses crt;
var st,st1:string;
d,i,d1:integer;
begin
clrscr;
write('Nhap xau ban dau:'); readln(st);
d:=length(st);
st1:='';
for i:=1 to d do
if st[i] in ['a'..'z'] then st1:=st1+st[i];
d1:=length(st1);
write('Xau ban vua tao la: ');
for i:=1 to d1 do
write(st1[i]);
readln;
end.
Bài 2:
uses crt;
var a:array[1..100]of real;
i,n:integer;
k:real;
begin
clrscr;
write('Nhap n='); readln(n);
for i:=1 to n do
begin
write('A[',i,']='); readln(a[i]);
end;
write('Nhap so k='); readln(k);
writeln('Cac boi cua ',k:4:2,' trong day la: ');
for i:=1 to n do
if a[i] mod k=0 then write(a[i]:4);
writeln;
writeln('Cac uoc cua ',k:4:2,' trong day la: ');
for i:=1 to n do
if k mod a[i]=0 then write(a[i]:4);
readln;
end.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Để phương trình \(\left(2x+1\right)^2\cdot\left(9x+2k\right)-5\left(x+2\right)=40\) có nghiệm là x=2 thì Thay x=2 vào phương trình \(\left(2x+1\right)^2\cdot\left(9x+2k\right)-5\left(x+2\right)=40\), ta được:
\(\left(2\cdot2+1\right)^2\cdot\left(9\cdot2+2k\right)-5\left(2+2\right)=40\)
\(\Leftrightarrow25\cdot\left(2k+18\right)-20=40\)
\(\Leftrightarrow25\left(2k+18\right)=60\)
\(\Leftrightarrow2k+18=\dfrac{12}{5}\)
\(\Leftrightarrow2k=-\dfrac{78}{5}\)
hay \(k=\dfrac{-39}{5}\)
Vậy: \(k=\dfrac{-39}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b1 \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)
\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c;\frac{c}{a}=1\Rightarrow c=a\)
\(\Rightarrow a=b=c\)
b2 \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}=k\)
=> \(k=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^2.9^8\)
= \(9^{2+8}\)
=\(9^{10}\)
\(k\) \(minh\)\(nha\)
\(minh\)\(k\)\(bn\)\(day\)
Ta có:
\(\frac{a}{k}=\frac{x}{a}\Rightarrow a^2=k.x\) (1)
\(\frac{b}{k}=\frac{y}{b}\Rightarrow b^2=k.y\) (2)
Chia (1) cho (2) ta được:
\(\frac{a^2}{b^2}=\frac{k.x}{k.y}=\frac{x}{y}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{x}{y}\left(đpcm\right).\)
Chúc bạn học tốt!