Lời giải:
a.

$\frac{x}{-2}=\frac{-8}{x}$

$\Rightarrow x^2=(-2)(-8)=16=4^2=(-4)^2$

$\Rightarrow x=4$ hoặc $x=-4$

b.

$\frac{-5}{-14}=\frac{20}{6-5x}$

$\frac{5}{14}=\frac{20}{6-5x}$

$\Rightarrow 6-5x=20.14:5=56$

$x=-10$

e: \(=\dfrac{-5}{4}:\dfrac{-5}{8}+\dfrac{3}{2}\cdot\dfrac{-1}{2}=\dfrac{5}{4}\cdot\dfrac{8}{5}-\dfrac{3}{4}=2-\dfrac{3}{4}=\dfrac{5}{4}\)

f: \(=\dfrac{15}{7}\cdot\left(-7\right)\cdot\dfrac{-1}{10}=15\cdot\dfrac{1}{10}=\dfrac{3}{2}\)

g: \(=\dfrac{7}{10}\cdot\dfrac{-8}{7}:\dfrac{8}{15}-\dfrac{1}{3}\cdot\dfrac{1}{4}=\dfrac{-4}{5}\cdot\dfrac{15}{8}-\dfrac{1}{12}=\dfrac{-3}{2}-\dfrac{1}{12}=\dfrac{-19}{12}\)

H đâu =))

\(d,\left(1-\dfrac{4}{7}\right)\left(\dfrac{-3}{8}-\dfrac{-5}{7}\right)=\dfrac{3}{7}.\dfrac{19}{56}=\dfrac{57}{392}\\ e,\left(-\dfrac{5}{6}\right)^2=\dfrac{25}{36}\\ c,\dfrac{4}{3}-\dfrac{1}{3}:5+\dfrac{3}{4}=\dfrac{4}{3}-\dfrac{1}{15}+\dfrac{3}{4}=\dfrac{19}{15}+\dfrac{3}{4}=\dfrac{121}{60}\)

a: \(=\dfrac{3}{11}+\dfrac{18}{66}=\dfrac{18}{66}+\dfrac{18}{66}=\dfrac{36}{66}=\dfrac{6}{11}\)

b: \(=\dfrac{2}{9}-\dfrac{5}{12}+\dfrac{3}{4}\)

\(=\dfrac{8}{36}-\dfrac{15}{36}+\dfrac{27}{36}=\dfrac{20}{36}=\dfrac{5}{9}\)

c: \(=\dfrac{3\cdot7-5}{35}\cdot\dfrac{-40-30}{16}\)

\(=\dfrac{16}{35}\cdot\dfrac{-70}{16}=-2\)

d: \(=\dfrac{1}{5}\cdot\dfrac{2-7}{4}-\dfrac{3}{4}\cdot\dfrac{4-11}{12}\)

\(=\dfrac{1}{5}\cdot\dfrac{-5}{4}-\dfrac{3}{4}\cdot\dfrac{-7}{12}\)

\(=\dfrac{-1}{4}+\dfrac{21}{48}=\dfrac{-12+21}{48}=\dfrac{9}{48}=\dfrac{3}{16}\)

a: \(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

b: \(\Leftrightarrow\dfrac{20}{6-5x}=\dfrac{5}{14}\)

=>6-5x=56

=>5x=-50

hay x=-10

c: =>7/17=x/119=-35/y

=>x=49; y=-85

d: =>x/27=1/9+1/9=2/9

hay x=6

\(a,\dfrac{x}{-2}=\dfrac{-8}{x}\)

\(\Leftrightarrow x^2=16\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(b,\dfrac{5}{14}=\dfrac{20}{6-5x}\)

\(\Leftrightarrow30-25x=280\)

\(\Leftrightarrow25x=30-280=-250\)

\(\Leftrightarrow x=-10\)

\(c,\dfrac{7}{17}=\dfrac{x}{119}=\dfrac{-35}{y}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{119.7}{17}=49\\y=\dfrac{-35.17}{7}=-85\end{matrix}\right.\)

\(d,\dfrac{x}{27}+\dfrac{-3}{27}=\dfrac{1}{9}\)

\(\dfrac{x-3}{27}=\dfrac{1}{9}\)

\(9x-27=27\)

\(9x=54\Rightarrow x=6\)

+ Quy ước: A: không sừng, a: có sừng

+ Bò không sừng có KG là: AA hoặc Aa

+ Bò có sừng có KG là: aa

a. Ptc: bò đực có sừng x bò cái ko sừng

aa x AA

F1: 100% Aa: ko sừng

b. F1 x F1: Aa x Aa

F2: 1AA : 2Aa : 1aa

KH: 3 ko sừng : 1 có sừng

c. Bò ko sừng F2 có KG là: AA hoặc Aa

Bò cái F1 có KG là Aa

+ P1: AA x Aa \(\rightarrow\) 1AA : 1Aa (100% bò ko sừng)

+ P2: Aa x Aa \(\rightarrow\) 1AA : 2Aa : 1aa (3 ko sừng : 1 có sừng)

không hiểu đang giảng hay hỏi

\(a,\dfrac{8}{35}\)

\(b,\dfrac{15}{8}\)

\(c,\dfrac{30}{21}\)

\(d,\dfrac{72}{175}\)

de qha

const fi='inp.inp';

       fo='out.inp';

var f1,f2:text;

st:array[1..100]of string;

i,n:integer;

begin

assign(f1,fi); reset(f1);

assign(f2,fo); rewrite(f2);

n:=0;

while not eof(f1) do 

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n:=n+1;

readln(f1,st[n]);

end;

for i:=1 to n do 

  writeln(f2,length(st[i]));

close(f1);

close(f2);

end.