1. 

 \(ƯCLN\left(a,b\right)=7\)

\(\Rightarrow a,b\)chia hết cho 7

\(\Rightarrow a,b\in B\left(7\right)\)

\(B\left(7\right)=\left(0;7;14;21;28;35;42;49;56;63;70;77;84;91;98;105...\right)\)

a, vì a+b=56 \(\Rightarrow\)\(a\le56;b\le56\)

\(\Rightarrow a=56;b=0.a=0;b=56\)

\(a=7;b=49.a=49;b=7\)

\(a=14;b=42.a=42;b=14\)

\(a=21;b=35.a=35;b=21\)

\(a=b=28\)

b, a.b=490 \(\Rightarrow a< 490;b< 490\)

\(\Rightarrow\) \(a=7;b=70-a=70;b=7\)

          \(a=14;b=35-a=35;b=14\)

c, BCNN (a,b) = 735

\(\Rightarrow a,b\inƯ\left(735\right)\)

\(Ư\left(735\right)=\left(1;3;5;7;15;21;35;49;105;147;245;735\right)\)

\(\Rightarrow\)\(a=7;b=105-a=105;b=7\)

2. 

a+b=27\(\Rightarrow\)\(a\le27;b\le27\)

ƯCLN(a,b)=3

\(\Rightarrow a,b\in B\left(_{ }3\right)\in\left(0;3;6;9;12;15;18;21;24;27;30;...\right)\)

BCNN(a,b)=60

\(\Rightarrow a,b\inƯ\left(60\right)\in\left(1;2;3;4;5;6;10;12;15;20;60\right)\)

\(\Rightarrow\)\(a=12;b=15-a=15;b=12\)

a) 

Thay x = -1 ( thỏa mãn ĐKXĐ ) vào biểu thức B , ta có :

\(B=\frac{2+1}{-1}=\frac{3}{-1}=-3\)

b) \(A=\frac{1}{x-2}-\frac{2x}{4-x^2}+\frac{1}{2+x}\)

\(A=\frac{1}{x-2}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{x+2}\)

\(A=\frac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{3x}{\left(x-2\right)\left(x+2\right)}\)

c) Ta có : 

\(P=A.B\)

\(P=\frac{3x}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)

Mà P = 1/2

\(\Leftrightarrow\frac{3x}{\left(x-2\right)\left(x+2\right)}.\frac{-\left(x-2\right)}{x}=\frac{1}{2}\)

\(\Leftrightarrow\frac{3}{x+2}.\frac{-1}{1}=\frac{1}{2}\)

\(\Leftrightarrow\frac{-3}{x+2}=\frac{1}{2}\)

\(\Leftrightarrow x+2=-6\Leftrightarrow x=-8\)( thỏa mãn )

d) P nguyên dương

\(\Leftrightarrow\frac{-3}{x+2}\)nguyên dương

<=> x + 2 thuộc Ư(3) { -1 ; -3 }

Bảng tìm x

Vậy ....................

cảm ơn bn nhé nhg mk hỏi sao x +2x+ x= 3x đc z mk tưởng là 4x

Cộng 3 ở 3 p/s đầu và trừ 4 ở p/s cuối . Nó sẽ xuất hiện tử chung thôi 

\(\frac{a+b-x}{b}+\frac{a+c-x}{b}+\frac{b+c-x}{a}+\frac{4x}{a+b+c}=1\)

\(\Leftrightarrow\left(\frac{a+b-x}{c}+1\right)+\left(\frac{a+c-x}{b}+1\right)+\left(\frac{b+c-x}{a}+1\right)+\left(\frac{4x}{a+b+c}-4\right)=0\)

\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{b}+\frac{a+b+c-x}{a}+\frac{4\left(x-a-b-c\right)}{a+b+c}=0\)

\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{b}+\frac{a+b+c-x}{a}-\frac{4\left(a+b+c-x\right)}{a+b+c}=0\)

\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}\right)=0\)

\(\Rightarrow a+b+c-x=0\)hoặc \(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0\)

Nếu \(a+b+c-x=0\Rightarrow x=a+b+c\)

Nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}=0\Rightarrow x\inℝ\)

mk ko bt 123

\(\frac{a}{c}=\frac{a-b}{b-c}\Rightarrow a\left(b-c\right)=c\left(a-b\right)\)           (1)

\(\frac{1}{c}+\frac{1}{a-b}=\frac{a-b+c}{c\left(a-b\right)}\)                  (2)

\(\frac{1}{b-c}-\frac{1}{a}=\frac{a-b+c}{a\left(b-c\right)}\)                  (3)

\(Từ\left(1\right),\left(2\right),\left(3\right)\Rightarrow\)điều phải chứng minh

dễ mk sao 0 biêy

Thế 

làm

đi ~.~

uk trong sgk cũng co

Ta có: a.b = 24 => a,b \(\in\)Ư(24)

Ư(24) ={1;2;3;4;6;8;12;24}

Vì a<b nên ta có: