Bài 1.

a,Vì \(\dfrac{a}{b}>1\)=>a<b

Với m∈N* Ta có

 \(am> bm\)=>\(am+ab> bm+ab\)=>\(a\left(b+m\right)> b\left(a+m\right)\)=>\(\dfrac{a}{b}>\dfrac{a+m}{b+m} \)

b, Vì \(\dfrac{a}{b}< 1\)=>a<b

Với m∈N* =>

 \(am< bm\)=>\(am+ab< bm+ab\)=>\(a\left(b+m\right)< b\left(a+m\right)\)=>\(\dfrac{a}{b}<\dfrac{a+m}{b+m} \)

Tự áp dụng cho bài 2 nhé bạn :)

\(A=\dfrac{\sqrt{20}-6}{\sqrt{14-6\sqrt{5}}}-\dfrac{\sqrt{20}-\sqrt{28}}{\sqrt{12-2\sqrt{35}}}=\dfrac{-2\left(3-\sqrt{5}\right)}{\sqrt{\left(3-\sqrt{5}\right)^2}}+\dfrac{2\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}}\)

\(=\dfrac{-2\left(3-\sqrt{5}\right)}{3-\sqrt{5}}+\dfrac{2\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}=-2+2=0\)

\(B=\sqrt{\dfrac{\left(9-4\sqrt{3}\right)\left(6-\sqrt{3}\right)}{\left(6-\sqrt{3}\right)\left(6+\sqrt{3}\right)}}-\sqrt{\dfrac{\left(3+4\sqrt{3}\right)\left(5\sqrt{3}+6\right)}{\left(5\sqrt{3}-6\right)\left(5\sqrt{3}+6\right)}}\)

\(=\sqrt{\dfrac{66-33\sqrt{3}}{33}}-\sqrt{\dfrac{78+39\sqrt{3}}{39}}=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\right)=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-1-\sqrt{3}-1\right)=-\sqrt{2}\)

a) Ta có: \(A=\dfrac{\sqrt{10}-3\sqrt{2}}{\sqrt{7-3\sqrt{5}}}-\dfrac{\sqrt{10}-\sqrt{14}}{\sqrt{6-\sqrt{35}}}\)

\(=\dfrac{2\sqrt{5}-6}{3-\sqrt{5}}-\dfrac{2\sqrt{5}-2\sqrt{7}}{\sqrt{7}-\sqrt{5}}\)

\(=\dfrac{\left(2\sqrt{5}-6\right)\left(3+\sqrt{5}\right)}{4}-\dfrac{\left(2\sqrt{5}-2\sqrt{7}\right)\left(\sqrt{7}+\sqrt{5}\right)}{2}\)

\(=\dfrac{\left(\sqrt{5}-3\right)\left(3+\sqrt{5}\right)-\left(2\sqrt{5}-2\sqrt{7}\right)\left(\sqrt{7}+\sqrt{5}\right)}{2}\)

\(=\dfrac{5-9-2\left(5-7\right)}{2}\)

\(=\dfrac{-4-2\cdot\left(-2\right)}{2}\)

\(=0\)

Bài 3:

a: \(A=5\cdot\left(-1\right)^3\cdot1^4=-5\)

b: \(B=9\cdot\left(-1\right)^5\cdot2^2=-36\)

b: $B=9\cdot {\left(-1\right)}^5\cdot 2^2=-36$

\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\\ P=\left(x-y-x-y\right)^2-4x^2\\ P=4y^2-4x^2=4\left(y-x\right)\left(x+y\right)\)

huhuhu phân tích cả buổi chả đc tí j

chừng có ai trả lời đc báo mình với nha

\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x-y\right)\left(x^2+8y^2\right)\)

\(=x^3-8y^3-\left(x^3-x^2y+8xy^2-8y^3\right)\)

\(=x^3-8y^3-x^3+x^2y-8xy^2+8y^3\)

\(=x^2y-8xy^2\)

\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x-y\right)\left(x^2+8y^2\right)\\ =x^3-8y^3-\left(x^3+8y^3-x^2y-8y^3\right)\\ =x^3-8y^3-x^3-8y^3+x^2y+8y^3\\ =-8y^3+x^2y\)

Ta có: \(\frac{x+2}{y+10}\)\(=\)\(\frac{1}{5}\)\(\Rightarrow\)\(5\left(x+2\right)=y+10\)(1)

             \(y-3x=2\)\(\Rightarrow\)\(y+2=3x\)                              (2)

Thay (2) vào (1) ta có:

\(5\left(x+2\right)=\left(y+2\right)+8\)

\(5x+10=3x+8\)

\(5x-3x=8-10\)

\(2x=-2\)

\(x=-2:2\)

\(x=-1\)

Vậy: x=-1

Chúc bạn làm bài tốt!