\(\frac{a+b}{b^2}\sqrt{\frac{a^2b^4}{a^2+2ab+b^2}}=\frac{a+b}{b^2}\cdot\frac{\sqrt{a^2b^4}}{\sqrt{\left(a+b\right)^2}}=\frac{a+b}{b^2}\cdot\frac{ab^2}{a+b}=a\)

à quên sửa chỗ này:)

\(=\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{a+b}=\left|a\right|\)xin lỗi nhé :v

Giải ra hộ mình nhé

Với \(a\ge0;a\ne1\)

\(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2}{\left(a-1\right)^2}=1\)

-> chọn B 

có bộ gõ kí hiệu Toán mà :))

ĐK : a >= 0 ; a khác 36

\(K=\left[\frac{a+14\sqrt{a}+100}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}+\frac{\left(\sqrt{a}+6\right)\left(\sqrt{a}-6\right)}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}-\frac{\left(\sqrt{a}-7\right)\left(\sqrt{a}+7\right)}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\right]\div\left(\frac{\sqrt{a}-6}{\sqrt{a}-6}-\frac{\sqrt{a}-7}{\sqrt{a}-6}\right)\)

\(=\frac{a+14\sqrt{a}+100+a-36-a+49}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\div\frac{1}{\sqrt{a}-6}\)

\(=\frac{a+14\sqrt{a}+113}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\cdot\left(\sqrt{a}-6\right)=\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}\)

Để K = 2 thì \(\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}=2\Rightarrow a+14\sqrt{a}+113=2\sqrt{a}+14\Leftrightarrow a+12\sqrt{a}+99=0\)

Với a >= 0 thì \(a+12\sqrt{a}+99\ge99>0\)=> Không có giá trị x thỏa mãn K = 2

Ta có : \(K=\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}=\frac{\left(a+14\sqrt{a}+49\right)+64}{\sqrt{a}+7}=\frac{\left(\sqrt{a}+7\right)^2+64}{\sqrt{a}+7}\)

\(=\left(\sqrt{a}+7\right)+\frac{64}{\sqrt{a}+7}\ge2\sqrt{\left(\sqrt{a}+7\right)\cdot\frac{64}{\sqrt{a}+7}}=16\)( bđt AM-GM )

Dấu "=" xảy ra <=> \(\sqrt{a}+7=\frac{64}{\sqrt{a}+7}\Rightarrow a=1\left(tm\right)\). Vậy MinK = 16

\(\frac{1}{\sqrt{5}+1}+\frac{1}{\sqrt{5}-2}-\frac{1}{3-\sqrt{5}}-\sqrt{5}\)

\(=\frac{\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}+\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\frac{3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}-\sqrt{5}\)

\(=\frac{\sqrt{5}-1}{5-1}+\frac{\sqrt{5}+2}{5-4}-\frac{3+\sqrt{5}}{9-5}-\sqrt{5}\)

\(=\frac{\sqrt{5}-1}{4}+\sqrt{5}+2-\frac{3+\sqrt{5}}{4}-\sqrt{5}\)

\(=\frac{1}{2}+2=\frac{5}{2}\)

\(\frac{1}{\sqrt{5}+1}+\frac{1}{\sqrt{5}-2}-\frac{1}{3-\sqrt{5}}-\sqrt{5}\)

\(=\frac{\sqrt{5}-1}{4}+\sqrt{5}+2-\frac{3+\sqrt{5}}{4}-\sqrt{5}\)

\(=\frac{\sqrt{5}-1-3-\sqrt{5}}{4}+2=-1+2=1\)

b14:

\(a,P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)

\(P=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)

\(P=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{1}{\sqrt{x}-1}\)    

sao ko gọn zị :v

\(M=\frac{3\left(\sqrt{x}+3\right)-8}{\sqrt{x}+3}=3-\frac{8}{\sqrt{x}+3}\)

Để M nguyên thì \(\frac{8}{\sqrt{x}+3}\)nguyên hay \(\sqrt{x}+3\inƯ\left(8\right)\)

bạn lập bảng xét nhé ;)

Bài 10 : 

a, Thay x = 16 vào P ta được : \(P=\frac{2\sqrt{16}-3}{\sqrt{16}+2}=\frac{2.4-3}{4+2}=\frac{5}{6}\)

b, \(P=\frac{2\sqrt{x}-3}{\sqrt{x}+2}=\frac{2\left(\sqrt{x}+2\right)-7}{\sqrt{x}+2}=2-\frac{7}{\sqrt{x}+2}\)

\(\Rightarrow\sqrt{x}+2\inƯ\left(7\right)=\left\{1;7\right\}\)

\(M=\frac{\sqrt{x}+1+4}{\sqrt{x}+1}=1+\frac{4}{\sqrt{x}+1}\)

Để M nguyên thì \(\frac{4}{\sqrt{x}+1}\)nguyên hay \(\sqrt{x}+1\inƯ\left(4\right)\)( tự lập bảng xét )

\(P=\frac{\sqrt{x}+2+5}{\sqrt{x}+2}=1+\frac{5}{\sqrt{x}+2}\)tương tự ý trên

10 Tú làm rồi không làm lại

Bài 4:

Để M nguyên \(\Leftrightarrow\sqrt{x}+2⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}+4⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}-3+7⋮2\sqrt{x}-3\)

Mà \(\Leftrightarrow2\sqrt{x}-3⋮2\sqrt{x}-3\)

\(\Rightarrow7⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Mà \(2\sqrt{x}-3\ge-3\left(x\ge0\right)\)

\(\Rightarrow2\sqrt{x}-3\in\left\{1;-1;7\right\}\)

\(\Rightarrow x\in\left\{4;1;25\right\}\)

Vậy...

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