\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)

\(=\frac{\sqrt{x}-3}{4}\)

\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)

Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)

\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)

\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)

thank

\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=4.\)

\(\Rightarrow\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}=4\)

\(\Rightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=4\)

\(\Rightarrow\sqrt{x-4}+2+\sqrt{x-4}-2=0\)

\(\Rightarrow2\sqrt{x-2}=0\)

\(\Rightarrow\sqrt{x-2}=0\Rightarrow x-2=0\Rightarrow x=2\)

\(x^2-2mx+3m-2=0\)

Thay m = -1 vào PT ta được:

\(x^2-2\left(-1\right)x+3\left(-1\right)-2=0\)

\(\Rightarrow x^2+2x-5=0\)

\(\Delta'=b'^2-ac=1^2-1.\left(-5\right)=6>0\)

Do \(\Delta'>0\Rightarrow\)PT có hai nghiệm phân biệt:

\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=-1+\sqrt{6}\)

\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=-1-\sqrt{6}\)

a) \(A=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{6}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

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

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

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

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

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

\(A=1\)

C1: Ta có: 49 chia 3 dư 1

=> 49^n chia 3 dư 1

13 chia 3 dư 1 

=> 13^n chia 3 dư 1

269 chia 3 dư 2

=>  \(49^n+296.13^n\)chia 3 dư 1+2.1=3  

=> \(49^n+296.13^n\)chia hết cho 3

C2: Hoặc bạn có thể làm theo cách đồng dư

\(49\equiv1\left(mod3\right)\)

=> \(49^n\equiv1^n\equiv1\left(mod3\right)\)

\(13\equiv1\left(mod3\right)\)

=> \(13^n\equiv1^n\equiv1\left(mod3\right)\)

\(296\equiv2\left(mod3\right)\)

=> \(49^n+296.13^n\equiv1+2.1\equiv3\equiv0\left(mod3\right)\)

=> \(49^n+296.13^n\)chia hết cho 3

Thêm đk n thuộc N*. Quy nạp thử xem nào:) (em ko chắc đâu nhá)

Với n = 1 thì nó đúng

Giả sử nó đúng với n = k tức là \(49^k+296.13^k⋮3\)

Ta chứng minh nó đúng với n = k + 1. Cần chứng minh \(49^k.49+296.13^k.13⋮3\)

\(\Leftrightarrow49\left(49^k+296.13^k\right)-296.13^k.36⋮3\)

Điều này hiển nhiên đúng do giả thiết quy nạp và \(296.13^k.36\) chia hết cho 3

Cho link nờ https://olm.vn/hoi-dap/detail/85790459247.html

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