hmm thật ra bạn ghi nhầm đề đúng hog , mk sửa lại nhá:(

Cho hàm số \(y=f\left(x\right)=\frac{1}{2}x+5\)

Tính f(0);    f(1);    f(2);   f(3);     f(-2);     f(-10)

\(f\left(0\right)=\frac{1}{2}.0+5=5\)

\(f\left(2\right)=\frac{1}{2}.2+5=6\)

\(f\left(3\right)=\frac{1}{2}.3+5=6,5\)

\(f\left(-2\right)=\frac{1}{2}.\left(-2\right)+5=4\)

\(f\left(-10\right)=\frac{1}{2}.\left(-10\right)+5=0\)

\(b,\sqrt{x-1}+\sqrt{4x-4}+\frac{1}{5}\sqrt{25x-25}=24\)

\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}+\frac{1}{5}\sqrt{25\left(x-1\right)}=24\)

\(\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=24\)

\(\sqrt{x-1}\left(1+2+1\right)=24\)

\(4\sqrt{x-1}=24\)

\(\sqrt{x-1}=6\)

\(x-1=36\)

\(x=37\left(TM\right)\)

\(\sqrt{x}=\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

\(B=\frac{\sqrt{x}-1}{2+\sqrt{x}}=\frac{\sqrt{5}+1-1}{2+\sqrt{5}+1}=\frac{\sqrt{5}}{\sqrt{5}+3}=\frac{\left(3-\sqrt{5}\right)\sqrt{5}}{\left(3^2-5\right)}=\frac{3\sqrt{5}-5}{4}\)

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

mà \(x\)nguyên nên \(\sqrt{x}+2\inƯ\left(3\right)\)mà \(\sqrt{x}+2\ge2\)nên \(\sqrt{x}+2=3\Leftrightarrow x=1\).

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

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

\(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\frac{2x+2}{\sqrt{x}}\)

\(A=\frac{2\sqrt{x}}{\sqrt{x}}+\frac{2x+2}{\sqrt{x}}\)

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

\(b,A=\frac{2\sqrt{x}+2x+2}{\sqrt{x}}=6\)

\(2\sqrt{x}+2x+2=6\sqrt{x}\)

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

\(\left(\sqrt{2}\sqrt{x}\right)^2-4\sqrt{x}+\sqrt{2}^2=0\)

\(\left(\sqrt{2x}-\sqrt{2}\right)^2=0\)

\(\sqrt{2x}=\sqrt{2}\)

\(x=1\)

Trả lời:

\(\sqrt{x^2-9}=\sqrt{x^2-3x}\)

\(\Leftrightarrow x^2-9=x^2-3x\)

\(\Leftrightarrow x^2-9-\left(x^2-3x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-x\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-x\right)=0\)

\(\Leftrightarrow\left(x-3\right)3=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt.

Ta có:\(\sqrt{n+1}-\sqrt{n}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)

\(2\sqrt{n+1}>\sqrt{n+1}+\sqrt{n}>0\Leftrightarrow\frac{1}{2\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}\)

\(0< 2\sqrt{n}< \sqrt{n+1}+\sqrt{n}\Leftrightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}< \frac{1}{2\sqrt{n}}\)

Ta có đpcm. 

Đk: \(x\ge3\); \(y\ge5\)

Ta có: \(x+y-2\sqrt{x-3}-4\sqrt{y-5}-3=0\)

<=> \(x-3-2\sqrt{x-3}+1+y-5-4\sqrt{y-5}+4=0\)

<=> \(\left(\sqrt{x-3}-1\right)^2+\left(\sqrt{y-5}-2\right)^2=0\)

<=> \(\hept{\begin{cases}\sqrt{x-3}-1=0\\\sqrt{y-5}-2=0\end{cases}}\) <=> \(\hept{\begin{cases}\sqrt{x-3}=1\\\sqrt{y-5}=2\end{cases}}\) <=> \(\hept{\begin{cases}x=4\\y=9\end{cases}}\)(tm)

Vậy (x;y) = {(4; 9)}