Bài 4:

a:ĐKXĐ: x>=0; x<>1

b: \(A=\frac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\frac{x+\sqrt{x}}{\sqrt{x}+1}\)

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

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

Bài 5:

\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}:\frac{x+16}{\sqrt{x}+2}\)

\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}\)

\(=\frac{x+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)

Bài 6:

Ta có: \(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\)

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

\(=\frac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\)

\(=\frac{3a-3\sqrt{ab}-2a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}=\frac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\)

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

Bài 3:

a: ĐKXĐ: a>0; b>0; a<>b

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

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

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

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

Bài 4:

a:ĐKXĐ: x>=0; x<>1

b: \(A=\frac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\frac{x+\sqrt{x}}{\sqrt{x}+1}\)

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

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

Bài 5:

\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}:\frac{x+16}{\sqrt{x}+2}\)

\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}\)

\(=\frac{x+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)

Bài 6:

Ta có: \(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\)

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

\(=\frac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\)

\(=\frac{3a-3\sqrt{ab}-2a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}=\frac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\)

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

1.

a. Em tự giải

b.

\(\left\{{}\begin{matrix}2x+y=4m-1\\3x-2y=-m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=8m-2\\3x-2y=-m+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+7\\y=\dfrac{3x+m-9}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+1\\y=2m-3\end{matrix}\right.\)

Để \(x+y=7\Rightarrow m+1+2m-3=7\)

\(\Rightarrow3m=9\Rightarrow m=3\)

2.

a. Em tự giải

b.

Phương trình có 2 nghiệm khi:

\(\Delta'=\left(m+1\right)^2-\left(2m+10\right)=m^2-9\ge0\)

\(\Rightarrow\left[{}\begin{matrix}m\ge3\\m\le-3\end{matrix}\right.\)

Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m+10\end{matrix}\right.\)

Ta có:

\(P=x_1^2+x_2^2+8x_1x_2=\left(x_1+x_2\right)^2+6x_1x_2\)

\(=4\left(m+1\right)^2+6\left(2m+10\right)=4m^2+20m+64\)

\(=4\left(m^2+5m+6\right)+40=4\left(m+2\right)\left(m+3\right)+40\)

Do \(\left[{}\begin{matrix}m\ge3\\m\le-3\end{matrix}\right.\) \(\Rightarrow\left(m+2\right)\left(m+3\right)\ge0\)

\(\Rightarrow P\ge40\)

Vậy \(P_{min}=40\) khi \(m=-3\)

(Nếu bài này giải là \(4m^2+20m+64=\left(2m+5\right)^2+39\ge39\) là sai vì dấu = khi đó xảy ra tại \(m=-\dfrac{5}{2}\) ko thỏa mãn điều kiện \(\Delta\) để pt có nghiệm)

b) \(\sqrt{x^2}=\left|-8\right|\)

\(\Rightarrow\left|x\right|=8\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

d) \(\sqrt{9x^2}=\left|-12\right|\)

\(\Rightarrow\sqrt{\left(3x\right)^2}=12\)

\(\Rightarrow\left|3x\right|=12\)

\(\Rightarrow\left[{}\begin{matrix}3x=12\\3x=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{12}{3}\\x=-\dfrac{12}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)