Bài 2: 

c: Để hai đường thẳng song song thì m-1=2

hay m=3

a.

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x+3}=a>0\\\sqrt{x^2+x+2}=b>0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=x+1\)

Pt trở thành:

\(a+b=2\left(a^2-b^2\right)\)

\(\Leftrightarrow a+b=\left(2a-2b\right)\left(a+b\right)\)

\(\Leftrightarrow2a-2b=1\) (do \(a+b>0\))

\(\Leftrightarrow2a=2b+1\)

\(\Leftrightarrow2\sqrt{x^2+2x+3}=2\sqrt{x^2+x+2}+1\)

\(\Leftrightarrow4\left(x^2+2x+3\right)=4\left(x^2+x+2\right)+1+4\sqrt{x^2+x+2}\)

\(\Leftrightarrow4x+3=4\sqrt{x^2+x+2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{3}{4}\\16\left(x^2+x+2\right)=\left(4x+3\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{3}{4}\\8x=23\end{matrix}\right.\) \(\Rightarrow x=\dfrac{23}{8}\)

b.

ĐKXĐ: \(x\ge3\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-3}=a\ge0\\\sqrt{x+2}=b>0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=-5\)

Phương trình trở thành:

\(\left(a-b\right)\left(ab+1\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(vô-nghiệm\right)\\ab+1=a+b\end{matrix}\right.\)

\(\Rightarrow ab-a-b+1=0\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=1\\\sqrt{x+2}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\left(ktm\right)\end{matrix}\right.\)

a)

=15(can6-1)/(6-1)+4/(4-2)-12/(3-4)

=3(can6-1)+2+12

=3\(\sqrt{6}\)-3+2+12

=17+3can6

các câu còn lại tương tự liên hợp mẫu 

làm cho e luôn vs đc k anh

e đg cần gấp á

VD1:

a: \(5\cdot\sqrt{25a^2}-25a\)

\(=5\cdot\left|5a\right|-25a\)

=-25a-25a(a<0)

=-50a

b: \(\sqrt{49a^2}+3a\)

\(=\sqrt{\left(7a\right)^2}+3a\)

=7a+3a

=10a

c: Đặt A=\(\sqrt{64a^2}-8a\)

\(=\sqrt{\left(8a\right)^2}-8a\)

=8|a|-8a

TH1: a>=0

=>A=8a-8a=0

TH2: a<0

=>A=-8a-8a=-16a

d: Đặt \(A=\sqrt{9a^6}-3a^3\)

\(=3\cdot\sqrt{a^6}-3a^3\)

\(=3\cdot\left|a^3\right|-3a^3\)

TH1: a>=0

=>\(A=3a^3-3a^3=0\)

TH2: a<0

=>\(A=-3a^3-3a^3=-6a^3\)

VD2:

a: \(4x-\sqrt{x^2-4x+4}\)

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

=4x-|x-2|

=4x-(x-2)(x>=2)

=4x-x+2

=3x+2

b: \(3x+\sqrt{x^2+6x+9}\)

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

=3x+|x+3|

=3x+(-x-3)(x<-3)

=2x-3

c: \(\frac{x+6\sqrt{x}+9}{x-9}\)

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

d: \(\frac{\sqrt{x^2+4x+4}}{x+2}\)

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

\(=\frac{\left|x+2\right|}{x+2}=\pm1\)

Bài nào?

5.

\(A=B\Leftrightarrow\left\{{}\begin{matrix}x-5\ge0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x>-2\end{matrix}\right.\Leftrightarrow x\ge5\)

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

Ta có A max <=> \(1+a+a^2min\)

Mà 1+a+a^2=\(\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ \)

Dấu bằng xảy ra <=> a=-1/2

=> \(A=\dfrac{2}{1+a+a^2}\le\dfrac{2}{\dfrac{3}{4}}=\dfrac{8}{3}\)

Vậy max A=8/3 <=> a=-1/2

=)) mỏi tay quá đê

Hì thanks bạn nhiều nhé