a,\(\Delta=3^2-4\left(-2\right).6=9+48=57\)

\(x_1=\dfrac{-3+\sqrt{57}}{-4}=\dfrac{3-\sqrt{57}}{4}\)

\(x_2=\dfrac{-3-\sqrt{57}}{-4}=\dfrac{3+\sqrt{57}}{4}\)

b, \(\Delta=6^2-4.3.3=36-36=0\)

\(\Rightarrow x_1=x_2=\dfrac{-6}{2.3}=\dfrac{-6}{6}=-1\)

c, \(\Delta=1^2-4.6.5=1-120=-119< 0\)

Vậy pt vô nghiệm

\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)

\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)

\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)

1: \(\Leftrightarrow x\left(2-3x\right)=0\)

=>x=0 hoặc x=2/3

2: =>(2x-3)(2x+3)=0

=>x=3/2 hoặc x=-3/2

3: =>(x-6)(x-5)=0

=>x=6 hoặc x=5

Bài 2: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(\Leftrightarrow\dfrac{6}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-9+x+3}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(x^2+x-12=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)

=>x=-4(nhận) hoặc x=3(loại)

Bài 1:

a,ĐKXĐ:\(\left\{{}\begin{matrix}\sqrt{a}+1\ne0\left(luôn.đúng\right)\\\sqrt{a}-5\ne0\end{matrix}\right.\Leftrightarrow\sqrt{a}\ne5\Leftrightarrow a\ne25\)

\(b,A=\left(3+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(3-\dfrac{a-5\sqrt{a}}{\sqrt{a}-5}\right)\)

\(\Rightarrow A=\left(3+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(3-\dfrac{\sqrt{a}\left(\sqrt{a}-5\right)}{\sqrt{a}-5}\right)\)

\(\Rightarrow A=\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)\)

\(\Rightarrow A=9-a\)

\(tan\alpha=\dfrac{1}{3}\Rightarrow\dfrac{sin\alpha}{cos\alpha}=\dfrac{1}{3}\Rightarrow cos\alpha=3sin\alpha\)

Thay cosa=3sina vào A, được:

\(A=\dfrac{sin^2a+9sin^2a}{sin^2a+9sin^2a+6sin^2a}=\dfrac{10sin^2a}{16sin^2a}=\dfrac{5}{8}\)

\(B=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

Với x >= 0 ; x khác 4 

\(B=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x-3y=4\\12x+3y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=19\\7x-3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\cdot\dfrac{x-4}{4}=\dfrac{4}{4}=1\)

Với x >= 0 ; x khác 4 

\(\left(\dfrac{\sqrt{x}+2-\sqrt{x}+2}{x-4}\right)\dfrac{x-4}{4}=1\)