Bạn xem lại câu \(2,\) với lại hai trường hợp sao lại dùng ngoặc nhọn?

b) \(PT\Leftrightarrow-4u+6u-u-3u=-3-24+27\)

\(\Leftrightarrow-2u=0\)

\(\Leftrightarrow u=0\)

Vậy: \(S=\left\{0\right\}\)

e) \(PT\Leftrightarrow0,1-t+0,2=2t-1-0,7\)

\(\Leftrightarrow3t=2\)

\(\Leftrightarrow t=\dfrac{2}{3}\)

Vậy: \(S=\left\{\dfrac{2}{3}\right\}\)

d) \(PT\Leftrightarrow-3+4x=2x-15\Leftrightarrow2x=-12\Leftrightarrow x=-6\)

Vậy: \(S=\left\{-6\right\}\)

f) \(PT\Leftrightarrow3x-5=2x\Leftrightarrow x=5\)

Vậy: \(S=\left\{5\right\}\)

Bài 2:

Câu 1:

\(PT\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{1}{6}\\x-\dfrac{1}{4}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{5}{12};\dfrac{1}{12}\right\}\)

Câu 2:

\(PT\Leftrightarrow\left|3x-5\right|=\dfrac{10}{21}\Leftrightarrow\left[{}\begin{matrix}3x-5=\dfrac{10}{21}\\3x-5=-\dfrac{10}{21}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{115}{63}\\x=\dfrac{95}{63}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{115}{63};\dfrac{95}{63}\right\}\)

Câu 3: 

\(PT\Leftrightarrow-\dfrac{16x}{15}.\dfrac{1}{7}=-\dfrac{5}{21}\Leftrightarrow-16x=-25\Leftrightarrow x=\dfrac{25}{16}\)

Vậy: \(x=\dfrac{25}{16}\)

a) \(f\left(x\right)-g\left(x\right)\) hay \(x^3-2x^2+3x+1-x^3-x+1=-2x^2+2x+2\)

b) \(f\left(x\right)-g\left(x\right)+h\left(x\right)=0\) hay \(-2x^2+2x+2+2x^2-1=2x+1\Rightarrow2x+1=0\Rightarrow x=-\dfrac{1}{2}\)

a) \(\left\{{}\begin{matrix}M=x^2y-2xy+6-xy=x^2y-3xy+6\\N=-2x^2y+2xy+x^2y-3=-x^2y+2xy-3\end{matrix}\right.\)

b) \(x=1;y=2\Rightarrow M=1^2.2-2.1.2+6-1.2=2\)

c) \(M+N\Rightarrow x^2y-3xy+6+\left(-x^2y\right)+2xy-3=-xy+3\)

Câu 1

Thay \(y=\dfrac{1}{2}\) vào phương trình đã cho ta có:

\(\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]-2.\dfrac{1}{2}\left(m-1\right)=2m^2-8\)

\(\Rightarrow1-m+1=2m^2-8\)

\(\Rightarrow2m^2+m-10=0\)

\(\Rightarrow\left(2m+5\right)\left(m-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2m+5=0\\m-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m=-\dfrac{5}{2}\\m=2\end{matrix}\right.\)

Câu 2:

Thay \(x=2\) vào phương trình đã cho ta có:

\(\left(2^2+1\right)=m\left(2+3\right)\)

\(\Rightarrow5m=5\)

\(\Rightarrow m=1\)

a) \(PT\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{2};\dfrac{4}{3}\right\}\)

b) \(PT\Leftrightarrow2x-2x^2+1-x+3x=2\)

\(\Leftrightarrow2x^2-4x+1=0\)

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

Vậy: \(S=\left\{1-\dfrac{1}{\sqrt{2}};1+\dfrac{1}{\sqrt{2}}\right\}\)

c) Chắc đề bài là \(2\left(x-5\right)\left(x+2\right)=x^2-5x\)

\(\Leftrightarrow2x^2+4x-10x-20=x^2-5x\)

\(\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

Vậy: \(S=\left\{5;-4\right\}\)

d) \(PT\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{3;2\right\}\)

Ta có: \(y=\left(2-a\right)x+a\) (1)

Thay điểm \(M\left(3;1\right)\) vào \(\left(1\right)\) ta có:

\(\left(2-a\right).3+a=1\)

\(\Leftrightarrow-2a=-5\)

\(\Leftrightarrow a=\dfrac{5}{2}\)

\(\Rightarrow2-a=\dfrac{5}{2}\)

\(\Leftrightarrow a=-\dfrac{1}{2}\)

Mà \(-\dfrac{1}{2}< 0\)

Vậy hàm số nghịch biến trên \(R\).

d) \(PT\Leftrightarrow x\left(2x-7\right)-4\left(x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)

e) \(PT\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{7;1\right\}\)

f) \(PT\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{1;3\right\}\)

ĐKXĐ: \(x\ge0;x\ne9\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{3}{x-9}\right):\dfrac{1}{\sqrt{x}-3}\)

\(=\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{1}{\sqrt{x}-3}\)

\(=\dfrac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{1}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)