Câu 3: 

\(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\left(m^2-3m+4\right)\)

\(=\left(2m-2\right)^2-4\left(m^2-3m+4\right)\)

\(=4m^2-16m+4-4m^2+12m-16=-4m-12\)

Để phương trình có hai nghiệm phân biệt thì -4m-12>0

=>-4m>12

hay m<-3

Áp dụng hệ thức Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=m^2-3m+4\end{matrix}\right.\)

Theo đề, ta có: \(x_1+x_2=x_1x_2\)

\(\Leftrightarrow m^2-3m+4-2m+2=0\)

=>(m-2)(m-3)=0

hay \(m\in\varnothing\)

Câu 1:

\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)

Câu 2:

\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)

\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)

6: \(=x^3\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(x-1\right)\left(x^2+x+1\right)\)

7: =(x-4)(x+2)

2/

\(2x^3-8x=2x\left(x^2-4\right)=2x\left(x-2\right)\left(x+2\right)\)

3/

\(9x^2-\left(x-1\right)^2=\left(3x\right)^2-\left(x-1\right)^2=\left(3x-x+1\right)\left(3x+x-1\right)\)

4/

\(x^2-3x+6y-4y^2=x^2-4y^2-3x+6y=\left(x^2-4y^2\right)-\left(3x-6y\right)\)

\(=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x+2y-3\right)\)

7: =(x-4)(x+2)

4: \(=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+2y-3\right)\)

\(1,=5x\left(1-4x+4x^2\right)=5x\left(2x-1\right)^2\\ 2,=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-3\right)\left(x-2y\right)\\ 3,=4x^2-\left(y+3\right)^2=\left(2x+y+3\right)\left(2x-y-3\right)\)

Bài 3: 

a: Tuổi mẹ hiện nay là: \(36\cdot\dfrac{7}{9}=28\left(tuổi\right)\)

Tuổi con hiện nay là: 36-28=8(tuổi)

MgO: Mg có điện hóa trị 2+, O có điện hóa trị 2-

FeF₃: Fe có điện hóa trị 3+, F có điện hóa trị 1-

BaCl₂: Ba có điện hóa trị 2+, Cl có điện hóa trị 1-

Ca₃N₂: Ca có điện hóa trị 2+, N có điện hóa trị 3-

Bài 4:

\(P=\dfrac{4x^2-2x+7}{2x-1}=\dfrac{2x\left(2x-1\right)+7}{2x-1}=2x+\dfrac{7}{2x-1}\in Z\\ \Leftrightarrow2x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-3;0;1;4\right\}\\ Q=\dfrac{4x^2-2x+3}{2x-1}=\dfrac{2x\left(2x-1\right)+3}{2x-1}=2x+\dfrac{3}{2x-1}\in Z\\ \Leftrightarrow2x-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-1;0;1;2\right\}\)

Bài 5:

\(M=\dfrac{\left(5x-1\right)\left(5x+1\right)}{1-5x}+\dfrac{\left(y-3\right)\left(5x+1\right)}{y-3}=-\left(5x+1\right)+5x+1=0\)

Bài 6:

\(VT=\dfrac{a\left(a+3b\right)}{\left(a+3b\right)\left(a-3b\right)}-\dfrac{\left(2a+b\right)\left(a-3b\right)}{\left(a-3b\right)^2}=\dfrac{a}{a-3b}-\dfrac{2a+b}{a-3b}=\dfrac{-a-b}{a-3b}\)

\(VP=\dfrac{\left(a+b\right)\left(a+c\right)}{\left(a+c\right)\left(3b-a\right)}=\dfrac{a+b}{3b-a}=\dfrac{-a-b}{a-3b}\)

Vậy ta đc đpcm