Câu 1 : A

Câu 2 : D

chúc bạn học giỏi

a: \(\Leftrightarrow x^2+6x+9+x^2-4-2x-2=7\)

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

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

\(\Leftrightarrow x^2+2x+1-3=0\)

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

hay \(x\in\left\{-\sqrt{3}-1;\sqrt{3}-1\right\}\)

b: \(\Leftrightarrow2x^2-x-\left(2x^2+3x-4x-6\right)=0\)

\(\Leftrightarrow2x^2-x-2x^2+x+6=0\)

=>6=0(vô lý)

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

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

=>x=-2 hoặc x=2

đ: \(\Rightarrow2x^2-2x-5x+5=0\)

=>(x-1)(2x-5)=0

=>x=1 hoặc x=5/2

a/

\(\Leftrightarrow x-2x^2+2x^2-3x-4x+6=0\)

\(\Leftrightarrow-6x+6=0\)

\(\Leftrightarrow x=1\)

b/

\(\Leftrightarrow2x^2-4x-2x^2-6x=0\)

\(\Leftrightarrow-10x=0\)

\(\Leftrightarrow x=0\)

c/

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

\(\Leftrightarrow3x\left(2x+3\right)=0\)

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

c/

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(9y^2+30y+25\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(3y+5\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\3x+5=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=-\frac{5}{3}\)

d/

\(\Leftrightarrow4x^2-4x+1+4x^2+4x+1-2\left(4x^2-2x-2\right)+x=12\)

\(\Leftrightarrow8x^2+x+2-8x^2+4x+4=12\)

\(\Leftrightarrow5x=6\)

\(\Leftrightarrow x=\frac{6}{5}\)

\(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)\left(2x+1\right)=7\)

\(\Leftrightarrow4x^2-8x+x-2-\left(4x^2+2x-6x-3\right)=7\)

\(\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\)

\(\Leftrightarrow-3x+1=7\)

\(\Leftrightarrow-3x=6\Leftrightarrow x=-2\)

\(\left(x-1\right)\left(x+2\right)-x-2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)

\(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-\left(4x^2-25\right)=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+25=0\)

\(\Leftrightarrow4x+26=0\)

\(\Leftrightarrow4x=26\Leftrightarrow x=\frac{13}{2}\)

1/ ( x-1) (2x+1) =0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-0,5\end{matrix}\right.\)

2/ x (2x-1) (3x+15) =0

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-5\end{matrix}\right.\)

3/ (2x-6) (3x+4).x=0

\(\Rightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)

4/ (2x-10)(x²+1)=0

\(\Rightarrow\left[{}\begin{matrix}2x-10=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x^2=-1\left(loại\right)\end{matrix}\right.\)

5/ (x²+3) (2x-1) =0

\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x^2=-3\left(loại\right)\\x=0,5\end{matrix}\right.\)

6/ (3x-1) (2x² +1)=0

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\2x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2=-0,5\left(loại\right)\end{matrix}\right.\)

1: Ta có: \(\left(x-1\right)\left(2x+1\right)=0\)

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

2: Ta có: \(x\left(2x-1\right)\left(3x+15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-5\end{matrix}\right.\)

3: Ta có: \(\left(2x-6\right)\left(3x+4\right)x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)

4: Ta có: \(\left(2x-10\right)\left(x^2+1\right)=0\)

mà \(x^2+1>0\forall x\)

nên 2x-10=0

hay x=5

5: Ta có: \(\left(x^2+3\right)\left(2x-1\right)=0\)

mà \(x^2+3>0\forall x\)

nên 2x-1=0

hay \(x=\dfrac{1}{2}\)

6: Ta có: \(\left(3x-1\right)\left(2x^2+1\right)=0\)

mà \(2x^2+1>0\forall x\)

nên 3x-1=0

hay \(x=\dfrac{1}{3}\)

\(a.x\left(x^2-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

\(b.\left(x-\frac{1}{2}\right)\left(2x+5\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-\frac{1}{2}=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{5}{2}\end{matrix}\right. \)

Câu \(b\) thấy hơi kì nên chắc đề như này.

\(c.x-2\left(\frac{2}{3}x-6\right)=0\\\Leftrightarrow x-\frac{4}{3}x+12=0\\\Leftrightarrow -\frac{1}{3}x+12=0\\\Leftrightarrow -\frac{1}{3}x=-12\\\Leftrightarrow x=36\)

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

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

\(f.x\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)

\(g.4x^2+4x+1=0\\ \Leftrightarrow4\left(x^2+x+\frac{1}{4}\right)=0\\\Leftrightarrow x^2+x+\frac{1}{4}=0\\\Leftrightarrow \left(x+\frac{1}{2}\right)^2=0\\\Leftrightarrow x+\frac{1}{2}=0\\ \Leftrightarrow x=-\frac{1}{2}\)

\(h.x^2-5x+6=0\\ \Leftrightarrow x^2-2x-3x+6=0\\\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

\(i.2x^2+3x=0\\ \Leftrightarrow x\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\frac{3}{2}\end{matrix}\right.\)

\(\begin{array}{l} a)x\left( {{x^2} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ {x^2} - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 1\\ x = - 1 \end{array} \right.\\ b)\left( {x - \dfrac{1}{2}} \right)\left( {2x + 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - \dfrac{1}{2} = 0\\ 2x + 5 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{1}{2}\\ x = - \dfrac{5}{2} \end{array} \right.\\ c)\left( {x - 2} \right)\left( {\dfrac{2}{3}x - 6} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\ \dfrac{2}{3}x - 6 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 9 \end{array} \right. \end{array}\)

a) \(x\left(x^2-1\right)=0\)

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

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

Vậy: x∈{-1;0;1}

d) \(x^2-2x=0\)

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

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

Vậy: x∈{0;2}

e) \(\left(x^2-2x+1\right)-4=0\)

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

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

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

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

Vậy: x∈{3;-1}

f) \(x\left(2x-1\right)=0\)

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

Vậy: \(x\in\left\{0;\frac{1}{2}\right\}\)

g) \(4x^2+4x+1=0\)

\(\Leftrightarrow\left(2x+1\right)^2=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

hay \(x=\frac{-1}{2}\)

Vậy: \(x=\frac{-1}{2}\)

h) \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

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

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

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

Vậy: x∈{2;3}

i) \(2x^2+3x=0\)

\(\Leftrightarrow x\left(2x+3\right)=0\)

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

Vậy: \(x\in\left\{0;\frac{-3}{2}\right\}\)