2 dễ rồi, suy nghĩ đy nhé

4) \(15-\left\{2\left[x-\left(2x-4\right)\cdot5\right]-3\left(x+1\right)\right\}=12-x\)

\(\Rightarrow15-\left[2\left(x-10+20\right)-3x-3\right]=12-x\)

\(\Rightarrow15-\left(2x-20+40-3x-3\right)=12-x\)

\(\Rightarrow15+x-17=12-x\)

\(\Rightarrow x+x=12-15+17\)

\(\Rightarrow2x=14\Rightarrow x=\dfrac{14}{2}=7\)

5) \(\left(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+\dfrac{3}{3\cdot4}+...+\dfrac{3}{89\cdot90}\right)-2x=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\)

\(\Rightarrow3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdot\cdot\cdot+\dfrac{1}{89\cdot90}\right)-2x=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{97}-\dfrac{1}{99}\)

\(\Rightarrow3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{89}-\dfrac{1}{90}\right)-2x=1-\dfrac{1}{99}\)

\(\Rightarrow3\left(1-\dfrac{1}{90}\right)-2x=\dfrac{98}{99}\)

\(\Rightarrow-2x=\dfrac{98}{99}-3\left(1-\dfrac{1}{90}\right)=\dfrac{98}{99}-\dfrac{89}{30}=-\dfrac{1957}{990}\)

\(\Rightarrow x=-\dfrac{1957}{990}:\left(-2\right)=\dfrac{1957}{1980}\)

Vậy.............

\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

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

\(\Leftrightarrow2x-5=0\Leftrightarrow x=\dfrac{5}{2}\)

Vậy x = 5/2

Cái tội lười làm bài tập nó thế đấy! Me, too!

\(\left(x-2\right)^4=\left(x-2\right)^2\)

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

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

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

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

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

Vậy x = 1, x = 2 hoặc x = 3

1) Thay x = 2 vào f(x) ta có:

\(f\left(x\right)\Leftrightarrow2\cdot2^2-\left(m+1\right)2-m=0\)

\(\Leftrightarrow8-2m-2-m=0\)

\(\Leftrightarrow6-3m=0\Leftrightarrow-3m=-6\)

\(\Leftrightarrow m=\dfrac{-6}{-3}=2\)

2) \(f\left(x\right)=2x^2-3x-2=0\)

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

\(\Leftrightarrow x\left(2x-2\right)-\dfrac{1}{2}\left(2x-2\right)-3=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)\left(2x-2\right)=3\)

Vậy được nghiệm \(x=-\dfrac{1}{2}\)

\(x^2+6x+8=0\)

\(\Leftrightarrow x^2+2x+4x+8=0\)

\(\Leftrightarrow x\left(x+2\right)+4\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+2\right)=0\)

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

Vậy đa thức có 2 nghiệm là x = -4 và x = -2

a) \(2x-10-\left[3x-13-\left(3-5x\right)-4\right]=7\)

\(\Leftrightarrow2x-10-\left(8x-20\right)=7\)

\(\Leftrightarrow-6x+10=7\Leftrightarrow-6x=-3\Leftrightarrow x=\dfrac{1}{2}\)

b) \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=-\dfrac{11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=-\dfrac{11}{2}+3=-\dfrac{5}{2}\) (vô lí)

Vậy k có giá trị x thỏa mãn đề