a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)

\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)

\(\Leftrightarrow9x^2=16\)

\(\Leftrightarrow x^2=\dfrac{16}{9}\)

hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)

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

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

\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

d: ta có: \(\dfrac{2x^3-8x^2+10x}{2x}=0\)

\(\Leftrightarrow x^2-4x+5=0\)

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

(2x² + 1)(x-3)=0

\(\Rightarrow\orbr{\begin{cases}2x^2+1=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x^2=-1\Rightarrow x^2=-\frac{1}{2}\left(vl\right)\\x=3\end{cases}}\)

Vậy x=3

48-(15-x)⁵=48

     (15-x)⁵=48-48

     (15-x)⁵=0

=>  15-x   =0

           x    =15-0

           x    =15

Vậy x=15

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

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

3) \(x^2+2x-48=\left(x-6\right)\left(x+8\right)\)

4) \(x^2-2x-48=\left(x-8\right)\left(x+6\right)\)

5) \(x^2+x-42=\left(x-6\right)\left(x+7\right)\)

6) \(x^2-x-42=\left(x-7\right)\left(x+6\right).\)

bài này tìm x hay tìm cực trị vậy

1) \(x^2+x-6\)

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

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

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

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

2) \(x^2-x-6\)

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

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

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

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

3) \(x^2+2x-48\)

\(=x^2+8x-6x-48\)

\(=\left(x^2+8x\right)-\left(6x+48\right)\)

\(=x\left(x+8\right)-6\left(x+8\right)\)

\(=\left(x+8\right)\left(x-6\right)\)

4) \(x^2-2x-48\)

\(=x^2-8x+6x-48\)

\(=\left(x^2-8x\right)+\left(6x-48\right)\)

​\(=x\left(x-8\right)+6\left(x-8\right)\)​

\(=\left(x-8\right)\left(x+6\right)\)

5) \(x^2+x-42\)

\(=x^2+7x-6x-42\)

\(=\left(x^2+7x\right)-\left(6x+42\right)\)

\(=x\left(x+7\right)-6\left(x+7\right)\)

\(=\left(x+7\right)\left(x-6\right)\)

6) \(x^2-x-42\)

\(=x^2-7x+6x-42\)

\(=\left(x^2-7x\right)+\left(6x-42\right)\)

\(=x\left(x-7\right)+6\left(x-7\right)\)

\(=\left(x-7\right)\left(x+6\right)\)

a: Ta có: \(x\left(2-x\right)+x^2+x=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

\(2^{2x+1}+4^x=48\)

\(\Leftrightarrow2^{2x+1}+2^{2x}=48\)

\(\Leftrightarrow2^{2x}\left(2+1\right)=48\)

\(\Leftrightarrow2^{2x}=16\)'

\(\Leftrightarrow2^{2x}=2^4\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

\(2^{2x+1}+4^x=48\)

\(\Rightarrow2^{2x+1}+2^{2x}=48\)

\(\Rightarrow2^{2x}.\left(2+1\right)=48\)

\(\Rightarrow2^{2x}.3=48\)

\(\Rightarrow2^{2x}=16\)

\(\Rightarrow2^{2x}=2^4\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

\(2^{2x+1^{ }}+4^{x^{ }}=48\)

theo như mk lm thì ko ra kết quả

đề bài đâu bn x

vs x là số tự nhiên ak

ko cs dữ kiện đề bài thì lm s mak lm đc