\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)

\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)

Vậy \(x=\dfrac{-255}{2}\)

\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)

b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

- Gửi lẻ câu hỏi ra nha bạn 2 3 câu 1 lần thôi .

a) (x-3)²-4=0

⇒ (x-3)²=4

⇒ hoặc x-3=2⇒x=5

hoặc x-3=-2⇒x=1

c) (x+4)²-(x+1)(x-1)=16

⇒ x²+8x+16-x²+1=16

⇒ 8x+17=16

⇒ 8x=-1

⇒ x=-1/8

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

\(\Leftrightarrow x^3-1-x^3-x=4\)

\(\Leftrightarrow-x=5\)

hay x=-5

c: Ta có: \(\left(2x-1\right)^3+\left(x+2\right)^3-9x\left(x+1\right)\left(x-1\right)=7\)

\(\Leftrightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8-9x^3+9x=7\)

\(\Leftrightarrow-6x^2+27x=0\)

\(\Leftrightarrow-3x\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{9}{2}\end{matrix}\right.\)

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

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

\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)

Bài 1.

a: =>2x^2=4

=>x^2=2

=>\(x=\pm\sqrt{2}\)

b: =>(x+1)^2-4=0

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

=>(x+3)(x-1)=0

=>x=1 hoặc x=-3

c: =>(2x-1)^2-3^2=0

=>(2x-1-3)(2x-1+3)=0

=>(2x-4)(2x+2)=0

=>x=2 hoặc x=-1

d: x^2-x=0

=>x(x-1)=0

=>x=0 hoặc x=1