\(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

\(\Leftrightarrow\dfrac{\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\)

\(\Leftrightarrow\dfrac{16-2x+x^2-9+2x-x^2}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\)

\(\Leftrightarrow\dfrac{7}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\Leftrightarrow\dfrac{7}{A}=1\Rightarrow A=7\)

Có: \(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

\(\Leftrightarrow\sqrt{\left(x-1\right)^2+15}-\sqrt{\left(x-1\right)^2+8}=1\)

\(\Leftrightarrow2\left(x-1\right)^2+23-2\sqrt{\left(x-1\right)^4+23\left(x-1\right)^2+120}=1\)

Đặt \(t=\left(x-1\right)^2\left(t\ge0\right)\)

\(\Rightarrow2t+23-2\sqrt{t^2+23t+120}=1\)

\(\Leftrightarrow t+11=\sqrt{t^2+23t+120}\)

\(\Leftrightarrow t^2+22t+121=t^2+23t+120\)

\(\Leftrightarrow t=1\left(TM\right)\)

\(\Rightarrow x\in\left\{0;2\right\}\)

Thay x=0 vào A, ta có:

\(A=\sqrt{16-2.0+0^2}+\sqrt{9-2.0+0^2}=7\)

Thay x=2 vào A, ta có:

\(A=\sqrt{16-2.1+1^2}+\sqrt{9-2.1+1^2}=\sqrt{15}+2\sqrt{2}\)

Ta có \(\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)=16-2x+x^2-\left(9-2x+x^2\right)=16-2x+x^2-9+2x-x=7\Leftrightarrow\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)=7\Leftrightarrow1.A=7\Leftrightarrow A=7\)

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

Sửa đề: \(16\left(x-1\right)^2-9\left(2x+1\right)^2=0\)

=>\(\left[4\left(x-1\right)\right]^2-\left[3\left(2x+1\right)\right]^2=0\)

=>\(\left(4x-4\right)^2-\left(6x+3\right)^2=0\)

=>\(\left(4x-4-6x-3\right)\left(4x-4+6x+3\right)=0\)

=>\(\left(-2x-7\right)\left(10x-1\right)=0\)

=>\(\left(2x+7\right)\left(10x-1\right)=0\)

=>\(\left[{}\begin{matrix}2x+7=0\\10x-1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=-7\\10x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=\dfrac{1}{10}\end{matrix}\right.\)

\(\left(2x+1\right)^2=25=5^2=\left(-5\right)^2\\ \left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

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

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

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

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

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

c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

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

d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

Chọn đáp án D.

Ta có:

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)