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

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

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

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

ta có:

\(\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow\left(16-2x+x^2-9+2x-x^2\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow7=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

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

Ta có:

\(\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)=7\)

\(\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow\left(16-2x+x^2-9+2x-x^2\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

\(\Leftrightarrow7=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)

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

Ủng hộ nha

cau tra loi duoc olm lua chon

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

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 \(x=-\dfrac{13}{24}\)