\(\left(x-3\right)^2+7x=9^2\)

=> \(x^2-6x+9+7x=81\)

=> \(x^2+x+9=81\)

=> \(x.\left(x+1\right)=72\)

=> \(x.\left(x+1\right)=8.9\)

=> \(x=8\)

Vậy x = 8

\(\left(x-3\right)^2+7x=9^2\)

 \(x^2-6x+3^2+7x=9^2\)

 \(x^2+\left(-6x+7x\right)+3^2=9^2\)

 \(x^2+x+3^2=9^2\)

 \(\left(x+3\right)^2=9^2\)

\(\Rightarrow\)\(x+3=9\) hoặc  \(x+3=-9\)

TH1:  \(x+3=9\Rightarrow x=9-3=6\)

TH2:  \(x+3=-9\Rightarrow x=-9-3=-12\)

Vậy \(x=6\)hoặc  \(x=-12\).

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rap ng bn 4 chan

thiếu đề bài bạn ơi

Chậc :))) T còn cách khác đây =)))

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

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

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

\(\Leftrightarrow2\sqrt{x-2}-1=2\sqrt{x-1-2\sqrt{x-2}}-2\sqrt{x-2}\)

\(\Leftrightarrow4x-4\sqrt{x-2}-7=-8\sqrt{x-2}-8\sqrt{x-2}.\sqrt{x-2\sqrt{x-2}-1}+8x-12\)

\(\Leftrightarrow5-4\sqrt{x-2}-4x=-8\sqrt{x-2}-8\sqrt{x-2}.\sqrt{x-2\sqrt{x-2}-1}\)

\(\Leftrightarrow x=\frac{9}{4}\) (tmyk)