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xin lỗi, chưa học tới lớp 9

đề đau bạn?????

Cho tui xin cái đề thì tui ms giúp đc chứ !!!

\(5,A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(A=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(A=\left|2x-1\right|+\left|2x-3\right|\)

\(A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)

\(A\ge2\)

\(< =>MIN:A=2\)dấu = xảy khi \(\frac{1}{2}\le x\le\frac{3}{2}\)

\(7:a,\sqrt{2-x}=3\)

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

\(\orbr{\begin{cases}2-x=9\\2-x=-9\end{cases}\orbr{\begin{cases}x=-7\left(KTM\right)\\x=11\left(TM\right)\end{cases}}}\)

\(b,\sqrt{4-4x+x^2}=3\)

\(\sqrt{\left(2-x\right)^2}=3\)

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

\(\orbr{\begin{cases}2-x=3\\2-x=-3\end{cases}\orbr{\begin{cases}x=-1\left(TM\right)\\x=5\left(TM\right)\end{cases}}}\)

\(c,\sqrt{4+x^2}+x=3\)

\(\sqrt{4+x^2}=3-x\)

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

\(4+x^2=9-6x+x^2\)

\(x=\frac{5}{6}\left(TM\right)\)

\(d,\frac{1}{2}\sqrt{16x-32}-2\sqrt{4x-8}+\sqrt{9x-18}=5\)

\(2\sqrt{x-2}-4\sqrt{x-2}+3\sqrt{x-2}=5\)

\(\sqrt{x-2}\left(2-4+3\right)=5\)

\(\sqrt{x-2}=5\)

\(\left|x-2\right|=25\)

\(\orbr{\begin{cases}x-2=25\\x-2=-25\end{cases}\orbr{\begin{cases}x=27\left(TM\right)\\x=-23\left(KTM\right)\end{cases}}}\)

thank

cần cái gì

tên nick hay đấy :v

Em ở Hà Nội :v

Sau thi chuyên Nguyễn Huệ :33