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2(x+1)^2=(x-1)( 1+ căn (2x+3) )^2
mong cac2 ban giup
Đề bài là :
[TEX]{2\left(x+1 \right)}^{2}=\left(x-1 \right){\left(1+\sqrt{2x+3} \right)}^{2}[/TEX]
Giải :
Điều kiện :
[TEX]x\geq\frac{-3}{2}[/TEX]
[TEX]{2\left(x+1 \right)}^{2}=\left(x-1 \right){\left(1+\sqrt{2x+3} \right)}^{2}[/TEX]
[TEX]\Leftrightarrow 2\left({x}^{2}+2x+1 \right)=\left(x-1 \right)\left(1+2x+3+2\sqrt{2x+3} \right)[/TEX]
[TEX]\Leftrightarrow{2x}^{2}+4x+2=x+{2x}^{2}+3x+2x\sqrt{2x+3}-1-2x-3-2\sqrt{2x+3} [/TEX]
[TEX]\Leftrightarrow 2x+6+2\sqrt{2x+3}-2x\sqrt{2x+3}=0[/TEX]
[TEX]\Leftrightarrow \left[ \left(2x+3 \right)+2\sqrt{2x+3}+1\right]+2-2x\sqrt{2x+3}=0[/TEX]
[TEX]\Leftrightarrow {\left(\sqrt{2x+3}+1 \right)}^{2}+ \left( {x}^{2}-2x\sqrt{2x+3}+2x+3\right)-\left({x}^{2} +2x+1\right) =0[/TEX]
[TEX]\Leftrightarrow {\left(\sqrt{2x+3}+1 \right)}^{2}+ {\left(-x-3 \right)}^{2}-{\left(x+1 \right)}^{2} =0[/TEX]
[TEX]\Leftrightarrow{\left(\sqrt{2x+3}+1 \right)}^{2}+ {\left(x+3 \right)}^{2}-{\left(x+1 \right)}^{2} =0 [/TEX]
Đến đây là được rồi.Bạn tự giải tiếp nhé!