Câu hỏi của Thỏ Ngọc

Question

$\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=9\%:4,5\text{%}$

nguyễn thị minh châu · Accepted Answer

$\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=9\%:4,5\%$$\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=2$$\left(x+\frac{3}{5}\right)^2=2-1\frac{16}{25}$$\left(x+\frac{3}{5}\right)^2=\frac{9}{25}=\frac{3}{5}^2$$x+\frac{3}{5}=\frac{3}{5}$$x=\frac{3}{5}-\frac{3}{5}$$x=0$Câu trả lời: $\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=9\%:4,5\%$$\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=2$$\left(x+\frac{3}{5}\right)^2=2-1\frac{16}{25}$$\left(x+\frac{3}{5}\right)^2=\frac{9}{25}=\frac{3}{5}^2$$x+\frac{3}{5}=\frac{3}{5}$$x=\frac{3}{5}-\frac{3}{5}$$x=0$

Tran Thi Thu Hien · Answer

$\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=9\%:4,5\%$$\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=2$$\left(x+\frac{3}{5}\right)^2=2-1\frac{16}{25}$$\left(x-\frac{3}{5}\right)^2=\frac{3^2}{5}$$x=\frac{3}{5}-\frac{3}{5}$$x=0$Câu trả lời: $\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=9\%:4,5\%$$\left(x+\frac{3}{5}\right)^2+1\frac{16}{25}=2$$\left(x+\frac{3}{5}\right)^2=2-1\frac{16}{25}$$\left(x-\frac{3}{5}\right)^2=\frac{3^2}{5}$$x=\frac{3}{5}-\frac{3}{5}$$x=0$

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