Số giá trị của x là 2.

\(\frac{6\frac{1}{4}}{x}=\frac{x}{1,96}\)

\(\Leftrightarrow x.x=6\frac{1}{4}.1,96\)

\(\Leftrightarrow x^2=12,25\)

\(\Leftrightarrow x^2=3,5^2\)

\(\Leftrightarrow x=3,5\)

Vậy \(x=3,5\)

Số giá trị của x là @2

\(\frac{6\frac{1}{4}}{x}=\frac{x}{1,96}\)

\(\Rightarrow x^2=6\frac{1}{4}.1,96\)

\(\Rightarrow x^2=\frac{25}{4}.\frac{49}{25}\)

\(\Rightarrow x^2=\frac{49}{4}\)

\(\Rightarrow x^2=\left(\frac{7}{2}\right)^2\)

\(\Rightarrow x=\frac{7}{2}\)

vậy \(x=\frac{7}{2}\)

Ta có : \(\frac{6\frac{1}{4}}{x}=\frac{x}{1,96}\)

\(\Rightarrow\frac{6,25}{x}=\frac{x}{1,96}\)

\(\Rightarrow\frac{6,25.1,96.x}{x.1,96}=\frac{x.x}{1,96.x}\)

\(\Rightarrow6,25.1,96.x=x.x\)

\(\Rightarrow12,25.x=x.x\)

Vì \(x=x\)nên để \(12,25.x=x.x\)thì \(12,25=x\)

Vậy \(x=12,25\)

ta có:\(\frac{6\frac{1}{4}}{x}=\frac{x}{1,96}\)

=>\(6\frac{1}{4}.1,96=x.x\)

=>6,25.1,96=2x

=>12,25=2x

=>x=12,25:2

Vậy x= 6,125

=>x.x=\(6\frac{1}{4}.1,96\)

=>x^2=12,25

=>x=3,5 hoặc x=-3,5

số giá trị là 2

\(2\)

\(\frac{6\frac{1}{4}}{x}=\frac{x}{1,96}\)

=> \(x^2=6\frac{1}{4}.1,96\)

=> \(x^2=12,25\)

=> \(x^2=\left(\pm3,5\right)^2\)

=> \(x=\pm3,5\)

Vậy \(x\in\left\{3,5;-3,5\right\}\)

\(\frac{6\frac{1}{4}}{x}=\frac{x}{1,96}\)

\(\frac{6,25}{x}=\frac{x}{1,96}\)

\(\Rightarrow6,25.1,96=x^2\)

\(\Rightarrow12.25=x^2\)

\(\Rightarrow\sqrt{12,25}=x=3,5\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3,5\\x=-3,5\end{array}\right.\)

2

Ta có: \(\frac{6\frac{1}{4}}{x}=\frac{x}{1,96}\)

\(\left(=\right)\frac{\frac{25}{4}}{x}=\frac{x}{1,96}\)

\(\left(=\right)x^2=12,25\)

\(=>\orbr{\begin{cases}x=3,5\\x=-3,5\end{cases}}\)

học tốt

Thanks!!!!!!!!!!!!

đk \(x\ne0\)

\(\frac{6\frac{1}{4}}{x}=\frac{x}{1,96}\)

\(\Leftrightarrow x^2=12,25\)

\(\Leftrightarrow x=\pm3,5(Tm)\)

\(\frac{6+\frac{1}{4}}{x}=\frac{x}{1,96}\)

=> \(x^2=\left(6+\frac{1}{4}\right)x1,96\)

=> \(x^2=\frac{25}{4}x1,96\)

=> \(x^2=12,25\)

=> \(x=3,5\)

Điều kiện: x - 5 \(\ne\) 0 <=> x \(\ne\) 5

phương trình <=> \(\frac{\left(x-5\right)+\left(x-6\right)+\left(x-7\right)+...+1}{x-5}=4\)

tính \(\left(x-5\right)+\left(x-6\right)+\left(x-7\right)+...+1=\left[\left(x-5\right)+1\right].\left(x-5\right):2=\frac{\left(x-4\right)\left(x-5\right)}{2}\)

pt <=> \(\frac{\left(x-4\right)\left(x-5\right)}{2.\left(x-5\right)}=4\) <=> x - 4 = 8 <=> x = 12 (thoả mãn)

Thi vòng 12 à bạn!!! Để mk chép đề mà làm