tìm x\(\left(x-\frac{2}{3}\right)^2=\frac{25}{36}\)
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a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)
\(\Rightarrow x=100\)
Vậy x = 100
b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)
\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
2)
Ta có:
\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)
( vì \(75⋮25\) )
\(\Rightarrowđpcm\)
\(F\left(x\right)=x^2-3x^3-\sqrt{25}+\frac{1}{2}x-\left(-3x^3+\frac{5x}{2}-\sqrt{36}\right)\)
=> \(F\left(x\right)=x^2-3x^3-5+\frac{1}{2}x+3x^3-\frac{5x}{2}+6\)
=> \(F\left(x\right)=x^2+\left(3x^3-3x^3\right)+\left(6-5\right)+\left(\frac{x}{2}-\frac{5x}{2}\right)\)
=> \(F\left(x\right)=x^2+1-2x\)
mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn
Đặt \(a=24-x,b=x-25\)
Khi đó pt ban đầu trở thành :
\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow30a^2+68ab+30b^2=0\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3a=-5b\\5a=-3b\end{cases}}\)
Đến đây bạn thay vào là dễ rồi nhé ! Chúc bạn học tốt !
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
\((x-\frac{2}{3})^2=\frac{25}{36}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{5}{6}\\x-\frac{2}{3}=\frac{-5}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{-1}{6}\end{cases}}\)
Vậy x=...( bạn tự kết luận nhé )
tính 2phan 3 mũ 2 trước rồi tinh như bình thường