tìm x, biết
(x-2/5)^2=16
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a, xem lại đề , sửa rồi thì báo cho tui
b, \(\left(x+\frac{2}{5}\right)^5=\left(x+\frac{2}{5}\right)^3\)
\(\Rightarrow\left(x+\frac{2}{5}\right)^5-\left(x+\frac{2}{5}\right)^3=0\)
\(\Rightarrow\left(x+\frac{2}{5}\right)^3.\left[\left(x+\frac{2}{5}\right)^2-1\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+\frac{2}{5}\right)^3=0\\\left(x+\frac{2}{5}\right)^2-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{2}{5}\\\left(x+\frac{2}{5}\right)^2=1\end{cases}}}\)
Ta có \(\left(x+\frac{2}{5}\right)^2=1\)
\(\Rightarrow\hept{\begin{cases}x+\frac{2}{5}=1\\x+\frac{2}{5}=-1\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3}{5}\\x=-\frac{7}{5}\end{cases}}}\)
Vậy \(x\in\text{{}-\frac{2}{5};\frac{3}{5};-\frac{7}{5} \)}
Bài làm:
Ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)
\(\Rightarrow\hept{\begin{cases}x^2=25\\y^2=9\end{cases}}\Rightarrow\hept{\begin{cases}x=\pm5\\y=\pm3\end{cases}}\)
Đặt \(\frac{x}{5}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=5k\\y=3k\end{cases}}\)
x2 - y2 = 16 <=> ( 5k )2 - ( 3k )2 = 16
<=> 25k2 - 9k2 = 16
<=> 16k2 = 16
<=> k2 = 1
<=> k = ±1
Với k = 1 => \(\hept{\begin{cases}x=5\\y=3\end{cases}}\)
Với k = -1 => \(\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)
Vậy ( x ; y ) = { ( 5 ; 3 ) , ( -5 ; -3 ) }
a: \(\Leftrightarrow x^2-7x^2+28x=16\)
\(\Leftrightarrow-6x^2+28x-16=0\)
\(\Leftrightarrow3x^2-14x+8=0\)
\(\text{Δ}=\left(-14\right)^2-4\cdot3\cdot8=100\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-10}{6}=\dfrac{4}{6}=\dfrac{2}{3}\\x_2=\dfrac{14+10}{6}=\dfrac{24}{6}=4\end{matrix}\right.\)
a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)
\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)
b) \(\left|2x+3\right|=x+2\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-\dfrac{5}{3}\end{matrix}\right.\)
Đính chính
Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)
\(b,\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow x-2015>0\Leftrightarrow x>2015\\ d,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-x>0\\x+6>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-x< 0\\x+6< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-6< x< 3\)
Giải:
1. Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{-21}{7}=-3\)
+) \(\frac{x}{2}=-3\Rightarrow x=-6\)
+) \(\frac{y}{5}=-3\Rightarrow y=-15\)
Vậy x = -6
y = -15
2. Ta có:
\(7x=3y\Rightarrow\frac{7x}{21}=\frac{3y}{21}=\frac{x}{3}=\frac{y}{7}\)
Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)
+) \(\frac{x}{3}=-4\Rightarrow x=-12\)
+) \(\frac{y}{7}=-4\Rightarrow y=-28\)
Vậy x = -12
y = -28
1/ \(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=-\frac{21}{7}=-3\)
\(\frac{x}{2}=-3\Rightarrow x=-6\)
\(\frac{x}{5}=-3\Rightarrow x=-15\)
2/ \(7x=3y\Rightarrow\frac{x}{7}=\frac{y}{3}\)
\(\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{16}{4}=4\)
\(\frac{x}{7}=4\Rightarrow x=28\)
\(\frac{y}{3}=4\Rightarrow y=12\)
à thôi, mình biết làm rồi :>
\(\left(x-\frac{2}{5}\right)^2=16\)
\(\Rightarrow\left(x-\frac{2}{5}\right)^2=\left(\pm4\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{5}=4\\x-\frac{2}{5}=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{5}\\x=\frac{-18}{5}\end{cases}}\)
Vậy \(x\in\left\{\frac{-18}{5};\frac{22}{5}\right\}\)