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Ta có x-y=2(x+y)
<=> x-y=2x+2y
<=> x=-3y (1)
=> x:y=-3y:y=-3
=> x-y=-3
<=> x=-3+y (2)
Từ (1) và (2) suy ra
-3y=-3+y
<=> -3y+3-y=0
<=> -4y=-3
<=> y=\(\frac{3}{4}\)
=> x=-3+\(\frac{3}{4}\)=\(\frac{-9}{4}\)
\(M=\left(x^4+2x^2y^2+y^4\right)+x^4+x^2y^2+y^2\)
\(M=\left(x^2+y^2\right)^2+x^2\left(x^2+y^2\right)+y^2\)
\(M=1^2+x^2.1+y^2\)
\(M=1+1=2\)
\(M=2x^4+3x^2y^2+y^4+y^2\)
\(M=2x^4+2x^2y^2+x^2y^2+y^4+y^2\)
\(M=2x^2\left(x^2+y^2\right)+y^2\left(x^2+y^2\right)+y^2\)
\(M=\left(2x^2+y^2\right)\left(x^2+y^2\right)+y^2\)
\(M=\left(2x^2+y^2\right).1+y^2\)
\(M=2x^2+2y^2=2\left(x^2+y^2\right)=2.1=2\)
Vậy M = 2
d, \(\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left[\frac{3}{5}+\left(-\frac{1}{4}\right)\right]:\frac{3}{7}\)
\(=\frac{7}{20}:\frac{3}{7}+\frac{7}{20}:\frac{3}{7}\)
\(=0\)
3/5/7 là hỗn số nhé ba năm phần bảy
tương tự như mấy câu khác nhé
a) \(\left(x-1,3\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x-1,3=3\\x-1,3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)
b) 24-x = 32
⇔ 24-x = 25
⇔ 4-x=5
⇔ x=-1
c) (x+1,5)2+(y-2,5)10=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,5\end{matrix}\right.\)
\(a,\left(x-1,3\right)^2=9\\ \Leftrightarrow\left(x-1,3+9\right)\left(x-1,3-9\right)=0\\ \Leftrightarrow\left(x-7,7\right)\left(x-10,3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7,7=\dfrac{77}{10}\\x=10,3=\dfrac{103}{10}\end{matrix}\right.\)
\(b,2^{4-x}=32=2^5\\ \Leftrightarrow4-x=5\\ \Leftrightarrow x=-1\)
\(c,\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1,5=-\dfrac{3}{2}\\y=2,5=\dfrac{5}{2}\end{matrix}\right.\)
3,5 + /x + \(\frac{3}{2}\) / = -1,5(-\(\sqrt{9}\))
=> 3,5 +/ x +\(\frac{3}{2}\) / = -1,5 ( -3 )
=> 3,5 + / x + \(\frac{3}{2}\) / =4,5
=> / x + \(\frac{3}{2}\) / = 4,5 - 3,5
=> / x + \(\frac{3}{2}\) / = 1
=> \(\hept{\begin{cases}x+\frac{3}{2}=1\\x+\frac{3}{2}=-1\end{cases}}\)
=> \(\hept{\begin{cases}x=1-\frac{3}{2}\\x=-1-\frac{3}{2}\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=\frac{-5}{2}\end{cases}}\)
vậy x = \(\frac{-1}{2}\)hay x = \(\frac{-5}{2}\)
\(3,5+\left|x+\frac{3}{2}\right|=-1,5.\left(-\sqrt{9}\right)\) \(3,5+\left|x+\frac{3}{2}\right|=-1,5.\left(-3\right)\) \(3,5+\left|x+\frac{3}{2}\right|=4,5\) \(\left|x+\frac{3}{2}\right|=4,5-3,5\) \(\left|x+\frac{3}{2}\right|=1\) \(\Rightarrow\orbr{\begin{cases}x+\frac{3}{2}=1\\x+\frac{3}{2}=-1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\) Vậy x=\(-\frac{1}{2}\) hoặc x=\(-\frac{5}{2}\)