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\(\sqrt{2x+1}-\sqrt{3x}=x-1\)
ĐK: \(x\ge0\)
\(\sqrt{2x+1}-\sqrt{3x}=3x-\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{2x+1}-\sqrt{3x}=\left(\sqrt{3x}-\sqrt{2x+1}\right)\left(\sqrt{3x}+\sqrt{2x+1}\right)\)
\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{3x}\right)\left(1+\sqrt{3x}+\sqrt{2x+1}\right)=0\)
\(\Leftrightarrow\sqrt{2x+1}=\sqrt{3x}\Rightarrow x=1\left(tm\right)\)
1. Ta có: \(x^2-2xy-x+y+3=0\)
<=> \(x^2-2xy-2.x.\frac{1}{2}+2.y.\frac{1}{2}+\frac{1}{4}+y^2-y^2-\frac{1}{4}+3=0\)
<=> \(\left(x-y-\frac{1}{2}\right)^2-y^2=-\frac{11}{4}\)
<=> \(\left(x-2y-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=-\frac{11}{4}\)
<=> \(\left(2x-4y-1\right)\left(2x-1\right)=-11\)
Th1: \(\hept{\begin{cases}2x-4y-1=11\\2x-1=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-3\end{cases}}\)
Th2: \(\hept{\begin{cases}2x-4y-1=-11\\2x-1=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)
Th3: \(\hept{\begin{cases}2x-4y-1=1\\2x-1=-11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)
Th4: \(\hept{\begin{cases}2x-4y-1=-1\\2x-1=11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=3\end{cases}}\)
Kết luận:...
a. \(x^2\left(y-1\right)+y^2\left(x-1\right)=1\)
<=> \(x^2y+y^2x-\left(x^2+y^2\right)=1\)
<=> \(xy\left(x+y\right)-\left(x+y\right)^2+2xy=1\)
Đặt: x + y = u; xy = v => u; v là số nguyên
Ta có: uv - \(u^2+2v=1\)
<=> \(u^2-uv-2v+1=0\)
<=> \(u^2+1=v\left(2+u\right)\)
=> \(u^2+1⋮2+u\)
=> \(u^2-4+5⋮2+u\)
=> \(5⋮2-u\)
=> 2 - u = 5; 2 - u = -5; 2- u = 1; 2- u = -1
Mỗi trường hợp sẽ tìm đc v
=> x; y
Ta có: \(x\left(x+1\right)=\frac{\sqrt{5}-1}{2}.\frac{\sqrt{5}+1}{2}=1\)
Ta có: x5 + x4 - x3 + 1 = (x5 + x4) - x3 + 1 = x3 - x3 + 1 = 1
x2 + x - 3 = x(x + 1) - 3 = - 2
x5 + x4 - x3 - 22016 = - 22016
Từ đó ta có
\(=1^{2017}+\frac{\left(-2\right)^{2016}}{-2^{2016}}=1-1=0\)
Ta có: \(x^2\text{+}x-1=...=0 \)
\(=>x^3\left(x^2\text{+}x-1\right)=0\)
=> \(x^5\text{+}x^4-x^3=0\)
=> A=\(\left(\left(x^5\text{+}x^4-x^3\right)\text{+}1\right)^{2017}\text{+}\frac{\left(\left(x^2\text{+}x-1\right)-2\right)^{2016}}{\left(x^5\text{+}x^4-x^3\right)-2^{2016}}\)
=\(1^{2017}\text{+}\frac{2^{2016}}{-2^{2016}}=1-1=0\)