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a. Ta có: \(x^2-10x+26+y^2+2y=0\Leftrightarrow\left(x^2-10x+25\right)+\left(y^2+2y+1\right)=0\\ \)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\Rightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)
b. \(\left(2x+5\right)^2-\left(x-7\right)^2=0\Leftrightarrow\left(2x+5+x-7\right).\left(2x+5-x+7\right)=0\)
\(\Leftrightarrow\left(3x-2\right).\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-12\end{cases}}}\)
c. \(25.\left(x-3\right)^2=49.\left(1-2x\right)^2\Leftrightarrow\left(5x-15\right)^2=\left(7-14x\right)^2\Leftrightarrow\left(5x-15\right)^2-\left(7-14x\right)^2=0\)
\(\Leftrightarrow\left(5x-15-7+14x\right).\left(5x-15+7-14x\right)=0\Leftrightarrow\left(19x-22\right).\left(-9x-8\right)=0\)
\(\Leftrightarrow\left(19x-22\right).\left(9x+8\right)=0\Leftrightarrow\orbr{\begin{cases}19x-22=0\\9x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{19}\\x=-\frac{8}{9}\end{cases}}}\)
d. \(\left(x+2\right)^2=\left(3x-5\right)^2\Leftrightarrow\left(x+2\right)^2-\left(3x-5\right)^2=0\Leftrightarrow\left(x+2+3x-5\right).\left(x+3-3x+5\right)=0\)
\(\Leftrightarrow\left(4x-3\right).\left(8-2x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-3=0\\8-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=4\end{cases}}}\)
e. \(x^2-2x+1=16\Leftrightarrow\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1-4\right).\left(x-1+4\right)=0\)
\(\Leftrightarrow\left(x-5\right).\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
xy + 3x - 2y - 7 = 0
\(\Rightarrow\) x(y + 3) - 2(y + 3) - 1 = 0
\(\Rightarrow\) (x - 2)(y + 3) = 1
\(\Rightarrow\) \(\orbr{\begin{cases}x-2=y+3=1\\x-2=y+3=-1\end{cases}}\)
\(\Rightarrow\) \(\orbr{\begin{cases}x=3;y=-2\\x=1;y=-4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5x^2+10y^2-6xy-4x-2y+3\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\) (đpcm)
p/s: chúc bạn học tốt
Hi hi mik cảm ơn bn và mik cx chúc bn học tốt nha!!!^-^!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(xy+3x-2y-7=0\)
\(\Leftrightarrow x\left(y+3\right)-2y-6=1\)
\(\Leftrightarrow\left(x-2\right)\left(y+3\right)=1\)
mà \(x,y\)nguyên nên \(x-2,y+3\)là ước của \(1\)nên ta có bảng giá trị:
x-2 | 1 | -1 |
y+3 | 1 | -1 |
x | 3 | -1 |
y | -2 | -4 |
Vậy phương trình có nghiệm là: \(\left(3,-2\right),\left(-1,-4\right)\).
b) \(5y-2x^2-2y^2+2=0\)
\(\Leftrightarrow16x^2+16y^2-40y-16=0\)
\(\Leftrightarrow\left(4x\right)^2+\left(4y-5\right)^2=41\)
Vì \(x,y\)nguyên nên \(\left(4x\right)^2,\left(4y-5\right)^2\)là các số chính phương.
Phân tích \(41\)thành tổng hai số chính phương có cách duy nhất bằng \(41=16+25\)
mà \(\left(4x\right)^2⋮16\)nên ta có:
\(\hept{\begin{cases}\left(4x\right)^2=16\\\left(4y-5\right)^2=25\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\pm1\\y=0\end{cases}}\)(vì \(y\)nguyên)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 3(2x - 1)2 \(\ge0\forall x\)
7(3y + 5)2 \(\ge0\forall x\)
Mà : 3(2x - 1)2 + 7(3y + 5)2 = 0
Nên : 3(2x - 1)2 = 7(3y + 5)2 = 0
\(\Leftrightarrow\hept{\begin{cases}3\left(2x-1\right)^2=0\\7\left(3y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^2=0\\\left(3y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)=0\\\left(3y+1\right)=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x=1\\3y=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{3}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(3x\left(x-3\right)+4x-12=0\)
=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)
=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(3x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)
\(\Leftrightarrow x^3+1-x^3+2x=17\)
=>2x+1=17
=>2x=17-1=16
=>\(x=\dfrac{16}{2}=8\)
c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)
=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)
=>\(15x=-14\)
=>\(x=-\dfrac{14}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)