Tim x,y biet
a)(2x +1)(3y-2)=55
b)(x+1)(xy-1)=3
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a: \(\left(x+5\right)^2>=0\forall x\)
\(\left(2y-8\right)^2>=0\forall y\)
Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)
b: \(\left(x+3\right)\left(2y-1\right)=5\)
=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)
a, x + xy + y = 9
=>xy + x+y+1=10
=>x.(y+1)+(y+1)=10
=>(x+1).(y+1)= 10.1 = 1.10 = 2.5 = 5.2 = (-10).(-1) = (-1).(-10) = (-2).(-5) = (-5).(-2)
ta có bảng các trường hợp sau
x+1 | 1 | 10 | 2 | 5 | -10 | -1 | -2 | -5 |
y+1 | 10 | 1 | 5 | 2 | -1 | -10 | -5 | -2 |
x | 0 | 9 | 1 | 4 | -11 | -2 | -3 | -6 |
y | 9 | 0 | 4 | 1 | -2 | -11 | -6 | -3 |
vậy
bn tich cho mk nha
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
xy-2x-3y+6=5+6
x(y-2)-3(y-2)=5+6
(x-3)(y-2)=11
(x-3)(y-2)=1.11;11.1
nếu (x-3)(y-2)=1.11=> x=4 và y=13
nếu (x-3)(y-2)=11.1=>x=14 và y=3
câu b tương tự
xy-2x+5y-10=2-10
x(y-2)+5(y-2)=-8
.....
nhớ tick nha
Nguyễn Thùy Linh sai vì 3y+6=3(y+2) chứ ko phải 3y+6=3(y-2)
a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 \(M=x^3+x^2y-2x^2-xy-y^2+3y+x-1\)
\(M=x^3+x^2y-2x^2-xy-y^2+\left(2y+y\right)+x-\left(-2+1\right)\)
\(M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-2\right)+1\)
\(M=\left(x^2.x+x^2.y-2x^2\right)-\left(x.y+y.y-2y\right)+\left(x+y-2\right)+1\)
\(M=x^2.\left(x+y-2\right)-y.\left(x+y-2\right)+\left(x+y-2\right)+1\)
\(M=x^2.0+y.0+0+1\)
\(M=1\)
\(N=x^3+x^2y-2x^2-xy^2+x^2y+2xy+2y+2x-2\)
\(N=x^3+x^2y-2x^2-xy^2+x^2y+2xy+2y+2x-\left(-4+2\right)\)
\(N=\left(x^3+x^2y-2x^2\right)-\left(x^2y+xy^2-2xy\right)+\left(2x+2y-4\right)+2\)
\(N=\left(x^2x+x^2y-2x^2\right)-\left(xyx+xyy-2xy\right)+\left(2x+2y-4\right)+2\)
\(N=x^2\left(x+y-2\right)-xy\left(x+y-2\right)+2\left(x+y-2\right)+2\)
\(N=x^2.0-xy.0+2.0+2\)
\(N=2\)
\(P=x^4+2x^3y-2x^3+x^2y^2-2x^2y-x\left(x+y\right)+2x+3\)
\(P=\left(x^4+x^3y-2x^3\right)+\left(x^3y+x^2y^2-2x^2y\right)-\left(x^2+xy-2x\right)+3\)\(P=\left(x^3x+x^3y-2x^3\right)+\left(x^2y.x+x^2yy-2x^2y\right)-\left(xx+xy-2x\right)+3\)
\(P=x^3\left(x+y-2\right)+x^2y\left(x+y-2\right)-x\left(x+y-2\right)+3\)
\(P=x^3.0+x^2y.0-x.0+3\)
\(P=3\)
Tích mình nha!
\(A=2x^2+x-5y+4\)
Thay x = 1/2 ; y = -1/52 vào biểu thức trên ta được :
\(=2.\frac{1}{4}+\frac{1}{2}-5.\frac{-1}{52}+4=1+\frac{5}{52}+4\)
\(=5+\frac{5}{52}=\frac{260}{52}+\frac{5}{52}=\frac{265}{52}\)
\(B=2x^2-3y^2+4z^3\)
Thay x = 2 ; y = z = -23 vào biểu thức trên ta được :
\(=2.4-3.169+4.2197=8-507+8788=8289\)
tương tự với c, bài này ko khó, tại số to nên tính có khi nhầm lẫn vài chỗ thôi.
1) \(2\left(x-1\right)^3-\left(x-1\right)=\left(x-1\right)\left(2\left(x-1\right)^2-1\right)\)
2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)
3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)
4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)