(x+y-12)^2+(y-2)^2=0
Những câu hỏi liên quan
a \(\left(x-1\right)^2-\left(y+1\right)^2=0\)
\(x+3y-5=0\)
b \(xy-2x-y+2=0\)
3x+y=8
c \(\left(x+y\right)^2-4\left(x+y\right)=12\)
\(\left(x-y\right)^2-2\left(x-y\right)=3\)
d \(2x-y=1\)
\(2x^2+xy-y^2-3y=-1\)
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: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
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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:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
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c.
\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)
Xét pt:
\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)
TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)
\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)
\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)
TH2: \(y=6-x\) thế vào...
\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)
\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)
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Xem thêm câu trả lời
cho các số x,y thỏa mãn x^4 +x^2*y^2+y^4=0; x^8 +y^8+x^4*y^4=8 .Biểu thức A=x^12+x^2*y^2+y^12 có giá trị là
x^8+x^4*y^4+y^8=(x^4+y^4)^2-x^4*y^4=((x^2+y^2)^2-2x^2*y^2)^2-(x^2*y^2)^2=8x^4+x^2*y^2+y^4=(x^2+y^2)^2-x^2*y^2=0
Đặt x^2+y^2=a; x^2*y^2=b
nên hệ pt
a^2-b=0(a^2-2b)^2-b^2=8Giải ra tìm a,b rồi thay vô tìm x,y
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g) x^2+y^2+2(x+y)+2=0
h) 4x+y^2-4x-4y+6=0
i) x^2-7x+12=0
k) 1/2 × x+7/8x=11
giải hệ phương trình:
x y ( 4 x y + y + 4 ) = y 2 ( 2 y + 5 ) − 1
2 x y ( x − 2 y ) + x − 14 y = 0
4, tim x,y thuoc z
|y-42|+|12-y|=0
|x+5|+(y-3)^2=0
(x^2-16)^2+|y-4|<0
8/9 : ( 2 - 3 x y ) = 5/3
( 2 - 2/3 x y ) : 4 + 7/12 = 11/12
3 : ( 2 x y - 6/15 ) = 1 và 1/2 ( k biết ghi hỗn số nên ghi vậy cho dễ hiểu ạ )
2 - 1/5 x ( y : 7/2 + 1 ) = 1/2
2 và 3/5 x ( 5 : y ) - 3/4 = 0
7/12 : y + 4/9 x 5/8 = 0
4/15 + 2 : ( y + 2/5 ) = 1/5
\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{15}\)
3 \(\times\) y = 2 - \(\dfrac{8}{15}\)
3 \(\times\) y = \(\dfrac{22}{15}\)
y = \(\dfrac{22}{15}\) : 3
y = \(\dfrac{22}{45}\)
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|x|+|y|=0
|x-12+y|+|y-4-y|<_0
|x+5|+|y-2|=0|x-12+y
a) |x|+|y|=0 => x = 0 ; y = 0
ok nha!! 45434364565475675686875654645745745745745634564
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a: (x - 11 + y)mũ 2 + (x - 1 - y)mũ 2 = 0
B: x + (-31/12)mũ 2 + (49/12)mũ 2 = 0
C:(3x - 5)mũ 100 + (2y + 1) mũ 200 nhỏ hơn bằng 0
D: (1/2x - 5)mũ 20 + (y mũ 2 - 1/4) mũ 10 nhỏ hơn bằng 0
E: 2 mũ x - 1 nhân 3 mũ x = 12 mũ x
Câu 1: |x - y| + | y + 9/25| = 0 (*)
Câu 2: | 3x - 2 | - | 2 - 3x| = 0 (*)
Câu 3: | x - 12| = | 12 - x | +1 (*)
Câu 4: | 1/2 - 1/3 + x | = - 1/4 - | y| (*)
Câu 5: | x - 2 | < 3 (*)