Tìm x, y biết:
a) \(\left|-x+2\right|=-\left|y+9\right|\)
b) \(\left|3x+4\right|+\left|2y-10\right|\le0\)
c) \(\left|-x-3\right|+\left|y+7\right|< 0\)
a,\(\left|3x-4\right|+\left|3y+5\right|=0\) b,\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\) c,\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
d,\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\) e,\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)
Giups mình giải bài tìm x,y,z này nhé!!! Cảm ơn nhiều ạ!!!
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
Tìm các cặp số x,y
\(\left(x-3\right)^2+\left(2y-1\right)^2=0\)
\(\left(4x-3\right)^4+\left(y+2\right)^2\le0\)
(\(x-3\))2 + (2y - 1)2 = 0
(\(x\) - 3)2 ≥ 0 ∀ \(x\)
(2y - 1)2 ≥ 0 ∀ y
⇔ (\(x\) - 3)2 + (2y - 1)2= 0
⇔ \(\left\{{}\begin{matrix}x-3=0\\3y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{3}\end{matrix}\right.\)
(4\(x-3\))4 + (y + 2)2 ≤ 0
(4\(x\) - 3)4 ≥ 0 ∀ \(x\)
(y + 2)2 ≥ 0 ∀ y
⇔(4\(x\) - 3)4 + (y+2)2 ≥ 0
⇔ (4\(x\) - 3)4 + (y + 2)2 ≤ 0 ⇔
⇔\(\left\{{}\begin{matrix}4x-3=0\\y+2=0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-2\end{matrix}\right.\)
B4:Giải hệ pt:
a)\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;0\right\}\)
b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\-4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;1\right\}\)
d.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\) ta có hệ pt:
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}7b=\dfrac{1}{3}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+15\times\dfrac{1}{21}=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+\dfrac{5}{7}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a=\dfrac{2}{7}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{21}\\\dfrac{1}{x}=\dfrac{1}{28}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=21\\x=28\end{matrix}\right.\)
vậy hệ pt có ndn\(\left\{28;21\right\}\)
1,\(\left\{{}\begin{matrix}x^2+xy-3x+y=0\\x^4+3x^2y-5x^2+y^2=0\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\left(2x-1\right)^2+4\left(y-1\right)^2=22\\xy\left(x-1\right)\left(y-2\right)=1\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x+y+1\right)=25\left(y+1\right)\\x^2+xy+2y^2+x-8y=9\end{matrix}\right.\)
4,\(\left\{{}\begin{matrix}5x^2y-4xy^2+3y^2-2\left(x+y\right)=0\\xy\left(x^2+y^2\right)+2=\left(x+y\right)^2\end{matrix}\right.\)
Tìm x,y thỏa mãn:
a)\(^{\left|x+2y\right|+\left|4y-3\right|\le0}\)
b)\(\left|x-y-5\right|+2017\left(y-11\right)^{2018}\le0\)
c)\(^{\left(x+y\right)^{2020}+2018.\left|y-1\right|=0}\)
a)\(\left\{{}\begin{matrix}2\left|x-6\right|+3\left|y-1\right|=5\\5\left|x-6\right|-4\left|y+1\right|=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2\left|x+y\right|-\left|x-y\right|=9\\3\left|x+y\right|+2\left|x-y\right|+17\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}4\left|x+y\right|+3\left|x-y\right|=8\\3\left|x+y\right|-5\left|x-y\right|=6\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}x^2-xy=24\\2x-3y=1\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}3x-4y+1=0\\xy=3\left(x+y\right)-9\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}2x+3y=5\\3x^2-y^2+2y=4\end{matrix}\right.\)
a: Đặt |x-6|=a, |y+1|=b
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a+3b=5\\5a-4b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
=>|x-6|=1 và |y+1|=1
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
b: Đặt |x+y|=a, |x-y|=b
Theo đề, ta có: \(\left\{{}\begin{matrix}2a-b=19\\3a+2b=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{55}{7}\\b=-\dfrac{23}{7}\left(loại\right)\end{matrix}\right.\)
=>HPTVN
c: Đặt |x+y|=a, |x-y|=b
Theo đề ta có: \(\left\{{}\begin{matrix}4a+3b=8\\3a-5b=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)
=>|x+y|=2 và x=y
=>|2x|=2 và x=y
=>x=y=1 hoặc x=y=-1
Giải các hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
giải hệ pt :
a,\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3-30=0\\x^2y+x\left(1+y+y^2\right)+y-11=0\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}xy^2-2y+3x^2=0\\y^2+x^2y+2x=0\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}3xy+2y=5\\2xy\left(x+y\right)+y^2=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)
TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)
Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)
TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)
2 câu dưới hình như em hỏi rồi?
Tìm x,y,z
\(\left|x-3\right|+\left|y-2x\right|+\left|2z-x+y\right|=0\)
\(\left|x-y\right|+\left|2y+x-\frac{1}{2}\right|+\left|x+y+z\right|\le0\)
a) \(\Leftrightarrow\left|x-3\right|=0;\left|y-2x\right|=0;\left|2z-x+y\right|=0\)
\(\Leftrightarrow x=3;y=2x;2z=-y+x\)
Ta có : y = 2x => y = 2 . 3 = 6
và 2z = -y + x => 2z = -6 + 3 = -3 => z = \(-\frac{3}{2}\)
b) \(\Leftrightarrow\left|x-y\right|+\left|2y+x-\frac{1}{2}\right|+\left|x+y+z\right|=0\) (vĩ mỗi số hạng trong tổng đều lớn hơn hoặc bằng 0)
\(\Leftrightarrow\left|x-y\right|=0;\left|2y+x-\frac{1}{2}\right|=0;\left|x+y+z\right|=0\)
\(\Leftrightarrow x=y;2y+x=\frac{1}{2};x+y=-z\)
Vì x = y nên \(2y+x=3y=\frac{1}{2}\Rightarrow x=y=\frac{1}{2}:3=\frac{1}{6}\)
và \(-z=x+y=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\Rightarrow z=-\frac{1}{3}\)