Tính giá trị của biểu thức
a) \(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-\left|x\right|\right)+y\) với x = 3 và y = -2
b) \(B=\left|2x-1\right|+\left|3y+2\right|\) với x = 3 và y = -3
Tính giá trị của biểu thức
a) \(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-\left|x\right|\right)+y\) với x = 3 và y = -2
b) \(B=\left|2x-1\right|+\left|3y+2\right|\) với x = 3 và y = -3
a, Với x = 3 và y = -2 ta có:
\(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-\left|3\right|\right)+\left(-2\right)\)
\(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-3\right)-2\)
\(A=\dfrac{3}{2}+\dfrac{4}{9}.3-2\)
\(A=\dfrac{3}{2}+\dfrac{4}{3}-2\)
\(A=\dfrac{5}{6}\)
Với x = 3 và y = -3 ta có:
\(B=\left|2.3-1\right|+\left|3.\left(-3\right)+2\right|\)
\(B=\left|5\right|+\left|-7\right|\)
\(B=5+7=12\)
Hoctot ! ko hiểu chỗ nào cứ hỏi cj nhé
\(\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}+\dfrac{4}{\left|y\right|-2}=\dfrac{11}{6}\\\dfrac{5}{\left|x+3\right|}+\dfrac{2}{\left|y\right|-2}=\dfrac{11}{6}\end{matrix}\right.\)
\(Đặt:\left\{{}\begin{matrix}a=\dfrac{1}{\left|x+3\right|}\left(ĐK:x\ne-3\right)\\b=\dfrac{1}{\left|y\right|-2}\left(ĐK:y\ne\pm2\right)\end{matrix}\right.\\ Có:\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}+\dfrac{4}{\left|y\right|-2}=\dfrac{11}{6}\\\dfrac{5}{\left|x+3\right|}+\dfrac{2}{\left|y\right|-2}=\dfrac{11}{6}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\5a+2b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\10a+4b=\dfrac{22}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-9a=-\dfrac{11}{6}\\a+4b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{11}{54}\\b=\dfrac{\dfrac{11}{6}-\dfrac{11}{54}}{4}=\dfrac{11}{27}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}=a=\dfrac{11}{54}\\\dfrac{1}{\left|y\right|-2}=b=\dfrac{11}{27}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\left|x+3\right|=54\\11\left(\left|y\right|-2\right)=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+3\right|=\dfrac{54}{11}\\\left|y\right|=\dfrac{27}{11}+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+3=\dfrac{54}{11}\\x+3=\dfrac{-54}{11}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{27}{11}+2\\y=-\left(\dfrac{27}{11}+2\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{21}{11}\left(TM\right)\\x=\dfrac{-87}{11}\left(TM\right)\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{49}{11}\left(TM\right)\\y=-\dfrac{49}{11}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left\{\left(\dfrac{21}{11};\dfrac{49}{11}\right);\left(\dfrac{-87}{11};\dfrac{49}{11}\right);\left(\dfrac{21}{11};\dfrac{-49}{11}\right);\left(\dfrac{-87}{11};\dfrac{-49}{11}\right)\right\}\)
Giải các hệ phương trình sau:
1, \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\dfrac{1}{x-5}+\dfrac{6}{\sqrt{y}-2}=2\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x+7}}-\dfrac{4}{\sqrt{y-6}}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x+7}}+\dfrac{3}{\sqrt{y-6}}=\dfrac{13}{6}\end{matrix}\right.\)
Tìm điều kiện giúp mình nhé!
1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\)
3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}\left(xy-2\right)^2+6y=3\left(\dfrac{1}{x}-\dfrac{3}{x^2}\right)\\y^3-4y^2+\dfrac{6}{x}+\left(y-1\right)\sqrt{\left(3y-2\right)}=\dfrac{9}{x^2}\end{matrix}\right.\)
Đạo hàm của hàm số \(y=\left(x^2-\dfrac{2}{x}\right)^3\)là:
A. \(y'=6\left(x+\dfrac{1}{x^2}\right)\left(x^2-\dfrac{2}{x}\right)^2\)
B. \(y'=3\left(x^2-\dfrac{2}{x}\right)^2\)
C. \(y'=6\left(x-\dfrac{1}{x^2}\right)\left(x^2-\dfrac{2}{x}\right)^2\)
D. \(y'=6\left(x-\dfrac{1}{x}\right)\left(x^2-\dfrac{2}{x}\right)^2\)
\(y'=3\left(x^2-\dfrac{2}{x}\right)^2.\left(x^2-\dfrac{2}{x}\right)'=3\left(x^2-\dfrac{2}{x}\right)^2\left(2x+\dfrac{2}{x^2}\right)\)
\(=6\left(x+\dfrac{1}{x^2}\right)\left(x^2-\dfrac{2}{x}\right)^2\)
Rút gọn phân thức:
1, \(\dfrac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
2, \(\dfrac{x^4-y^4}{x^3+y^3}\)
3, \(\dfrac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\)
4, \(\dfrac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3}\)
5, \(\dfrac{x^3-7x+6}{x^2\left(x-3\right)^2+4x\left(3-x\right)^2+4\left(x-3\right)^2}\)
1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)
\(=\dfrac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}=\dfrac{x+y-1}{x-y+1}\)
2: \(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x^2-xy+y^2}\)
3: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
thực hiện phép tính
a)\(\dfrac{3}{5}\)-\(\dfrac{1}{2}\)\(\sqrt{1\dfrac{11}{25}}\)
b)(5+2\(\sqrt{6}\))(5-2\(\sqrt{6}\))
c)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
d)\(\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)(với x,y>0)
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
10. giải hpt bằng phương pháp thế:
6) \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}3x-2=y\\2x+3y=6\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}2x+3y=2\\4x-y-1=0\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-2y=3\\2x-\dfrac{4}{3}y=1\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}5x+y=3\\2x+0,4y=1,2\end{matrix}\right.\)
giúp mk vs ạ mai mk học rồi
6. \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\\dfrac{2+6y}{4}-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-2\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(1-\dfrac{y}{2}\right).3\\6\left(1-\dfrac{y}{2}\right)+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(1-\dfrac{y}{2}\right)\\y=\left(VNghiệm\right)\end{matrix}\right.\Leftrightarrow\) không tồn tại x, y
(Các câu khác tương tự nhé.)
câu 3: giải hệ phương trình
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{5x}{6}-y=\dfrac{-5}{6}\\\dfrac{2x}{2x+y}+3y=\dfrac{-2}{3}\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x\sqrt{3}+3y=1\\2x-y\sqrt{3}=\sqrt{3}\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.=17\)
giúp mk vs ạ mk cần gấp
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5a+b=5\\15a=24\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{8}{5}\\b=-3\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{6}{x}=30\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)