1, \(\dfrac{x}{3}=\dfrac{y}{2}\); \(\dfrac{x}{5}=\dfrac{x}{7}\) và x +2y = -112
2, 2x = 3y; 5y = 7z và 3x - 7y + 5z = 30
3, \(\dfrac{x}{y}=\dfrac{10}{9};\) \(\dfrac{y}{z}=\dfrac{3}{4}\) và x +2y - 3z = -48
Làm giúp, cần gấp
giải các hệ phương trình
a \(\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\)
\(\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\)
b \(\dfrac{5}{x+y-3}-\dfrac{2}{x-y+1}=8\)
\(\dfrac{3}{x+y-3}+\dfrac{1}{x-y+1}=\dfrac{3}{2}\)
c \(\sqrt{x-1}-3\sqrt{y+2}=2\)
\(2\sqrt{x-1}+5\sqrt{y+2}=15\)
d \(\dfrac{7}{\sqrt{x-7}}-\dfrac{4}{\sqrt{y+6}}=\dfrac{5}{3}\)
\(\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\)
e \(7x^2+13y=-39\)
\(5x^2-11y=33\)
f \(2\left(x-1\right)^2-3y^3=7\)
\(5\left(x-1\right)^2+6y^3=4\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
Tìm x, y, z
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)
+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)
+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)
\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)
Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!
Rút gọn:
a) A= \(\dfrac{x+y}{x-y}-\dfrac{x}{x+y}+\dfrac{2y^2}{x^2-y^2}\)
b) B= \(\dfrac{x}{x-2}-\dfrac{10}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-1}{x+3}\)
c) C= \(\dfrac{1}{x-1}-\dfrac{x-1}{x^2+x+1}-\dfrac{3}{x^3-1}\)
a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{3y^2+3xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{3y}{x-y}\)
Thực hiện các phép tính sau:
a,(\(\dfrac{x}{x+1}\)+\(\dfrac{x-1}{x}\)):(\(\dfrac{x}{x+1}\)-\(\dfrac{x-1}{x}\))
b,(1+\(\dfrac{x}{y}\)+\(\dfrac{x^2}{y^2}\)).(1-\(\dfrac{x}{y}\)).\(\dfrac{y^2}{x^3-y^3}\)
\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)
\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)
\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)
Rút gọn các biểu thức:
a) {\(\dfrac{1}{x^2}\) + \(\dfrac{1}{y^2}\) + \(\dfrac{2}{x+y}\)(\(\dfrac{1}{x}\) + \(\dfrac{1}{y}\))} : \(\dfrac{x^3+y^3}{x^2y^2}\)
b) {\(\dfrac{1}{\left(2x-y\right)^2}\) + \(\dfrac{2}{4x^2-y^2}\) + \(\dfrac{1}{\left(2x+y\right)^2}\)} . \(\dfrac{4x^2+4xy+y^2}{16x}\)
c) (\(\dfrac{x^2-xy}{x^2y+y^3}\) - \(\dfrac{2x^2}{y^3-xy^2+x^2y-x^3}\))(1 - \(\dfrac{y-1}{x}\) - \(\dfrac{y}{x^2}\))
Tìm y
a) 3 \(\dfrac{1}{5}\) : 2\(\dfrac{1}{3}\): y = \(\dfrac{12}{7}\)
b) 3 : y x 3 \(\dfrac{1}{2}\)= \(\dfrac{2}{3}x\dfrac{3}{4}\)
c) \(3\dfrac{2}{3}-y+1\dfrac{3}{4}=2\)
mình ko chép đề bài nha
a) \(\dfrac{16}{5}\): \(\dfrac{7}{3}\) : y =\(\dfrac{12}{7}\)
\(\dfrac{48}{35}\): y = \(\dfrac{12}{7}\)
y = \(\dfrac{48}{35}\): \(\dfrac{12}{7}\)
y = \(\dfrac{4}{5}\)
b) 3 : y x \(\dfrac{7}{2}\)= \(\dfrac{1}{2}\)
3 : y = \(\dfrac{1}{2}:\dfrac{7}{2}\)
3 : y = \(\dfrac{1}{7}\)
y = 3 : \(\dfrac{1}{7}\)
y = 21
Rút gọn các biểu thức:
a) (x - \(\dfrac{1}{1-x}\)) : \(\dfrac{x^2-x+1}{x^2-2x+1}\)
b) (1 + \(\dfrac{x}{y}\) + \(\dfrac{x^2}{y^2}\))(1 - \(\dfrac{x}{y}\))\(\dfrac{y^2}{x^3-y^3}\)
a) Ta có: \(\left(x-\dfrac{1}{1-x}\right):\dfrac{x^2-x+1}{x^2-2x+1}\)
\(=\left(x+\dfrac{1}{x-1}\right):\dfrac{x^2-x+1}{\left(x-1\right)^2}\)
\(=\dfrac{x^2-x+1}{x-1}\cdot\dfrac{\left(x-1\right)^2}{x^2-x+1}\)
\(=x-1\)
b) Ta có: \(\left(1+\dfrac{x}{y}+\dfrac{x^2}{y^2}\right)\left(1-\dfrac{x}{y}\right)\cdot\dfrac{y^2}{x^3-y^3}\)
\(=\left(\dfrac{y^2}{y^2}+\dfrac{xy}{y^2}+\dfrac{x^2}{y^2}\right)\cdot\left(\dfrac{y-x}{y}\right)\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{-\left(x-y\right)}{y}\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{-1}{y}\)
Giaỉ hệ phương trình sau bằng phương pháp thế
a)\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2};\dfrac{3}{x}-\dfrac{4}{y}=-1\)
b)\(\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1;\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\)
c)\(\dfrac{5x}{x+1}+\dfrac{y}{y-3}=27;\dfrac{2x}{x+1}-\dfrac{3y}{y-3}=4\)
d)\(\dfrac{7}{x+2}+\dfrac{3}{y}=2;\dfrac{4}{x+2}-\dfrac{1}{y}=\dfrac{5}{2}\)
e)\(\dfrac{2x}{x+4}+\dfrac{2y}{2y-3}=27;\dfrac{2x}{x+4}-\dfrac{6y}{2y-3}=4\)
Bạn nào biết thì giải giúp mình với ạ,mình xin cảm ơn ạ!!!
giải hệ pt
a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{7x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}+\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)
Rút gọn biểu thức:
A = \([(32)^{\dfrac{2}{3}}]^{\dfrac{-2}{5}}\)
B= \(\dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\)
C = \((a^{\dfrac{1}{3}}-b^{\dfrac{2}{3}})(a^{\dfrac{2}{3}}+a^{\dfrac{1}{3}}×b^{\dfrac{4}{3}}+b^{\dfrac{4}{9}})\)
D = \((x+y^\dfrac{3}{2}÷\sqrt{x})^\dfrac{2}{3}÷[\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}+\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}]^\dfrac{2}{3}\)
E = \([\dfrac{1}{x^{\dfrac{1}{2}}-4x^{\dfrac{-1}{2}}}-\dfrac{2\sqrt[3]{x}}{x\sqrt[3]{x}-4\sqrt[3]{x}}]^{-2}-\sqrt{x^2+8x+16}\)