giải hpt \(\int^{\frac{x-y}{7}+\frac{2x+y}{17}=7}_{\frac{4x+y}{5}+\frac{y-7}{19}=15}\)
\(\hept{\begin{cases}\frac{x-y}{7}+\frac{2x+y}{17}=7\\\frac{4x+y}{5}+\frac{y-7}{19}=15\end{cases}}\)
bài 2: Tính hai cạnh góc vuông của một tâm giác vuông có độ dài cạnh huyền = 37m và diện tích = 210\(m^2\)
bài 3: giải hệ pt sau:
a. \(\left\{{}\begin{matrix}\frac{x-y}{7}+\frac{2x+y}{17}=7\\\frac{4x+y}{5}+\frac{y-7}{19}=15\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}5\left(x+2y\right)-3\left(x-y\right)=99\\x-y=7\left(x-y\right)+3y=17\end{matrix}\right.\)
c. \(\frac{x}{x-1}-\frac{2\sqrt{2}}{1-x}-\frac{6+\sqrt{2}}{x^2-1}\)= 0
d. \(\left\{{}\begin{matrix}\frac{2}{x+3}-\frac{5}{y-2}=1\\\frac{x+4}{x+3}+\frac{y}{y-2}=2\end{matrix}\right.\)
e. \(\left\{{}\begin{matrix}\frac{x-y}{7}+\frac{2x+y}{17}=7\\\frac{4+y}{5}+\frac{y-7}{19}=15\end{matrix}\right.\)
giải hpt
\(\left\{{}\begin{matrix}\frac{4}{2x+y}+\frac{1}{3x-y}=2\\4x+12=7\left(2x+y\right)\left(3x-y\right)\end{matrix}\right.\)
Giải HPT
\(\hept{\begin{cases}\frac{2x-3y}{4}-\frac{x+y-1}{5}=2x-y-1\\\frac{4x+y-2}{4}=\frac{2x-y-3}{6}-\frac{x-y-1}{3}\end{cases}}\)
a)\(\int^{\frac{x}{y}-\frac{x}{y+12}=1}_{\frac{x}{y-12}-\frac{x}{y}=2}\)
b)\(\int^{4\left(x+y\right)=5\left(x-y\right)}_{\frac{40}{x+y}+\frac{40}{x-y}=9}\)
Giải các hpt
trừ 2 về đi bạn , cả 2 câu đều k khó đâu
a)x=144 , y=36
b)x=9 , y=1
cần lời giải thì nói mình
1.Giải hpt
\(\left\{{}\begin{matrix}\frac{7}{\sqrt{x+7}}-\frac{4}{\sqrt{y-6}}=\frac{-1}{4}\\\frac{5}{\sqrt{x+7}}+\frac{3}{\sqrt{y-6}}=\frac{11}{4}\end{matrix}\right.\)
giải hpt
a) \(\left\{{}\begin{matrix}\frac{5}{x-1}+\frac{1}{y-1}=10\\\frac{1}{x-1}-\frac{3}{y-1}=18\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{7}{\sqrt{x-7}}-\frac{4}{\sqrt{y+6}}=\frac{5}{2}\\\frac{5}{\sqrt{x-7}}+\frac{3}{\sqrt{y+6}}=\frac{13}{6}\end{matrix}\right.\)
a) Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=a\\\frac{1}{y-1}=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}5a+b=10\\a-3b=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15a+3b=30\\a-3b=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-3b=18\\16a=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=3\\\frac{1}{y-1}=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=\frac{4}{5}\end{matrix}\right.\)
Vậy...
b) Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=a\\\frac{1}{\sqrt{y+6}}=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\frac{5}{2}\\5a+3b=\frac{13}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}31a-12b=\frac{15}{2}\\20a+12b=\frac{26}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\frac{5}{2}\\51a=\frac{97}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{97}{306}\\b=\frac{-43}{612}\end{matrix}\right.\)( loại vì \(a,b>0\) )
Vậy hệ vô nghiệm
Is that true .-.
Cho xin solve lại câu b)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}21a-12b=\frac{15}{2}\\20a+12b=\frac{26}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5a+3b=\frac{13}{6}\\41a=\frac{97}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{97}{246}\\b=\frac{8}{123}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=\frac{97}{246}\\\frac{1}{\sqrt{y+6}}=\frac{8}{123}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{126379}{9409}\\y=\frac{14745}{64}\end{matrix}\right.\)
Vậy...
giải hpt \(\int^{\frac{x}{y}-\frac{x}{y+12}=1}_{\frac{x}{y-12}-\frac{x}{y}=2}\)
\(\int^{x\left(\frac{1}{y}-\frac{1}{y+12}\right)=1}_{x\left(\frac{1}{y-12}-\frac{1}{y}\right)=2}\Leftrightarrow\int^{\frac{1}{y}-\frac{1}{y+12}=\frac{1}{x}}_{\frac{1}{y-12}-\frac{1}{y}=\frac{2}{x}}\Leftrightarrow\int^{\frac{2}{y}-\frac{2}{y+12}=\frac{2}{x}\left(1\right)}_{\frac{1}{y-12}-\frac{1}{y}=\frac{2}{x}\left(2\right)}\)
Lấy vế trừ vế của pt (1) và (2) ta có
\(\frac{2}{y}-\frac{2}{y+12}-\frac{1}{y-12}+\frac{1}{y}=0\)
\(\Leftrightarrow\frac{3}{y}-\frac{2}{y+12}-\frac{1}{y-12}=0\Leftrightarrow3\left(y+12\right)\left(y-12\right)-2y\left(y-12\right)-y\left(y+12\right)=0\)
Rút gọn giải pt bậc 2 sau thay trở lại tìm x
Giải hpt: \(\left\{{}\begin{matrix}x+\frac{1}{y}=\frac{7}{2}\\y+\frac{1}{x}=\frac{7}{3}\end{matrix}\right.\)
ĐKXĐ: ....
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{xy+1}{y}=\frac{7}{2}\\\frac{xy+1}{x}=\frac{7}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}xy+1=\frac{7}{2}y\\xy+1=\frac{7}{3}x\end{matrix}\right.\)
\(\Rightarrow\frac{7}{2}y=\frac{7}{3}x\Rightarrow y=\frac{2}{3}x\)
Thay vào pt đầu:
\(x+\frac{3}{2x}=\frac{7}{2}\Leftrightarrow2x^2-7x+3=0\) (casio)