Giải hẹ phương trình:
\(\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\)
\(\dfrac{2x+6}{x-1}+\dfrac{3y+14}{y+3}=18\)
Giải các hệ phương trình sau
f.{ (2x - y) (x + 3y) = 4
{ (5x + y) (x + 3y) = 24
g.{ \(\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\)
{ \(\dfrac{9x+4y-13}{5}+\dfrac{3\left(x-2\right)}{4}=15\)
h.{\(\dfrac{1}{x}+\dfrac{1}{y}=2\)
{\(\dfrac{3}{x}-\dfrac{4}{y}=-1\)
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
Giải hệ phương trình sau:
1/ \(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=3\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=3\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)\(\left(Đk:x,y\ne-1\right)\)
\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=3\\\dfrac{2x}{x+1}+\dfrac{6y}{y+1}=-2\end{matrix}\right.\)
\(\Rightarrow\dfrac{5y}{y+1}=-5\)
\(\Leftrightarrow5y=-5y-5\)
\(\Leftrightarrow10y=-5\)
\(\Leftrightarrow y=-\dfrac{1}{2}\Rightarrow x=-2\)
<=>\(\dfrac{2x}{x+1}-\dfrac{x}{x+1}=4< =>x-4x=4< =>x=-\dfrac{4}{3}\Rightarrow y=-\dfrac{1}{4}\)
Giải các phương trình sau:
\(a.\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(b.\dfrac{7}{x+2}=\dfrac{3}{x-5}\)
\(c.\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
\(d.\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
Tìm x và y
\(\dfrac{x}{3}=\dfrac{y}{4}\)và\(x+y=14\)
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
\(\dfrac{x}{3}=\dfrac{y}{4}\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{14}{7}\)=2
* \(\dfrac{x}{3}=2=>x=6\)
*\(\dfrac{y}{4}=2=>y=8\)
Vậy( x, y) ∈{ 6, 8}
Kiểm tra lại nhaa
áp dụng tính chất dãy tỉ số bằng nhau
Ta có:x/3=y/4=x+y/3+4=14/7=2
Vậy x=2.3=6
y=2.4=8
1.\(\dfrac{x}{2}=\dfrac{y}{5}\)và x+y=21 2.7x=3y và x-y=16
3.\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}\)và x-y-z=36 4.x:y:z=3:5:7 và 2x+3y-z=-14
\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)
\(x=\dfrac{-45}{2}\)
\(y=-27\)
\(z=\dfrac{-63}{2}\)
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) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{6}{x}=30\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(\dfrac{7}{y+4}=\dfrac{14}{18}\) \(\dfrac{5xy-3}{54}=\dfrac{14}{18}\) \(\dfrac{6}{32:y}=\dfrac{12}{16}\)
TRÌNH BÀY CẢ LỜI GIẢI MIK ĐANG RẤT GẤP
giải hệ phương trình bằng pp sd bđt:
\(\left\{{}\begin{matrix}x+y^2+z^3=14\\\left(\dfrac{1}{2x}+\dfrac{1}{3y}+\dfrac{1}{6z}\right)\left(\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{6}\right)=1\end{matrix}\right.\)
giải các hệ phương trình
\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}=\dfrac{y+7}{3}-4\end{matrix}\right.\)
b2.
\(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
B3. Tìm ĐKXĐ
\(\dfrac{1}{x\sqrt{x}+1}-\dfrac{2}{\sqrt{x}+1}\)
b4. so sánh A với 1
A=\(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
b5.tính
a,\(\sin47+2\sin38-\cos43-\cos52\)
b, \(C=\dfrac{2\sin^2x-1}{\sin x-\cos x}\)
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)