GIẢI HPT
\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{6}\end{matrix}\right.\)
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: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
a.\(\sqrt{28a^4}\)
b. A=\(\left(\dfrac{\sqrt{21}-\sqrt{7}}{\sqrt{3-1}}+\dfrac{\sqrt{10}-\sqrt{5}}{\sqrt{2}-1}\right)\)\(\div\)\(\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
c.\(\left\{{}\begin{matrix}\dfrac{3}{2x}-y=6\\\dfrac{1}{x}+2y=-4\end{matrix}\right.\)
`a)sqrt{28a^4}`
`=sqrt{7.4.a^4}`
`=2sqrt7a^2`
`b)A=((sqrt{21}-sqrt7)/(sqrt3-1)+(sqrt{10}-sqrt5)/(sqrt2-1)):1/(sqrt7-sqrt5)`
`=((sqrt7(sqrt3-1))/(sqrt3-1)+(sqrt5(sqrt2-1))/(sqrt2-1)).(sqrt7-sqrt5)`
`=(sqrt7+sqrt5)(sqrt7-sqrt5)`
`=7-5=2`
`c)` $\begin{cases}\dfrac{3}{2x}-y=6\\\dfrac{1}{x}+2y=-4\end{cases}$
`<=>` $\begin{cases}\dfrac{3}{x}-2y=12\\\dfrac{1}{x}+2y=-4\end{cases}$
`<=>` $\begin{cases}\dfrac{4}{x}=8\\2y+\dfrac{1}{x}=-4\end{cases}$
`<=>` $\begin{cases}x=\dfrac12\\2y=-4-2=-6\end{cases}$
`<=>` $\begin{cases}x=\dfrac12\\y=-3\end{cases}$
Vậy HPT có nghiệm `(x,y)=(1/2,-3)`.
giải hpt:
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+x+1=3y\\y^2+y+1=3x\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
ĐK: \(x>\dfrac{1}{4};y>-\dfrac{1}{4}\), đặt \(a=\dfrac{1}{\sqrt{12x-3}};b=\dfrac{1}{\sqrt{4y+1}}\)với a,b>0
khi đó, ta có hệ phương mới \(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}80a+40b=8\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}45a=3\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35.\dfrac{1}{15}+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\b=\dfrac{1}{15}\end{matrix}\right.\)
thay \(\dfrac{1}{\sqrt{12x-3}}=a\) hay \(\dfrac{1}{\sqrt{12x-3}}=\dfrac{1}{15}\Rightarrow\sqrt{12x-3}=15\Leftrightarrow12x-3=225\Leftrightarrow12x=228\Leftrightarrow x=19\left(TMĐK\right)\) thay \(\dfrac{1}{\sqrt{4y+1}}=b\) hay
\(\dfrac{1}{\sqrt{4y+1}}=\dfrac{1}{15}\Rightarrow\sqrt{4y+1}=15\Leftrightarrow4y+1=225\Leftrightarrow4y=224\Leftrightarrow y=56\left(TMĐK\right)\)
Vậy (x;y)=(9;56) là nghiệm duy nhất của hệ phương trình đã cho.
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
ĐK: x,y#0, khi đó \(\dfrac{1}{x}+\dfrac{1}{y}=4\Rightarrow x+y=4xy\)
Do đó \(x\left(1+4y\right)+y=2\Leftrightarrow x+4xy+y=2\Leftrightarrow x+x+y+y=2\Leftrightarrow2\left(x+y\right)=2\Leftrightarrow x+y=1\)
Mà \(4xy=x+y\Leftrightarrow4xy=1\Leftrightarrow xy=\dfrac{1}{4}\)
Vậy \(x+y=1;xy=\dfrac{1}{4}\)
Do đó x,y là nghiệm của phương trình:
\(t^2-t+\dfrac{1}{4}=0\)
\(\Delta=b^2-4ac=1-4.1.\dfrac{1}{4}=0\)
Phương trình có nghiêm kép \(x_1=x_2=-\dfrac{b}{2a}=-\dfrac{-1}{2}=\dfrac{1}{2}\)
\(\Rightarrow x=y=\dfrac{1}{2}\left(nhận\right)\)
Vậy (x;y)=\(\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) là nghiệm duy nhất của hệ phương trình đã cho.
c)\(\left\{{}\begin{matrix}x^2+x+1=3y\\y^2+y+1=3x\end{matrix}\right.\)
Trừ vế đối vế hai phương trình, ta được:
\(x^2+x+1-y^2-y-1=3y-3x\\ \Leftrightarrow x^2-y^2+4x+4y=0\\ \Leftrightarrow\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\\ \Leftrightarrow\left(x-y\right)\left(x+y+4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x+y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-x-4\end{matrix}\right.\)
+Với x=y thế vào \(x^2+x+1=3y\) ta được
\(x^2+x+1=3x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Do đó (x;y)=(1;1) là một nghiệm của hệ phương trình đã cho.
+Với y=-x-4 thế vào \(x^2+x+1=3y\) ta được
\(x^2+x+1=3\left(-x-4\right)\Leftrightarrow x^2+4x+13=0\Leftrightarrow\left(x+2\right)^2+9=0\)(*)
Mặt khác \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+9\ge0\Rightarrow\left(x+2\right)^2\ge-9>0\), do đó phương trình (*) vô nghiệm
Vậy (x;y)=(1;1) là nghiệm duy nhất của hệ phương trình đã cho.
1. Giải các hpt sau:
a, \(\left\{{}\begin{matrix}x-y=4\\3x+4y=19\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}x-\sqrt{3y}=\sqrt{3}\\\sqrt{3x}+y=7\end{matrix}\right.\)
2. Giải các hpt sau:
a, \(\left\{{}\begin{matrix}2-\left(x-y\right)-3\left(x+y\right)=5\\3\left(x-y\right)+5\left(x+y\right)=-2\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x+y=24\\\dfrac{x}{9}+\dfrac{y}{27}=2\dfrac{8}{9}\end{matrix}\right.\) d, \(\left\{{}\begin{matrix}\sqrt{x-1}-3\sqrt{y+2}=2\\2\sqrt{x-1}+5\sqrt{y+2=15}\end{matrix}\right.\)
3. Cho hpt \(\left\{{}\begin{matrix}\left(m+1\right)x-y=3\\mx+y=m\end{matrix}\right.\)
a, Giải hpt khi m=\(\sqrt{2}\)
b, tìm giá trị của m để hpt có nghiệm duy nhất thỏa mãn: x+y>0
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
giải các hpt sau: a)\(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}y=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{3x}{4}+\dfrac{2y}{5}=2,3\\x-\dfrac{3y}{5}=0,8\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)cíu zới
a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x+\sqrt{2}\left(4\sqrt{5}-3\sqrt{2}\right)=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x=-1-4\sqrt{10}+6=5-4\sqrt{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{5}y=2,3\\x-\dfrac{3}{5}y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
c: ĐKXĐ: y>2
\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x-1\in\left\{2;-2\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)
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: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}4\left(2x-y+3\right)-3\left(x-2y+3\right)=48\\3\left(3x-4y+3\right)+4\left(4x-2y-9\right)=48\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-2\left(2x+1\right)+1,5=3\left(y-2\right)-6x\\11,5-4\left(3-x\right)=2y-\left(5-x\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\\\dfrac{9x+4y-13}{5}-\dfrac{3\left(x-2\right)}{4}=15\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\sqrt{3}x-\sqrt{5}y=2\sqrt{6}-\sqrt{15}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
Giải các hệ phương trình sau bằng cách đặt ẩn số phụ:
1) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{y+2x}=3\\\dfrac{4}{x+2y}-\dfrac{3}{y+2x}=1\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}x^2+y^2=13\\3x^2-2y^2=-6\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}|x|+4|y|=18\\3|x|+|y|=10\end{matrix}\right.\)
GIẢI GIÚP MÌNH VỚI M.N
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
ba cái đồ êu!!
câu số 6 (con số của quỷ sa tăng :v)
đặt \(\left\{{}\begin{matrix}a=\left|x\right|\\b=\left|y\right|\end{matrix}\right.\) (a,b >/ 0)
hpt trở thành : \(\left\{{}\begin{matrix}a+4b=18\\3a+b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x\right|=2\\\left|y\right|=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=4\\y=-4\end{matrix}\right.\end{matrix}\right.\)
Vậy hpt có các ng (x;y) là: (có 4 nghiệm tự kết luận)
1, \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\) (I) (ĐKXĐ: x, y \(\ne\)0)
Đặt \(\dfrac{1}{x}=a\) ; \(\dfrac{1}{y}=b\)
Hệ pt (I) trở thành :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\) \(\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}-7b=\dfrac{-1}{3}\\a+b=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a+\dfrac{1}{21}=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\left(tm\right)\\a=\dfrac{1}{28}\left(tm\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{28}\\\dfrac{1}{y}=\dfrac{1}{21}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}+\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=4\end{matrix}\right.\)
a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)