\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

các bn ơi giúp mình với

a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

vậy  hệ pt có ndn \(\left\{2;0\right\}\)

b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\-4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

vậy hệ pt có ndn \(\left\{2;1\right\}\)

d.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)

đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\) ta có hệ pt:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}7b=\dfrac{1}{3}\\8a+15b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+15\times\dfrac{1}{21}=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a+\dfrac{5}{7}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\8a=\dfrac{2}{7}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{21}\\\dfrac{1}{x}=\dfrac{1}{28}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}y=21\\x=28\end{matrix}\right.\)

vậy hệ pt có ndn\(\left\{28;21\right\}\)

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

2 câu dưới hình như em hỏi rồi?

a. Đề sai, hệ pt này không giải được

b.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2-2=-x^2\\x^2y^2+xy+1=3x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2y^2-6=-3x^2\\x^2y^2+xy+1=3x^2\end{matrix}\right.\)

Cộng vế:

\(4x^2y^2+xy-5=0\Rightarrow\left[{}\begin{matrix}xy=1\\xy=-\dfrac{5}{4}\end{matrix}\right.\)

Thế vào pt \(x^2y^2+xy+1=3x^2\) sẽ tìm được x \(\Rightarrow y\)

a, Ta có : \(\left\{{}\begin{matrix}3x+2y=-2\\-x+4y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}3\left(4y-3\right)+2y=-2\\x=4y-3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}12y-9+2y=-2\\x=4y-3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}14y=7\\x=4y-3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=\frac{1}{2}\\x=\frac{4.1}{2}-3=-1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-1;\frac{1}{2}\right)\)

b, Ta có : \(\left\{{}\begin{matrix}x+2y=11\\5x-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\5\left(11-2y\right)-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\55-10y-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\-13y=-52\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2.4=3\\y=4\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)

c, Ta có : \(\left\{{}\begin{matrix}10x-9y=1\\15x+21y=36\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}30x-27y=3\\30x+42y=72\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}10x-9y=1\\-69y=-69\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}10x-9=1\\y=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(1;1\right)\)

d, Ta có : \(\left\{{}\begin{matrix}2x+y=3\\x+y=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3-2x\\x+2-2x=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3-2x\\2-x=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3-2.0=3\\x=0\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(0;3\right)\)

e, Ta có : \(\left\{{}\begin{matrix}x+y=2\\2x-3y=9\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2-y\\2\left(2-y\right)-3y=9\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2-y\\4-2y-3y=9\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2-y\\-5y=5\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2+1=3\\y=-1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-1\right)\)

f, Ta có : \(\left\{{}\begin{matrix}x-2y=11\\5x+3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11+2y\\5\left(11+2y\right)+3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11+2y\\55+10y+3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11+2y\\13y=-52\end{matrix}\right.\)​

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-4\right)\)

g, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=9-5=4\\x=3\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)

h, Ta có : \(\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x+3\left(3x+8\right)=-7\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x+9x+24=-7\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}14x=-31\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=-\frac{31}{14}\\y=3.\left(-\frac{31}{14}\right)+8=\frac{19}{14}\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-\frac{31}{14};\frac{19}{14}\right)\)

...????

5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)

Thay từng TH rồi làm nha bạn

3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)

thay nhá

Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)

PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)

+) Với y = x - 1 thay vào pt (2):

\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))

Anh quy đồng lên đê, chắc cần vài con trâu đó:))

+) Với y = 2x + 3...