x+1/3 = y+3/4 và xy=3
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Bài 1:
Ta có:
[tex]\left\{\begin{matrix} xy^{2}+x+y+\frac{1}{y}=4 & \\ y^{2}+x+\frac{1}{y}=3 & \end{matrix}\right.(y\neq 0)[/tex]
Từ phương trình suy ra:
[tex]\left\{\begin{matrix} y(xy+1)+\frac{xy+1}{y}=4 & \\ y^{2}+\frac{xy+1}{y}=3 & \end{matrix}\right.[/tex]
Đặt [tex]xy+1=a,y=b(b\neq 0)[/tex] ta có:
[tex]\left\{\begin{matrix} b^{2}+\frac{a}{b}=3 & \\ ab+\frac{a}{b}=4 & \end{matrix}\right.[/tex]
[tex]\Rightarrow \left\{\begin{matrix} 3b-b^{3}=a & \\ ab^{2}+a=4b & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3b-b^{3}=a & \\ b\left ( 2b^{2}-b^{4}-1 \right )=0 & \end{matrix}\right.[/tex]
[tex]\Leftrightarrow \left\{\begin{matrix} b=0 & \\ a=0 & \end{matrix}\right.[/tex](Loại) hoặc [tex]\left\{\begin{matrix} b=1 & \\ a=2 & \end{matrix}\right.[/tex] hoặc [tex]\left\{\begin{matrix} b=-1 & \\ a=-2 & \end{matrix}\right.[/tex]
TH1: [tex]\left\{\begin{matrix} b=1 & \\ a=2 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1 & \\ y=1 & \end{matrix}\right.[/tex]
TH2: [tex]\left\{\begin{matrix} b=-1 & \\ a=-2 & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=3 & \\ y=-1 & \end{matrix}\right.[/tex]
Vậy hệ phương trình có hai nghiệm: [tex]\left\{\begin{matrix} x=1 & \\ y=1 & \end{matrix}\right.[/tex] hoặc [tex]\left\{\begin{matrix} x=3 & \\ y=-1 & \end{matrix}\right.[/tex]
1) Ta có: \(\dfrac{1}{7}x^2y^3\cdot\left(-\dfrac{14}{3}xy^2\right)\cdot\left(-\dfrac{1}{2}xy\right)\left(x^2y^4\right)\)
\(=\left(-\dfrac{1}{7}\cdot\dfrac{14}{3}\cdot\dfrac{-1}{2}\right)\left(x^2y^3\cdot xy^2\cdot xy\cdot x^2y^4\right)\)
\(=\dfrac{1}{3}x^6y^{10}\)
2) Ta có: \(\left(3xy\right)^2\cdot\left(-\dfrac{1}{2}x^3y^2\right)\)
\(=9xy^2\cdot\dfrac{-1}{2}x^3y^2\)
\(=-\dfrac{9}{2}x^4y^4\)
3) Ta có: \(\left(-\dfrac{1}{4}x^2y\right)^2\cdot\left(\dfrac{2}{3}xy^4\right)^3\)
\(=\dfrac{1}{16}x^4y^2\cdot\dfrac{8}{27}x^3y^{12}\)
\(=\dfrac{1}{54}x^7y^{14}\)
a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
a: \(A=\dfrac{4}{9}x^4y^2\cdot\dfrac{3}{2}x^2yz=\dfrac{2}{3}x^6y^3z\)
Hệ số; biến;bậc lần lượt là 2/3; x^6y^3z;10
b: \(B=\dfrac{-2}{3}\cdot\dfrac{1}{2}\cdot\left(-1\right)\cdot xy^2\cdot xy^3\cdot x^2y^2=\dfrac{1}{3}x^4y^7\)
Hệ số;biến;bậc lần lượt là 1/3;x^4y^7;11
c: \(C=\left(-\dfrac{8}{9}x^3y^4\right)^2\cdot x^6y^3=\dfrac{64}{81}x^6y^8\cdot x^6y^3=\dfrac{64}{81}x^{12}y^{11}\)
Hệ số;biến;bậc lần lượt là 64/81; x^12y^11; 23
(x+y)^2 =a^2
x^2 +2xy +y^2 =a^2
x^2+y^2 =a^2-2xy =a^2 -2b
x^3 +y^3 = (x+y)(x^2 -xy +y^2)
=a(a^2-2b-b)
=a(a^2-3b)
=a^3- 3ab
(x^2 +y^2)^2=(a^2-2b)^2 ( cái này tính cho x^4 + y^4)
tương tự như câu đầu tiên
x^5+ y^5 (cái đó mình không biết)
Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x.y}{2.3}=\dfrac{54}{6}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=81\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm9\end{matrix}\right.\)
b) \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x^2-y^2}{5^2-3^2}=\dfrac{4}{16}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{4}\\y^2=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{5}{2}\\y=\pm\dfrac{3}{2}\end{matrix}\right.\)
c: Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)
nên \(\dfrac{x}{10}=\dfrac{y}{15}\)
Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)
nên \(\dfrac{y}{15}=\dfrac{z}{21}\)
mà \(\dfrac{x}{10}=\dfrac{y}{15}\)
nên \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{92}{46}=2\)
Do đó: x=20; y=30; z=42
a: (x-2)(y-3)=5
=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)
b: (2x-1)*(y-4)=-11
=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)
=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)
c: xy-2x+y=3
=>\(x\left(y-2\right)+y-2=1\)
=>\(\left(x+1\right)\left(y-2\right)=1\)
=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)
=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)
có sai đề ko bạn
đặt x+1/3=k
y+3/4=k
=>x=1/3k
y=3/4k
ta có:xy=3
1/3k*3/4k=3
1/4k =3
k =12
với k=12=>x=35/3
y=45/4