1!+2!+3!...+x!=y^2
Giải các hệ phương trình sau:
c.{ 2(x - 2) + 3(1 + y) = 2
{ 3(x - 2) - 2(1 + y) = -3
d.{ (x - 5)(y - 2) = (x + 2)(y - 1)
{ (x - 4)(y + 7) = (x - 3)(y + 4)
e.{ 1/x - 1/y = 1
{ 3/x + 4/y = 5
e: \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{3}{y}=3\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7}{y}=-2\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\\dfrac{1}{x}=1+\dfrac{2}{7}=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\x=\dfrac{7}{9}\end{matrix}\right.\)
Rút gọn
1 (x-2)^2 + (x+3)^2-2.(x+1).(x-1)
2 ( x-y).(x+y).(x^2 +y^2).(x^4 + y^4)
3 3.(x-y)^2-2(x+y)^2+(x+y).(x-y)
4 (x-1)^2 -2(x-1)(x-3)+(x-3)^2
1. x/y-2=3/2 và x-y=4
2. x-4/y+2=1/2 và x+y=5
3. 3/x-2=2/y+2 và x+y=5
4.3/x-2=2/y+2 và x+y=1
5.x+2/y+3=5/6 và x-y=1
6. x-1/y+4=3/4 và 2x=3y
7. x-1/y+4=3/4 và 2x=3y+2
vd câu 1:
ta có x-y=4 =>x=4+y
ta có pt:
4+y/y-2=3/2
=>8+2y=3y-6
=>-y=-14
=>y=14
=>x=4+y=4+14=18
các bài khác cũng tương tự thôi bạn
dấu chéo có nghĩa là phân số hí
\begin{cases}
x+\sqrt{x(x^2-3x+3)}=\sqrt[3]{y+2}+\sqrt{y+3}+1 & \\
3\sqrt{x-1}-\sqrt{x^2-6x+6}=\sqrt[3]{y+2}+1
\end{cases}
\begin{cases}
y^2+x^3-x^2+2\sqrt[3]{y^4}+\sqrt[3]{y^2}=2x\sqrt{x-1}(y+\sqrt[3]{y}) & \\
y^4+\sqrt{y^3-y^2+1}=y(x-1)^3+1
\end{cases}
giải hệ pt (đặt ẩn phụ )
a) x+2/x+1 + 2/y-2 =6
5/x+1 -1/y-2 =3
b) 2/2x-y +3/x-2y =1/2
2/2x-y -1/x-2y =1/18
c) 2|x-6| +3|y+1| =5
5|x-6| -4|y+1| =1
d) |x| +|y-3| =1
y - |x| =3
a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
=>x+1=1 và y-2=1/2
=>x=0 và y=5/2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)
=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6
=>x-2y=9 và 2x-y=12
=>x=5; y=-2
c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)
=>|x-6|=1 và |y+1|=1
=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
làm phép chia
4x^2(y+z)^5:2x(y+z)^3
-x^2(y-1)^3(z+2)^2:1/2x^2(y-1)^2
x^m+1(y+2)^m:y(y+2)
3/4(x+2)^2m(x-3)^n-2:2/3(x+2)(x-3)^2
\(\dfrac{4x^2\left(y+z\right)^5}{2x\left(y+z\right)^3}=2x\left(y+z\right)^2\)
PP nhóm
1)x^2+x-y^2+y
2)4x^2-9y^2+4x-6y
3)x^2+x+y^2+y+2xy
4)-x^2+5x+2xy-5y-y^2
5)x^2-y^2+2x+1
6)x^2-1-y^2+2y
7)x^2+2xz-y^2+2uy+z^2-u^2
8)x^3+3x^2y+x+3xy^2+y+y^3
9)x^3+y(1-3x^2)+x(3y^2-1)-y^3
10)27x^3+27x^2+9x+1+x+1/3
1) x2 + x - y2 + y = (x2 - y2) + (x + y) = (x - y)(x + y) + (x + y) = (x - y + 1)(x + y)
2) 4x2 - 9y2 + 4x - 6y = (4x2 - 9y2) + (4x - 6y) = (2x - 3y)(2x + 3y) + 2(2x - 3y) = (2x - 3y)(2x + 3y + 2)
3) x2 + x + y2 + y + 2xy = (x2 + 2xy + y2) + (x + y) = (x + y)2 + (x + y) = (x + y)(x + y + 1)
4) -x2 + 5x + 2xy - 5y - y2 = -(x2 - 2xy + y2) + (5x - 5y) = -(x - y)2 + 5(x - y) = (x - y)(y - x + 5)
5) x2 - y2 + 2x + 1 = (x2 + 2x + 1) - y2 = (x + 1)2 - y2 = (x + 1 + y)(x - y + 1)
6) x2 - 1 - y2 + 2y = x2 - (y2 - 2y + 1) = x2 - (y - 1)2 = (x - y + 1)(x + y - 1)
7) x2 + 2xz - y2 + 2uy + z2 - u2 =(x2 + 2xz + z2) - (y2 - 2uy + u2) = (x + z)2 - (y - u)2 = (x + z - y + u)(x + z + y - u)
8) x3 + 3x2y + x + 3xy2 + y + y3 = (x3 + 3x2y + 3xy2 + y3) + (x + y) = (x + y)3 + (x + y) = (x + y)(x2 + 2xy + y2 + 1)
9) x3 + y(1 - 3x2) + x(3y2 - 1) - y3 = x3 + y - 3x2y + 3xy2 - x - y3 = (x3 - 3x2y + 3xy2 - y3) - (x - y) = (x - y)3 - (x - y) = (x - y)(x2 - 2xy+y2-1)
giải các hệ phương trình
a \(\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\)
\(\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\)
b \(\dfrac{5}{x+y-3}-\dfrac{2}{x-y+1}=8\)
\(\dfrac{3}{x+y-3}+\dfrac{1}{x-y+1}=\dfrac{3}{2}\)
c \(\sqrt{x-1}-3\sqrt{y+2}=2\)
\(2\sqrt{x-1}+5\sqrt{y+2}=15\)
d \(\dfrac{7}{\sqrt{x-7}}-\dfrac{4}{\sqrt{y+6}}=\dfrac{5}{3}\)
\(\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\)
e \(7x^2+13y=-39\)
\(5x^2-11y=33\)
f \(2\left(x-1\right)^2-3y^3=7\)
\(5\left(x-1\right)^2+6y^3=4\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
2*x^2*y^3+5*y^2*x^3+-1/2*x^3*y^2+-1/2*x^2*y^3
\(2x^2y^3+5y^2x^3+\left(-\dfrac{1}{2}x^3y^2\right)+\left(-\dfrac{1}{2}x^2y^3\right)\\ =\left[2x^2y^3+\left(-\dfrac{1}{2}x^2y^3\right)\right]+\left[5x^3y^2+\left(-\dfrac{1}{2}x^3y^2\right)\right]\\ =\dfrac{3}{2}x^2y^3+\dfrac{9}{2}x^3y^2\)