Giải và biện luận hệ phương trình:\(\hept{\begin{cases}2\left(m-1\right)\cdot x+y=2\\\left(m+2\right)\cdot x+\left(m-1\right)\cdot y=3\end{cases}}\)
Giải và biện luận hệ phương trình:\(\hept{\begin{cases}2\left(m-1\right)\cdot x+y=2\\\left(m+2\right)\cdot x+\left(m-1\right)\cdot y=3\end{cases}}\)
giải hệ phương trình :
a) \(\hept{\begin{cases}x\cdot\left(1+y-x\right)=-2\cdot y^2-y\\x\cdot\left(\sqrt{2\cdot y}-2\right)=y\cdot\left(\sqrt{x-1}-2\right)\end{cases}}\)
b) \(\hept{\begin{cases}1+x\cdot y+\sqrt{x\cdot y}=x\\\frac{1}{x\cdot\sqrt{x}}+y\cdot\sqrt{y}=\frac{1}{\sqrt{x}}+3\cdot\sqrt{y}\end{cases}}\)
Làm hộ mk nhé mk tick cho :))))))))))
câu 1: Giải và biện luận hệ phương trình:\(\hept{\begin{cases}2\left(m-1\right)\cdot x+y=2\\\left(m+2\right)\cdot x+\left(m-1\right)\cdot y=3\end{cases}}\)
câu 2: giải hệ phương trình \(\hept{\begin{cases}x+y=\sqrt{4z-1}\\y+z=\sqrt{4x-1}\\x+z=\sqrt{4y-1}\end{cases}}\)
\(\hept{\begin{cases}\left(2x-3\right)\cdot\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\cdot\left(3y-3\right)=3y\left(x+1\right)-12\end{cases}}\) giải hệ phương trình
\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{cases}}\)
\(\hept{\begin{cases}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{cases}}\)
\(\hept{\begin{cases}4xy-4xy+8x+12x-6y-12-54=0\\3xy-3xy-3x+3y-3y-3+12=0\end{cases}}\)
\(\hept{\begin{cases}20x-6y-66=0\\-3x+9=0\end{cases}}\)
\(\hept{\begin{cases}2\left(10x-3y\right)=66\\-3\left(x-3\right)=0\end{cases}}\)
\(\hept{\begin{cases}10x-3y=33\\x-3=0\end{cases}}\)
\(\hept{\begin{cases}10x-3y=33\\x=3\end{cases}}\)
\(\hept{\begin{cases}\left(x-3\right)\cdot\left(2y+5\right)=\left(2x+7\right)\cdot\left(y-1\right)\\\left(4x+1\right)\cdot\left(3y-6\right)=\left(6x-1\right)\cdot\left(2y+3\right)\end{cases}}\)
Tính tổng:
\(S=\frac{x+1}{x\cdot\left(x-y\right)\cdot\left(x-z\right)}+\frac{y+1}{y\cdot\left(y-z\right)\cdot\left(y-x\right)}+\frac{z+1}{z\cdot\left(z-x\right)\left(z-y\right)}\)
\(S=\frac{yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
+ \(yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)\)
\(=yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left[\left(y-z\right)+\left(x-y\right)\right]\)
\(+xy\left(z+1\right)\left(x-y\right)\)
\(=\left(y-z\right)\left[yz\left(x+1\right)-zx\left(y+1\right)\right]+\left(x-y\right)\left[xy\left(z+1\right)-zx\left(y+1\right)\right]\)
\(=\left(y-z\right)\left[z\left(y-x\right)\right]+\left(x-y\right)\cdot x\cdot\left(y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(\Rightarrow S=\frac{1}{xyz}\)
Thu gọn các đơn thức sau, xác định hệ số, phần biến và bậc của đơn thức
A=\(\left(\dfrac{-3}{7}\cdot x^3\cdot y^2\right)\cdot\left(\dfrac{-7}{9}\cdot y\cdot z^2\right)\cdot\left(6\cdot x\cdot y\right)\)
B= \(-4\cdot x\cdot y^3\cdot\left(-x^2\cdot y\right)^3\cdot\left(-2\cdot x\cdot y\cdot z^3\right)^2\)
HELP ME
\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)
\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)
\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)
\(A=2x^4y^4z^2\)
\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)
\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)
\(B=8x^7y^{y^8}z^6\)
Thu gọn biểu thức :
1, \(\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot\left(y-x\right)+\left(x-y\right)^2\)
2, \(\left(x-y+z\right)^2+2\cdot\left(x-y+z\right)\cdot\left(y-z\right)+\left(y-z\right)^2\)
1, đa thức đã cho \(\Leftrightarrow\left(2x-y\right)^2-2\left(2x-y\right)\left(x-y\right)+\left(x-y\right)^2=\left[\left(2x-y\right)-\left(x-y\right)\right]^2=\left(2x-y-x+y\right)^2=x^2\)
2, đa thức đã cho \(\Leftrightarrow\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2=\left[\left(x-y+z\right)+\left(y-z\right)\right]^2=\left(x-y+z+y-z\right)^2=x^2\)
--- giải chi tiết lắm rồi đó---
a, \(\left(2x-y\right)^2+2\left(2x-y\right)\left(y-x\right)+\left(x-y\right)^2\)
\(=4x^2-4xy+y^2+2\left(2xy-2x^2-y^2+xy\right)+x^2-2xy+y^2\)
\(=4x^2-4xy+y^2+4xy-4x^2-2y^2+2xy+x^2-2xy+y^2\)
\(=x^2\)
b, \(\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z\right)\left[1+2\left(y-z\right)\right]+y^2-2yz+z^2\)
\(=\left(x-y+z\right)\left(1+2y-2z\right)+y^2-2yz+z^2\)
\(=x+2xy-2xz-y-2y^2+2yz+z+2yz-2z^2+y^2-2yz+z^2\)
\(=x-y+z+2xy-2xz+2yz-y^2-z^2\)
Chúc bạn học tốt!!!
