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1: \(=\left(\dfrac{1}{2}:\dfrac{-2}{3}\right)\cdot\left(a^m:a^2\right)\cdot\left(b^n:b\right)\cdot\left(c^2:c\right)\)
\(=\dfrac{-3}{4}a^{m-2}b^{n-1}c\)
2: \(=\dfrac{2}{7}\left(x-2y\right)^{2m+1}\cdot\dfrac{-14}{5\left(2y-x\right)^{2m-1}}\)
\(=\dfrac{-4}{5}\cdot\dfrac{-\left(2y-x\right)^{2m+1}}{\left(2y-x\right)^{2m-1}}=\dfrac{4}{5}\cdot\left(2y-x\right)^2\)
a: =>3M+2x^4y^4=x^4y^4
=>3M=-x^4y^4
=>M=-1/3*x^4y^4
b: x^2-2M=3x^2
=>2M=-2x^2
=>M=-x^2
c: =>M=-x^2y^3-3x^2y^3=-4x^2y^3
d: =>M=7x^2y^2-3x^2y^2=4x^2y^2
a) \(\left(5xy^3\right)^2-2.5xy^3.6yz^2+\left(6yz^2\right)^2\)=\(\left(5xy^3-6yz^2\right)^2\)
b) \(\left(\frac{1}{3}u^2v^3\right)^2-2.\frac{1}{3}u^2v^3.\frac{1}{2}u^3v+\left(\frac{1}{2}u^3v\right)^2\)=\(\left(\frac{1}{3}u^2v^3-\frac{1}{2}u^3v\right)^2\)
a) (5x³y² - 3x²y + xy) : xy
= 5x³y² : xy + (-3x²y : xy) + xy : xy
= 5x²y - 3x + 1
b) A + 2M = P
A = P - 2M
= 3x³ - 2x²y - xy + 3 - 2.(x³ - x²y + 2xy + 3)
= 3x³ - 2x²y - xy + 3 - 2x³ + 2x²y - 4xy - 6
= (3x³ - 2x³) + (-2x²y + 2x²y) + (-xy - 4xy) + (3 - 6)
= x³ - 5xy - 3
Vậy A = x³ - 5xy - 3
a) \(A:xy\)
\(=\left(5x^3y^2-3x^2y+xy\right):xy\)
\(=5x^3y^2:xy-3x^2y:xy+xy:xy\)
\(=5x^2y-3x+1\)
b) \(A+2M=P\)
\(\Rightarrow A+2\cdot\left(x^3-x^2y+2xy\right)=3x^3-2x^2y-xy+3\)
\(\Rightarrow A+2x^3-2x^2y+4xy=3x^3-2x^2y-xy+3\)
\(\Rightarrow A=3x^3-2x^3-2x^2y+2x^2y-xy-4xy+3\)
\(\Rightarrow A=x^3-4xy+3\)
Câu 5:
a: Khi m=3 thì \(f\left(x\right)=\left(2\cdot3+1\right)x-3=7x-3\)
\(f\left(-3\right)=7\cdot\left(-3\right)-3=-21-3=-24\)
\(f\left(0\right)=7\cdot0-3=-3\)
b: Thay x=2 và y=3 vào f(x)=(2m+1)x-3, ta được:
\(2\left(2m+1\right)-3=3\)
=>2(2m+1)=6
=>2m+1=3
=>2m=2
=>m=1
c: Thay m=1 vào hàm số, ta được:
\(y=\left(2\cdot1+1\right)x-3=3x-3\)
*Vẽ đồ thị
d: Để hàm số y=(2m+1)x-3 là hàm số bậc nhất thì \(2m+1\ne0\)
=>\(2m\ne-1\)
=>\(m\ne-\dfrac{1}{2}\)
e: Để đồ thị hàm số y=(2m+1)x-3 song song với đường thẳng y=5x+1 thì \(\left\{{}\begin{matrix}2m+1=5\\-3\ne1\end{matrix}\right.\)
=>2m+1=5
=>2m=4
=>m=2
a) \(\frac{x+7}{2x+3}-\frac{5}{2x+3}=\frac{x+7-5}{2x+3}=\frac{x+2}{2x+3}\)
b) \(\frac{m^2}{3\left(m+3\right)}+\frac{2m+3}{m+3}=\frac{m^2}{3\left(m+3\right)}+\frac{\left(2m+3\right).3}{3.\left(m+3\right)}\)
\(=\frac{m^2+6m+9}{3\left(m+3\right)}=\frac{\left(m+3\right)^2}{3\left(m+3\right)}=\frac{m+3}{3}\)
c) \(\frac{x^2-4}{x+5}.\frac{2x+10}{x+2}=\frac{\left(x-2\right)\left(x+2\right).2.\left(x+5\right)}{\left(x+5\right).\left(x+2\right)}=\left(x-2\right).2=2x-4\)
d) \(\frac{3+6y}{y^2-2y+1}:\frac{2y+1}{y-1}=\frac{3\left(2y+1\right)}{\left(y-1\right)^2}.\frac{y-1}{2y+1}=\frac{3}{y-1}\)
\(a,\frac{x+7}{2x+3}-\frac{5}{2x+3}\)
\(=\frac{x+7-5}{2x+3}\)
\(=\frac{x+2}{2x+3}\)
\(b,\frac{m^2}{3\left(m+3\right)}+\frac{2m+3}{m+3}\)
\(=\frac{m^2}{3\left(m+3\right)}+\frac{3\left(2m+3\right)}{3\left(m+3\right)}\)
\(=\frac{m^2+6m+9}{3\left(m+3\right)}\)
\(=\frac{\left(m+3\right)^2}{3\left(m+3\right)}\)
\(=\frac{m+3}{3}\)
\(c,\frac{x^2-4}{x+5}.\frac{2x+10}{x+2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{x+5}.\frac{2\left(x+5\right)}{x+2}\)
\(=2\left(x-2\right)\)
d, nghịch đảo lên rồi làm tương tự nha
a) \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^6.\)
b) \(\left(-x\right)^7:\left(-x\right)^5\)
\(=\left(-x\right)^2\)
\(=x^2.\)
c) \(5x^3y^4:10x^2y\)
\(=\frac{1}{2}xy^3.\)
d) \(\frac{3}{4}x^3y^3:\left(-\frac{1}{2}xy^2\right)\)
\(=-\frac{3}{2}x^2y.\)
e) \(\left(-xy\right)^{14}:\left(-xy\right)^7\)
\(=\left(-xy\right)^7\)
\(=-x^7y^7.\)
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