1, Tìm x hoặc y biết:
a) 2x (x-5)-x(2x+3)=26
b) (3y^2-y+1)(y-1)+y^2(4-3y)=5/2
c) 2x^2+3(x-1)(x+1)=5x(x+1)
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\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
a, \(2x\left(x-5\right)-x\left(2x+3\right)=25\)
\(\Rightarrow2x^2-10x-2x^2-3x=25\)
\(\Rightarrow-13x=25\Rightarrow x=\dfrac{-25}{13}\)
b, \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\dfrac{5}{2}\)
\(\Rightarrow3y^3-y^2+y-3y^2+y-1+4y^2-3y^3=\dfrac{5}{2}\)
\(\Rightarrow2y-1=\dfrac{5}{2}\Rightarrow2y=\dfrac{7}{2}\Rightarrow y=\dfrac{7}{4}\)
c, \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2+x-x-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow-5x=3\Rightarrow x=\dfrac{-3}{5}\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
Bài 2:
a: \(A=\left(x+1\right)^3+5=20^3+5=8005\)
b: \(B=\left(x-1\right)^3+1=10^3+1=1001\)
Bài 1:
a) \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3+x-2\)
b) \(\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(2x-y\right)\)
\(=\dfrac{1}{2}x^2y^2\cdot\left(4x^2-y^2\right)\)
\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)
Bài 2:
a) \(2x\cdot\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Rightarrow2x^2-10x-2x^2-3x=26\)
\(\Rightarrow-13x=26\)
\(\Rightarrow x=2\)
b) \(\left(3y^2-y+1\right)\cdot\left(y-1\right)+y^2\cdot\left(4-3y\right)-\dfrac{5}{2}=0\)
\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3-\dfrac{5}{2}=0\)
\(\Rightarrow2y+\dfrac{7}{5}=0\)
\(\Rightarrow2y=-1,4\)
\(\Rightarrow y=-0,7\)
c) \(2x^2+3\left(x-1\right)\cdot\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow5x^2-5x^2-5x=3\)
\(\Rightarrow-5x=3\)
\(\Rightarrow x=0,6\)