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1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)
\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )
2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)
\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)
\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )
Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))
1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
=> \(-4x^2+28x+4x^3-20x=28x^2-13\)
=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)
=> \(-4x^2+4x^3+8x-28x^2+13=0\)
=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)
=> \(-32x^2+4x^3+8x+13=0\)
=> vô nghiệm
2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)
=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)
=> \(-14x^2-56x+12=0\)
=> .... tự tìm
Câu c dấu bằng chỗ nào ?
a) \(\left(1+x\right)^2+\left(1-x\right)^2\)
\(=1+2x+x^2+1-2x+x^2\)
\(=2x^2+2\)
b) \(\left(x+2\right)^2+\left(1+x\right)\left(1-x\right)\)
\(=x^2+4x+4+1-x^2\)
\(=4x+5\)
c) \(\left(x-3\right)^2+3\left(x+1\right)^2\)
\(=x^2-6x+9+3x^2+6x+3\)
\(=4x^2+12\)
d)\(\left(2+3x\right)\left(3x-2\right)-\left(3x+1\right)^2\)
\(=9x^2-4-9x^2-6x-1\)
\(=-6x-5\)
e) \(\left(x+5\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=x^2-2x+5x-10-x^2-4x-4\)
\(=-x-14\)
f) \(\left(x+3\right)\left(2x-5\right)-2\left(1+x\right)^2\)
\(=2x^2-5x+6x-15-2-4x-2x^2\)
\(=-3x-17\)
g) \(\left(4x-1\right)\left(4x+1\right)-4\left(1-2x\right)^2\)
\(=16x^2-1-4+16x-16x^2\)
\(=16x-5\)
#Học tốt!
a) \(A=\dfrac{\left(-2\right)^5}{\left(-2\right)^3}=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)
b) \(y\ne0:B=\dfrac{\left(-y\right)^7}{\left(-y\right)^3}=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)
c) \(x\ne0:C=\dfrac{\left(x\right)^{12}}{\left(-x\right)^{10}}=\left(x\right)^{12-10}=\left(x\right)^2=x^4\)
d) \(x\ne0:D=\dfrac{2x^6}{\left(2x\right)^3}=\dfrac{2x^6}{8x^3}=\dfrac{1}{4}\left(x\right)^{6-3}=\dfrac{1}{4}\left(x\right)^3\)
e) \(x\ne0:E=\dfrac{\left(-3x\right)^5}{\left(-3x\right)^2}=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)
f) \(x,y\ne0:F=\dfrac{\left(xy^2\right)^4}{\left(xy^2\right)^2}=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2=x^2y^4\)
i) \(x\ne-2:I=\dfrac{\left(x+2\right)^9}{\left(x+2\right)^6}=\left(x+2\right)^{9-6}=\left(x+2\right)^3\)
1) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
2) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
3) \(\left(x+\frac{1}{3}\right)^4=\left[\left(x+\frac{1}{3}\right)^2\right]^2=\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)^2=x^4+\frac{4}{9}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x+\frac{2}{9}x^2=x^4+\frac{2}{3}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x\)
4) \(\left(2x+y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
5) Sửa đề: \(\left(\frac{x}{2}-2y\right)^3=\frac{x^3}{8}-\frac{3x^2}{2}+6xy^2-8y^3\)
6) \(\left(\sqrt{2x-y}\right)^4=\left(2x-y\right)^2=4x^2-4xy+y^2\)
7) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
8) \(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)