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Bài 1 :
a) \(x^4-4x^2-4x-1\)
\(=x^4-\left(4x^2+4x+1\right)\)
\(=x^4-\left(2x+1\right)^2\)
\(=\left(x^2-2x-1\right)\left(x^2+2x+1\right)\)
b) \(x^2+2x-15\)
\(=x^2+2x+1-16\)
\(=\left(x+1\right)^2-4^2\)
\(=\left(x+1+4\right)\left(x+1-4\right)=\left(x+5\right)\left(x-3\right)\)
c) \(x^3y-2x^2y^2+5xy\)
\(=xy\left(x^2-2xy+5\right)\)
B2:
a) \(2\left(x-1\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=2\left(x^2-2x+1\right)-\left(4x^2-9\right)\)
\(=2x^2-4x+2-4x^2+9\)
\(=-2x^2-4x+11\)
b) \(\left(x+3\right)^2-2\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(x+3-x+3\right)^2=6^2=36\)
c) \(4\left(x-1\right)\left(x+3\right)+5\left(2x+1\right)^2-2\left(5-3x\right)^2\)
\(=4\left(x^2+2x-3\right)+5\left(4x^2+4x+1\right)-2\left(9x^2-30x+25\right)\)
\(=4x^2+8x-12+20x^2+20x+5-18x^2+60x-50\)
\(=6x^2+88x-57\)
B1:
a) \(9x^2+90x+225-\left(x-7\right)^2\)
= \(9x^2+90x+225-x^2+14x-49\)
= \(8x^2+104x+176\)
= \(\left(x+2\right)\left(x+11\right)\)
b) \(49\left(y-4\right)^2-9y^2-36y+36\)
= \(49\left(y^2-8y+16\right)-9y^2-36y+36\)
= \(49y^2-392y+784-9y^2-36y+36\)
= \(40y^2-428y+820\)
= \(\left(5y-41\right)\left(8y-20\right)\)
B2:
a) A = \(xy-4y-5y+20=xy-9y+20\)
A = \(y\left(x-9\right)+20\)
Với x = 14, y = \(\dfrac{11}{2}\)
A = \(\dfrac{11}{2}\left(14-9\right)+20=47,5\)
b) B = \(x^2+xy-5x-5y\)
B = \(x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)
Với x = -5, y = -8
B = \(\left(-5-8\right)\left(-5-5\right)=130\)
B3:
a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\left(2x-5\right)\left(-2\right)=0\)
\(x=\dfrac{5}{2}\)
b) \(\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\left(x+3\right)x\left(x-2\right)=0\)
\(\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
c) \(\left(2x^3+2x^2\right)+\left(3x^2+3\right)=0\)
\(2x^3+5x^2+3=0\)
\(\Rightarrow\) Đề sai rồi, nghiệm khủng bố lắm.
BÀI 1:
a) \(ĐKXĐ:\) \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
b) \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}\)
\(=\frac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{x-2}\)
c) \(A=0\) \(\Rightarrow\)\(\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\) \(x+2=0\)
\(\Leftrightarrow\)\(x=-2\) (loại vì ko thỏa mãn ĐKXĐ)
Vậy ko tìm đc x để A = 0
p/s: bn đăng từng bài ra đc ko, mk lm cho
Sửa đề: \(B=\left(x+4\right)^2+\left(x+5\right)^2-\left(x-5\right)^2-2x\left(x+1\right)\)
\(=x^2+8x+16+x^2+10x+25-x^2+10x-25-2x^2-2x\)
\(=-x^2+26x+16\)
b: \(Q=y\left(x-5\right)-4\left(x-5\right)=\left(x-5\right)\left(y-4\right)\)
=10*0,5
=5