Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=x-3-\) \(\left(x+4\right)\)\(\)
<=> \(4x^2-4x-3x^2+15-x^2=x-3-x-4\)
<=> \(-4x+15=-7\)
<=> \(x=\frac{11}{2}\)
\(\left(2x^2-3x+1\right)\left(x^2-5\right)-\left(x^2-x\right)\left(2x^2-x-10\right)=5\)
<=> \(2x^4-10x^2-3x^3+15x+x^2-5-\left(2x^4-x^3-10x^2-2x^3+x^2+10x\right)=5\)
<=> \(2x^4-10x^2-3x^3+15x+x^2-5-2x^4+x^3+10x^2+2x^3-x^2-10x=5\)
<=> \(5x-5=5\)
<=> \(5x=10\)
<=> \(x=2\)
Bài 1
Em xem lại đề nhé
a. Ta có VP=\(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x^3+xy^2-x^2y-y^3\right)\)
\(=VT\)
b.
1.\(\left(x-3\right)\left(x-2\right)-\left(x+10\right)\left(x-5\right)=0\)
\(\Leftrightarrow x^2-5x+6-\left(x^2+5x-50\right)=0\)
\(\Leftrightarrow-10x=-56\Rightarrow x=\frac{56}{10}\)
2.\(\left(2x-1\right)\left(3-x\right)+\left(x-2\right)\left(x+3\right)=\left(1-x\right)\left(x-2\right)\)
\(=-2x^2+7x-3+x^2+x-6=-x^2+3x-2\)
\(\Leftrightarrow5x=7\Leftrightarrow x=\frac{7}{5}\)
Ta có:
\(E=x^3-y^3-36xy\)
\(E=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy\)
\(E=12\left(x^2+xy+y^2\right)-36xy\) ( vì x - y =12 )
\(E=12\left(x^2+y+y^2-3xy\right)\)
\(E=12\left(x^2-2xy+y^2\right)\)
\(E=12\left(x-y\right)^2\)
\(E=12\cdot12^2\) ( vì x - y =12 )
\(E=12^3=1728\)
Hok tốt!
a) \(\left(x-3\right).\left(x^2+3x+9\right)-x.\left(x+4\right)\left(x-4\right)=21\)
\(\Leftrightarrow x^3-27-x.\left(x^2-16\right)=21\) \(\Leftrightarrow x^3-27-x^3+16x=21\)
\(\Leftrightarrow16x=21+27\) \(\Leftrightarrow16x=48\) \(\Leftrightarrow x=3\)
b) \(\left(x+2\right)\left(x^2-2x+4\right)-x.\left(x^2+2\right)=4\)
\(\Leftrightarrow x^3+8-x^3-2x=4\) \(\Leftrightarrow-2x=4-8\) \(\Leftrightarrow-2x=-4\) \(\Leftrightarrow x=2\)