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a, \(x^4-5x^3+2x^2+10x+2=0\)
\(\Rightarrow x^4+x^3-6x^3-6x^2+8x^2+8x+2x+2=0\)
\(\Rightarrow x^3\left(x+1\right)-6x^2\left(x+1\right)+8x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3-6x^2+8x+2\right)=0\)
Vì \(x^3-6x^2+8x+2>0\) nên \(x+1=0\Rightarrow x=-1\)
Các câu còn lại tương tự!
Chúc bạn học tốt!!!
a) \(x^3-3x^2-x+3\) \(=\left(x^3-x\right)-\left(3x^2-3\right)=x\left(x^2-1\right)-3\left(x^2-1\right)\)
\(=\left(x-3\right)\left(x^2-1\right)\)
b) \(x^3-4x^2-x+4=\left(x^3-x\right)-\left(4x^2-4\right)=x\left(x^2-1\right)+4\left(x^2-1\right)\)
\(=\left(x-4\right)\left(x^2-1\right)\)
c) \(2x^3-x^2-2x+1=\left(2x^3-2x\right)-\left(x^2-1\right)=2x\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(2x-1\right)\left(x^2-1\right)\)
d) \(5x^3-x^2-5x+1=\left(5x^3-5x\right)-\left(x^2-1\right)=5x\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(5x-1\right)\left(x^2-1\right)\)
Bài 1 :
a) (3a+4b)3+(3a-4b)3-48a2b2
=27a3+108a2b+144ab2+64b3+27a3-108a2b+144ab2-64b3-48a2b2
=54a3+288ab2-48a2b2
=2a(27a2+144b2-24ab)
b) (5x+2y)(5x-2y)+(2x-y)3+(2x+y)3
=25x2-4y2+8x3-12x2y+6xy2-y3+8x3+12x2y+6xy2+y3
=16x3+25x2-y2+12xy2
=x2(16x+25)-y2(1-12x)
Bài 2 :
\(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
b)\(x^3-4x^2+3x=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{3}\\x=1\end{cases}}\)
c)Nếu đề đổi thành =1 thì có vẻ hợp lí hơn
d)\(\left(3x-1\right)^3-3\left(3x+2\right)^2+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-3\left(9x^2+12x+4\right)+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-27x^2-36x-12+13=0\)
\(\Leftrightarrow27x^3-54x^2-27x=0\)
\(\Leftrightarrow27x\left(x^2-2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}27x=0\\x^2-2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\-\left(x^2+2x+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\-\left(x+1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
#H
1) \(B=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)+2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+\left(4x-4\right)\cdot\left(x+3\right)+2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+12x-4x-12+50-60+18x^2\)
\(=42x^2-72x+43\)
2) \(C=\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a+1\right)^2\)
\(=4a^4-4a^3+2a^2+4a^3-4a^2+2a+2a^2-2a+1-\left(4a^2+4a+1\right)\)
\(=4a^4+2a^2-4a^2+2a^2+1-4a^2-4a-1\)
\(=4a^4-4a^2-4a\)
3) Sky Sơn Tùng làm đúng rồi nhé.
4) \(E=\left(x^2-5x+1\right)^2+2\left(5x-1\right)\left(x^2-5x+1\right)\left(5x-1\right)^2\)
\(=x^4+27x^2+1-10x^3+250x^5-1400x^4+1030x^3-302x^2+40x-2\)
\(=-1399x^4-275x^2-1+1020x^3+250x^5+40x\)
5) \(F=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2\)
\(=\left[a^2+b^2-c^2-\left(a^2-b^2+c^2\right)\right]\cdot\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\)
\(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\cdot2a^2\)
\(=\left(2b^2-2c^2\right)\cdot2a^2\)
\(=2\left(b^2-c^2\right)\cdot2a^2\)
\(=2\left(b-c\right)\left(b+c\right)\cdot2a^2\)
\(=2\cdot2a^2\cdot\left(b-c\right)\left(b+c\right)\)
\(=4a^2\cdot\left(b-c\right)\left(b+c\right)\)
6) \(G=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+\left(-c\right)^2+2ab-2ac-2bc-2\left(a^2+2ab+b^2\right)\)
\(=a^2+b^2+c^2+2ab+a^2+b^2+\left(-c\right)^2+2ab-2a^2-4ab-2b^2\)
\(=0+0+c^2+0+c^2\)
\(=2c^2\)
7) \(H=\left(a+c\right)\left(a-c\right)-\left(a-b-c\right)\left(a-b+c\right)+b\left(b-2x\right)\)
\(=a^2-c^2-\left[\left(a-b\right)^2-c^2\right]+b^2-2bx\)
\(=a^2-c^2-\left(a^2-2ab+b^2-c^2\right)+b^2-2bx\)
\(=a^2-b^2-a^2+2ab-b^2+c^2+b^2-2bx\)
\(=2ab-2bx\)
\(D=\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)=\left(9x-1+1-5x\right)^2=\left(4x\right)^2=16x^2\)
a) \(x^2-y^2-5x-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
b) \(5x^3-5x^2y-10x^2+10xy\)
\(=\left(5x^3-5x^2y\right)-\left(10x^2-10xy\right)\)
\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x^2-10x\right)\)
\(=5x\left(x-y\right)\left(x-2\right)\)
c) \(x^3-2x^2-x+2\)
\(=\left(x^3-2x^2\right)-\left(x-2\right)\)
\(=x^2\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(-y^2+2xy-x^2+3x-3y\)
\(=-\left(y^2-2xy+x^2\right)+\left(3x-3y\right)\)
\(=-\left(y-x\right)^2+3\left(x-y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)
\(=\left(x-y\right)\left(-x+y+3\right)\)
g) \(4x^2-8x+3\)
\(=4x^2-6x-2x+3\)
\(=\left(4x^2-6x\right)-\left(2x-3\right)\)
\(=2x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-3\right)\left(2x-1\right)\)
h) \(2x^2-5x-7\)
\(=2x^2+2x-7x-7\)
\(=\left(2x^2+2x\right)-\left(7x+7\right)\)
\(=2x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left(2x-7\right)\)
k) \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
a, 15x5 - 10x4 + 5x3 + 10x2
b, -2a5x4 + 10a3x2 - 6a2x
c, 6x4 - 2x3 - 15x2 + 23x - 6
d, a5 - b5
mk rút gọn luôn ko làm từng bc vì dài nhé :D