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8 tháng 9 2021
a: Ta có: \(x^2-4-\left(x+2\right)^2\)
\(=x^2-4-x^2-4x-4\)
=-4x-8
b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-4-x^2+2x+3\)
=2x-1
c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)
\(=\left(x-2\right)\left(x+2-x-5\right)\)
\(=-3x+6\)
d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
=4
8 tháng 9 2021
e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)
\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)
\(=29a^2-45a-3-36a^2+24a-4\)
\(=-7a^2-21a-7\)
g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)
\(=25y^2-9-25y^2+40y-16\)
=40y-25
h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)
\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)
\(=35x^3+15x^2+15x\)
i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=16x^2\)
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
PN
1
26 tháng 9 2021
a: Ta có: \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x^2=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2=5\)
\(\Leftrightarrow13x=13\)
hay x=1
29 tháng 11 2022
1: \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
2: \(=8x^2+12x-10x-15-4\left(2x^2-x+4x-2\right)+10x+7\)
\(=8x^2+12x-8-8x^2+4x-16x+8\)
\(=0\)
3: \(=\left(6x+1-6x+1\right)^2=4\)
5: \(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)
\(=3x^3+6x-3x^3+3x=9x\)
DH
25 tháng 6 2017
a,\(A=\left(x^4-3x^2+9\right)\left(x^2+3\right)+\left(3-x^2\right)^2\)
\(A=x^6-3x^4+9x^2+3x^4-9x^2+27+9-6x^2+x^4\)
\(A=x^6+x^4-6x^2+36\)
b, \(M=5\left(x+2y\right)^2-\left(3y+2x\right)^2+\left(4x-y\right)^2+3\left(x-2y\right)\left(x+2y\right)\)
\(M=5\left(x^2+4xy+4y^2\right)-\left(9y^2+12xy+4x^2\right)+\left(16x^2-8xy+y^2\right)+3\left(x^2-4y^2\right)\)
\(M=5x^2+20xy+20y^2-9y^2-12xy-4x^2+16x^2-8xy+y^2+3x^2-12y^2\)
\(M=20x^2\)
Các câu còn lại làm tương tự! Chúc bạn học tốt!!!
25 tháng 6 2017
E=\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(\Leftrightarrow\left(6x+1\right)^2-2\left(1+6x\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(\Leftrightarrow\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\)
\(\Leftrightarrow\left(6x+1-6x+1\right)^2=2^2=4\)
19 tháng 12 2021
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
TL :
x3−6x+5=x3−x2+x2−x−5x+5=x2(x−1)+x(x−1)−5(x−1)=(x−1)(x2+x−5)
ht