(9+8x3)x2 =?
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NG
a,x3+3x2+3x+1
b,x2+6x+9
c,-x3+9x2-27x+27
d,x2+4x+4
k,10x-25-x2
f,(x+y)2-9x2
g,8x3+42x2y+16xy2+6xy+y3
2
H9
HT.Phong (9A5)
CTVHS
3 tháng 7 2023
a) \(x^3+3x^2+3x+1=x^2+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x-1\right)^3\)
b) \(x^2+6x+9=x^2+2\cdot3\cdot x+3^2=\left(x+3\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-\left(x^3-9x^2+27x-27\right)\)
\(=-\left(x^3-3\cdot3\cdot x^2+3\cdot3^2\cdot x-3^3\right)=-\left(x-3\right)^3\)
d) \(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
k) \(10x-25-x^2=-x^2+10x-25=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2\cdot5\cdot x+5^2\right)=-\left(x-5\right)^2\)
f) \(\left(x+y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left[\left(x-y\right)-3x\right]\left[\left(x-y\right)+3x\right]\)
\(=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
31 tháng 7 2021
1. x2 - 6x + 9=(x-3)2
2. 25 + 10x + x2=(x+5)2
3. \(\dfrac{1}{4}a^2+2ab^2+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
4.\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)
5.x3 + 8y3=(x+8y)(x2-8xy+64y2)
6.8y3 -125=(2y-5)(4y2+10y+25)
7.a6-b3=(a2-b)(a4+a2b+b2)
8 x2 - 10x + 25=(x-2)2
31 tháng 7 2021
1) \(x^2-6x+9=\left(x-3\right)^2\)
2) \(25+10x+x^2=\left(5+x\right)^2\)
3) \(\dfrac{1}{4}a^2+2ab+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
4) \(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)
5) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
6) \(8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)
7) \(a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
8) \(x^2-10x+25=\left(x-5\right)^2\)
9) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
29 tháng 10 2021
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
10 tháng 10 2021
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
19 tháng 8 2021
a: Ta có: \(y\left(x^2-y^2\right)\cdot\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
=0
b: Ta có: \(\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\left(8x^3-\dfrac{1}{27}\right)\)
\(=8x^3+\dfrac{1}{27}-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c: Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
31 tháng 10 2021
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
31 tháng 10 2021
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
AH
Akai Haruma
Giáo viên
2 tháng 10 2021
a.
$x^4-6x^2+9=0$
$\Leftrightarrow (x^2-3)^2=0$
$\Leftrightarrow x^2-3=0$
$\Leftrightarrow x^2=3$
$\Leftrightarrow x=\pm \sqrt{3}$
b.
$8x^3+12x^2+6x-63=0$
$\Leftrightarrow (8x^2+12x^2+6x+1)-64=0$
$\Leftrightarrow (2x+1)^3=64=4^3$
$\Leftrightarrow 2x+1=4$
$\Leftrightarrow x=\frac{3}{2}$
c. $(3-2x)^2-25=0$
$\Leftrightarrow (3-2x)^2-5^2=0$
$\Leftrightarrow (3-2x-5)(3-2x+5)=0$
$\Leftrightarrow (-2-2x)(8-2x)=0$
$\Leftrightarrow -2-2x=0$ hoặc $8-2x=0$
$\Leftrightarrow x=-1$ hoặc $x=4$
AH
Akai Haruma
Giáo viên
2 tháng 10 2021
d.
$6(x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1$
$\Leftrightarrow (x+1)^2[6-2(x+1)]+2(x^3-1)=1$
$\Leftrightarrow (x+1)^2(4-2x)+2x^3-3=0$
$\Leftrightarrow 6x+1=0$
$\Leftrightarrow x=\frac{-1}{6}$
e. $(x-2)^2-(x-2)(x+2)=0$
$\Leftrightarrow (x-2)[(x-2)-(x+2)]=0$
$\Leftrightarrow (x-2)(-4)=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
f. $x^2-4x+4=25$
$\Leftrightarrow (x-2)^2=5^2=(-5)^2$
$\Leftrightarrow x-2=5$ hoặc $x-2=-5$
$\Leftrightarrow x=7$ hoặc $x=-3$
( 9 + 8 * 3 ) * 2 = 66
HT
=33x2
=66