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MA
1
27 tháng 9 2021
Bài 1:
a) \(=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1=\left(x+1\right)\left(x+3\right)\)
f) \(=\left(x^2-4x+4\right)-9=\left(x-2\right)^2-3^2=\left(x-5\right)\left(x+1\right)\)
h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
k) \(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^3-x-1\right)\)
m) \(=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-9=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
Bài 2:
a) \(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
e) \(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
f) giống câu a
g) \(=x^2-2xy=x\left(x-2y\right)\)
i) \(=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
k) \(=\left(x+1\right)^3-27z^3=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
l) \(=\left(2x+1\right)^2-9y^2=\left(2x+1-3y\right)\left(2x+1+3y\right)\)
KK
13 tháng 9 2021
a. (3x - 1)2 - 16
= (3x - 1)2 - 42
= (3x - 1 - 4)(3x - 1 + 4)
= (3x - 5)(3x + 3)
= 3(3x - 5)(x + 1)
b. (5x - 4)2 - 49x2
= (5x - 4)2 - (7x)2
= (5x - 4 - 7x)(5x - 4 + 7x)
= \(\left(-2x-4\right)\left(2x-4\right)\)
= 2(-x - 2) . 2(x - 2)
= 4(-x - 2)(x - 2)
c. (2x + 5)2 - (x - 9)2
= (2x + 5 - x + 9)(2x + 5 + x - 9)
= (x + 14)(3x - 4)
F
13 tháng 9 2021
2.
a)(3x-1)2-16 = (3x-1)2-42
=(3x-1-4).(3x-1+4)= (3x-5).(3x+3)
b)(5x-4)2-49x2 =(5x-4)2-(7x)2
= (5x-4-7x).(5x-4+7x) = (-2x-4).(12x-4)
c)(2x+5)2-(x-9)2 =(2x+5-x+9).(2x+5+x-9) =(x+14).(3x-4)
d)(3x+1)2-4.(x-2)2 =(3x+1)2-[2.(x-2)]2 = [3x+1-2.(x-2)].[3x+1+2.(x-2)]
= (3x+1-2x+4).(3x+1+2x-4)
= (x+5).(5x-3)
e)9.(2x+3)2-4.(x+1)2 =[3.(2x+3)]2-[2.(x+1)]2
=[3.(2x+3)-2.(x+1)].[3.(2x+3)+2.(x+1)]
=(6x+9-2x-2).(6x+9+2x+2)
=(4x+7).(8x+11)
giúp em với e sắp đến hạn rùi ạ
4 tháng 9 2021
a: Ta có: \(\left(4x-6\right)^2-16\left(x-3\right)^2\)
\(=16x^2-48x+36-16x^2+96x-144\)
\(=48x-108\)
\(=48\cdot\left(-7\right)-108\)
\(=-444\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=x^3-1+x^3+27\)
\(=2x^3+26\)
\(=-2+26=24\)
11 tháng 9 2021
a: \(4x^2-6x=2x\left(2x-3\right)\)
b: \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c: \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
d: \(3x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(3x+5\right)\)
e: \(2x^2\left(x+1\right)+4\left(x+1\right)=2\left(x+1\right)\left(x^2+2\right)\)
f: \(-3x-6xy+9xz=-3x\left(1+2y-3z\right)\)
MA
2
11 tháng 11 2021
\(a,=\left(6x+1-6x+1\right)^2=4\\ b,=3x^2-6x-5x+5x^2-8x^2-24=-11x-24\\ c,=14x^2+x-3-5x^2-18x+8-9x^2+17x=5\\ d,=6x^2+43x-40-6x^2-7x+3-36x+27=-10\)
11 tháng 11 2021
a) \(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2=\left(6x+1-6x+1\right)^2=2^2=4\)
b) \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
c) \(\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x=\left(7x-3\right).2x+\left(7x-3\right)-\left[\left(5x-2\right).x+4\left(5x-2\right)\right]-9x^2+17x=14x^2-6x+7x-3-\left(5x^2-2x+20x-8\right)-9x^2+17x=5x^2+18x-3-\left(5x^2+18x-8\right)=5x^2+18x-3-5x^2-18x+8=5\)
d) \(\left(6x-5\right)\left(x+8\right)-\left(3x-1\right)\left(2x+3\right)-9\left(4x-3\right)=\left(6x-5\right).x+8\left(6x-5\right)-\left[\left(3x-1\right).2x+3\left(3x-1\right)\right]-36x+27=6x^2-5x+48x-40-\left(6x^2-2x+9x-3\right)-36x+27=6x^2+7x-13-\left(6x^2+7x-3\right)=6x^2+7x-13-6x^2-7x+3=-10\)
10 tháng 9 2021
a: Ta có: \(8x\left(x-2017\right)-2x+4034=0\)
\(\Leftrightarrow\left(x-2017\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
11 tháng 9 2021
c: Ta có: \(4-x=2\left(x-4\right)^2\)
\(\Leftrightarrow2\left(x-4\right)^2+x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)
giúp với em sắp phải nộp deadline r ạ:((
31 tháng 8 2021
a: Ta có: \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=-7\)
\(\Leftrightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=-7\)
\(\Leftrightarrow-9x=-18\)
hay x=2
b: ta có: \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2-5x+3\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+x-6x^2+5x-3=0\)
\(\Leftrightarrow17x=-5\)
hay \(x=-\dfrac{5}{17}\)
c: Ta có: \(\left(2x-1\right)^3+12\left(x-1\right)\left(x+1\right)=14x-13\)
\(\Leftrightarrow8x^3-12x^2+6x-1+12x^2-12-14x+13=0\)
\(\Leftrightarrow8x^3-8x=0\)
\(\Leftrightarrow8x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
31 tháng 8 2021
a) \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=-7\)
\(\Rightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=-7\)
\(\Rightarrow-9x=-18\)
\(\Rightarrow x=2\)
b) \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2-5x+3\)
\(\Rightarrow x^3+6x^2+12x+8-x^3+x=6x^2-5x+3\)
\(\Rightarrow18x=-5\)
\(\Rightarrow x=-\dfrac{5}{18}\)
c) \(\left(2x-1\right)^3+12\left(x-1\right)\left(x+1\right)=14x-13\)
\(\Rightarrow8x^3-12x^2+6x-1+12x^2-12=14x-13\)
\(\Rightarrow8x^3-8x=0\)
\(\Rightarrow8x\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
30 tháng 11 2021
Xét tứ giác MNKP có
E là trung điểm của NP
E là trung điểm của MK
Do đó: MNKP là hình bình hành
mà \(\widehat{PMN}=90^0\)
nên MNKP là hình chữ nhật
a: \(4x^2-12x+9=\left(2x-3\right)^2\)
b: \(4x^2+4x+1=\left(2x+1\right)^2\)
c: \(36x^2+12x+1=\left(6x+1\right)^2\)
d: \(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)
a)\(=\left(2x\right)^2-2.2x.3+3^2\)
\(=\left(2x+3\right)^2\)
b)\(=\left(2x\right)^2+2.2x.1+1^2\)
\(=\left(2x+1\right)^2\)