(5x + 2) ⋮ (x +...
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(5x + 2) ⋮ (x + 1)
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TG
0
16 tháng 7 2020
Bài 1:
a) Ta có: \(\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=4x^2-4x+1+4\left(x^2+2x-3\right)-2\left(25-30x+9x^2\right)\)
\(=4x^2-4x+1+4x^2+8x-12-50+60x-18x^2\)
\(=-10x^2+64x-61\)
b) Ta có: \(\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1\right)^2-\left(2a\right)^2-\left(2a^2+1\right)^2\)
\(=-4a^2\)
c) Ta có: \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=\left(9x-1+1-5x\right)^2\)
\(=\left(4x\right)^2=16x^2\)
d)
Sửa đề: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)
Ta có: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)
\(=\left(x^2+5x-1+5x-1\right)^2\)
\(=\left(x^2+10x-2\right)^2\)
\(=x^4+100x^2+4+20x^3-40x-4x^2\)
\(=x^4+20x^3+96x^2-40x+4\)
e) Ta có: \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x\left(x^2-1\right)-\left(x^3+1\right)\)
\(=x^3-x-x^3-1\)
=-x-1
f) Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
NQ
0
NL
1
13 tháng 10 2017
(x2-5x+1)2+2(5x-1)(x2-5x+1)+(5x-1)2
= [(x2-5x+1)+(5x-1)]2
= (x2-5x+1+5x-1)2
= (x2)2
= x4
24 tháng 8 2019
\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
24 tháng 8 2019
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
18 tháng 9 2020
(x+2)^2= 9
=> (x+2)^2= 3^2=(-3)^2
TH1: x+2=3
=> x=3-2=1
TH2: x+2=-3
=> x=(-3)-2=-5
18 tháng 9 2020
Bài làm :
\(a,\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\left(x+2\right)^2=9\)
\(\Leftrightarrow\left(x+2\right)^2=3^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x = 1 hoặc x = -5 .
\(b,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-\left(25x^2-3^2\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow\left(25x^2-25x^2\right)+10x=30-9-1\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy x = 2 .
\(c,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3+x^2+x-x^2-x-1+\left(x^2+2x\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3-1+2x^2-x^3+4x-2x^2=5\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(2x^2-2x^2\right)+4x=5+1\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy x = 3/2 .
Học tốt nhé
HD
1
13 tháng 7 2015
(x2 - 5x + 1)2 + 2(5x - 1)(x2 - 5x + 1) + (5x - 1)2 = (x2 - 5x + 1 + 5x - 1)2 = (x2)2 = x4
Ta có 5x + 2 \(⋮\)x + 1
<=> 5x + 5 - 3 \(⋮\)x + 1
<=> 5(x + 1) - 3 \(⋮\) x + 1
Vì 5(x + 1) \(⋮\)x + 1
=> 3 \(⋮\)x + 1
=> x + 1 \(\inƯ\left(3\right)\)
=> x + 1 \(\in\left\{1;3;-1;-3\right\}\)
=> \(x\in\left\{0;2;-2;-4\right\}\)
Vậy \(x\in\left\{0;2;-2;-4\right\}\)