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N
0
26 tháng 7 2021
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
30 tháng 10 2023
4:
\(P=\left(x+4\right)\left(x^2-4x+16\right)-\left(64-x^3\right)\)
\(=x^3+64-64+x^3=2x^3\)
Khi x=-20 thì \(P=2\cdot\left(-20\right)^3=-16000\)
=>Chọn C
2: Đề khó hiểu quá bạn ơi
HA
1
KK
2 tháng 5 2022
\(\dfrac{x+9}{x+4}-\dfrac{5}{x-4}=\dfrac{-20}{x^2-16}\)ĐKXĐ:x khác +-4
=>\(\dfrac{\left(x+9\right).\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{5.\left(x+4\right)}{\left(x+4\right).\left(x-4\right)}=\dfrac{-20}{\left(x+4\right).\left(x-4\right)}\)
=> khử mẫu:
=>(x+9).(x-4)-5.(x+4)=-20
=>x2-4x+9x-36-5x-20=-20
=>x2-56=-20
=>x2=62
=>x=6
L
15 tháng 11 2019
1)
=a^4+2a^2+1-a^2
=(a^2+1)^2-a^2
=(a^2-a+1)(a^2+a+1)
2)
=a^4+4b^4-4a^2b^2
=(a^2+2b^2)^2-4a^2b^2
=(a^2-2ab+2b^2)(a^2+2ab+2b^2)
3)
=(8x^2+1)^2-16x^2
=(8x^2-4x+1)(8x^2+4x+1).
4)
=x^5+x^4+x^3-x^3+1
=x^2(x^2+x+1)-(x-1)(x^2+x+1)
=(x^2-x+1)(x^2+x+1)
5).
=x^7-x+x^2+x+1
=x(x^6-1)+x^2+x+1
=x(x^3-1)(x^3+1)+x^2+x+1
=x(x-1)(x^2+x+1)(x^3+1)+x^2+x+1
=(x^2+x+1)[(x^2-x)(x^3+1)+1]
6)
=x^8-x^2+x^2+x+1
=x^2(x-1)(x^2+x+1)(x^3+1)+x^2+x+1
Xong nhóm x^2+x+1 vào.
7)
=x^4-(2x-1)^2
=(x^2-2x+1)(x^2+2x-1)
8)
=(a^8+b^8)^2-a^8b^8
=(a^8-a^4b^4+b^8)(a^8+a^4b^4+b^8).
TD
3
27 tháng 12 2017
a)\(\frac{2x}{x+5}+\frac{10}{x+5}=\frac{2x+10}{x+5}=\frac{2\left(x+5\right)}{x+5}=2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{8\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{x-2}\)
PD
27 tháng 12 2017
a) \(\frac{2x}{x+5}+\frac{10}{x+5}\)=\(\frac{2x+10}{x+5}\)=\(\frac{2\left(x+5\right)}{x+5}\)=\(2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}\)=\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{\left(x+2-x+2\right)\left(x+2+x-2\right)+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4\times2x+16}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8}{x-2}\)
MD
0
19 tháng 2 2022
đk : x khác -4 ; 4
\(\Rightarrow96+\left(1-3x\right)\left(x+4\right)=\left(2x+1\right)\left(x-4\right)-5\left(x^2-16\right)\)
\(\Leftrightarrow96x+x+4-3x^2-12x=2x^2-7x-4-5x^2+80\)
\(\Leftrightarrow92x=72\Leftrightarrow x=\dfrac{72}{92}=\dfrac{18}{23}\)(tm)