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20 tháng 10 2021
a: \(\left(x-4\right)^2-\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^2-8x+16-x^2+9=5\)
\(\Leftrightarrow-8x=-20\)
hay \(x=\dfrac{5}{2}\)
25 tháng 8 2021
a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)
\(\Leftrightarrow9x+7=16\)
\(\Leftrightarrow9x=9\)
hay x=1
ND
Nguyễn Đức Trí
VIP
8 tháng 7 2023
a) P(x)=4x2-6x+a; Q(x)=x-3
Lấy P(x):Q(x)=4x-6 dư a+30
Vậy để P(x)⋮Q(x) ⇒ a+30=0 ⇒ a=-30
b) P(x)=2x2+x+a; Q(x)=x+3
Lấy P(x):Q(x)=2x-7 dư a+21
Vậy để P(x)⋮Q(x) ⇒ a+21=0 ⇒ a=-21
c) P(x)=x3+ax2-4; Q(x)=x2+4x+4
Lấy P(x):Q(x)=x+a-4 dư -4(a-5)x+12
Vậy để P(x)⋮Q(x) ⇒ -4(a-5)x+12=0 ⇒ (a-5)x=3
⇒ a-5 ϵ {-1;1;-3;3} (a ϵ Z)
⇒ a ϵ {4;6;2;8}
d) P(x)=2x2+ax+1; Q(x)=x-3
Lấy P(x):Q(x)=2x+a+6 dư 3a+19
Vậy để P(x)⋮Q(x) ⇒ 3a+19=0 ⇒ a=-19/3
e) P(x)=ax5+5x4-9; Q(x)=x-1
Lấy P(x):Q(x)=ax4+(a-5)x3+(a-5)x2+(a-5)x+1 dư a-4
Vậy để P(x)⋮Q(x) ⇒ a-4=0 ⇒ a=4
f) P(x)=6x3-x2-23x+a; Q(x)=2x+3
Lấy P(x):Q(x)=3x2-5x-4 dư a+12
Vậy để P(x)⋮Q(x) ⇒ a+12=0 ⇒ a=-12
g) P(x)=x3-6x2+ax-6 Q(x)=x-2
Lấy P(x):Q(x)=x2-2x+a-4 dư 2(a-4)-6
Vậy để P(x)⋮Q(x) ⇒ 2(a-4)-6=0 ⇒ a=7
Bài h có a,b bạn xem lại đề
12 tháng 7 2019
\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x^2+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+2x+x+2-\left(x^2-2x-x+2\right)=2x^2+4\)
\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)
\(\Leftrightarrow6x=2x^2+4\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
12 tháng 7 2019
\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)
\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)
13 tháng 8 2021
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)
\(\Leftrightarrow39x=-34\)
hay \(x=-\dfrac{34}{39}\)
b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)
8 tháng 1 2022
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^4-10x^3+6x^2\)
c: \(=-x^5+2x^3-\dfrac{3}{2}x^2\)
d: \(=2x^3+10x^2-8x-x^2-5x+4=2x^3+9x^2-13x+4\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 9 2023
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
a)
\(\left(x-3\right)\left(x+3\right)-2x=2x\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=2x-2x\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow x-3=0\) hoặc \(x+3=0\)
\(\Leftrightarrow x=3\) \(x=-3\)