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14 tháng 6 2022
a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)
\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)
\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)
c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)
d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)
5 tháng 12 2017
1)⇔x2+1x-3x+3=0
⇔x(x+1)-3(x+1)=0
⇔(x+1)(x-3)=0
⇔x+1=0 hoặc x-3=0
⇔x=-1 hoặc x=3
5 tháng 12 2017
4)⇔x(1+5x)=0
⇔x=0 hoặc 1+5x=0
⇔x=0 hoặc 5x=-1
⇔x=0 hoặc x=-0.2
H
12 tháng 12 2018
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
19 tháng 6 2019
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right),\left(x+4\right)\)
KT
1
3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
1 tháng 1 2018
a)\(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}:\dfrac{2\left(x-3\right)}{3\left(x+1\right)}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)
\(=\dfrac{-\left(x-3\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)
\(=\dfrac{-\left(9+3x+x^2\right)3}{10}\)
b)\(4x^2-16:\dfrac{3x+6}{7x-2}\)
\(=4\left(x^2-4\right):\dfrac{3\left(x+2\right)}{7x-2}\)
\(=4\left(x-2\right)\left(x+2\right)\cdot\dfrac{7x-2}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)
c)\(\dfrac{3x^3+3}{x-1}:x^2-x+1\)
\(=\dfrac{3\left(x^3+1\right)}{x-1}:x^2-x+1\)
\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{x-1}\cdot\dfrac{1}{x^2-x+1}\)
\(=\dfrac{3\left(x+1\right)}{x-1}\)
d)\(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)
\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{-\left(x-1\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)
\(=\dfrac{-2\left(1+x+x^2\right)}{2x+3y}\)
HN
1 tháng 1 2018
a) \(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)
\(=\dfrac{27-x^3}{5x+5}.\dfrac{3x+3}{2x-6}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}.\dfrac{3\left(x+1\right)}{2\left(x-3\right)}\)
\(=-\dfrac{3\left(x-3\right)\left(x^2+3x+9\right)\left(x+1\right)}{10\left(x+1\right)\left(x-3\right)}\)
\(=-\dfrac{3\left(x^2+3x+9\right)}{10}\)
b) \(4x^2-16:\dfrac{3x+6}{7x-2}\)
\(=4x^2-16.\dfrac{7x-2}{3x+6}\)
\(=\dfrac{4\left(x^2-4\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)
c) \(\dfrac{3x^3+3}{x-1}:x^2-x+1\)
\(=\dfrac{3x^3+3}{x-1}.\dfrac{1}{x^2-x+1}\)
\(=\dfrac{3\left(x^3+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3\left(x+1\right)}{x-1}\)
d) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
\(=\dfrac{4x+6y}{x-1}.\dfrac{1-x^3}{4x^2+12xy+9y^2}\)
\(=\dfrac{2\left(2x+3y\right)\left(1-x\right)\left(1+x+x^2\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)
\(=-\dfrac{2\left(2x+3y\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)
\(=-\dfrac{2\left(x^2+x+1\right)}{2x+3y}\)
14 tháng 1 2018
\(a,\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{2}{3};-1;\dfrac{1}{2}\right\}\)
\(b,\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x^2\right)=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)\left(1+x\right)-\left(1-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1-x-1-x-x-3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(-3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\-3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-1\right\}\)
\(c,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\-5x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{7}{5}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-2;\dfrac{7}{5}\right\}\)
\(d,x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
Vậy phương trình có nghiệm duy nhất x = -1
\(e,x^3-7x+6=0\)
\(\Leftrightarrow x^3-4x-3x+6=0\)
\(\Leftrightarrow x\left(x^2-4\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;2;-3\right\}\)
\(f,x^4-4x^3+12x-9=0\)
\(\Leftrightarrow\left(x^4-9\right)-\left(4x^3-12x\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)-4x\left(x^2+3\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x^2-3-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3>0\forall x\\x^2-4x-3>0\forall x\end{matrix}\right.\)
Vậy phương trình vô nghiệm
\(g,x^5-5x^3+4x=0\)
\(\Leftrightarrow x\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^4-4x^2-x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^2-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\) hoặc x = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\\x=-1\end{matrix}\right.\) hoặc x =0
Vậy tập nghiệm của pt \(S=\left\{0;1;-1;2;-2\right\}\)
\(h,x^4-4x^3+3x^2+4x-4=0\)
\(\Leftrightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Leftrightarrow\left(x^4-x^2\right)-\left(4x^3-4x\right)+\left(4x^2-4\right)=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4x\left(x^2-1\right)+4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{1;-1;2\right\}\)
5 tháng 7 2018
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
5 tháng 7 2018
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
24 tháng 8 2018
\(A=x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1>1\)(dương)
\(B=x^2+4x+6=x^2+2.x.2+2^2+2=\left(x+2\right)^2+2>2\)(dương)
\(C=x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>\frac{3}{4}\)(dương)
\(D=x^2+x+1=x^2+2x\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>\frac{3}{4}\)(dương)
\(E=x^2+3x+3=x^2+2.x.\frac{3}{2}+\frac{9}{4}+\frac{3}{4}=\left(x+\frac{3}{4}\right)^2+\frac{3}{4}>\frac{3}{4}\)(dương)
Bạn làm tương tự nhé
\(\left(2x+1\right)^2=x^2\Leftrightarrow\left[{}\begin{matrix}2x+1=x\\2x+1=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(3x-4x^2+6-8x=x^2+4x+6\Leftrightarrow5x^2+9x=0\Leftrightarrow x=0;x=-\dfrac{9}{5}\)
đk : x khác 0 ; -1
\(\Rightarrow x^2+3x+x^2-x-2=2x\left(x+1\right)\Leftrightarrow2x-2=2x\left(voli\right)\)
Vậy pt vô nghiệm