(2X+1)2(X+1)(X+2)+2=0
Tìm x biết:
a)(x+3)^2+(x-2)(x+2)-2(x+1)=7
b)x(2x-1)-(x-2)(2x+3)=0
c)(x-1)(x+2)-x-2=0
d)x[(3x+2)+(x+1)^2-(2x-5)(2x+5)]=0
đ) 2x^2-7x+5=0
e) (2x+3)(x-5)=(2x+1)(2×+3)
chúc bạn học giỏi
a: \(\Leftrightarrow x^2+6x+9+x^2-4-2x-2=7\)
\(\Leftrightarrow2x^2+4x-4=0\)
\(\Leftrightarrow x^2+2x-2=0\)
\(\Leftrightarrow x^2+2x+1-3=0\)
\(\Leftrightarrow\left(x+1\right)^2=3\)
hay \(x\in\left\{-\sqrt{3}-1;\sqrt{3}-1\right\}\)
b: \(\Leftrightarrow2x^2-x-\left(2x^2+3x-4x-6\right)=0\)
\(\Leftrightarrow2x^2-x-2x^2+x+6=0\)
=>6=0(vô lý)
c: \(\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=0\)
=>x=-2 hoặc x=2
đ: \(\Rightarrow2x^2-2x-5x+5=0\)
=>(x-1)(2x-5)=0
=>x=1 hoặc x=5/2
Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0 d) x2.(x+1)-x.(x+1)+x.(x-1)=0
b) 9.(2x+1)2-4.(x+1)2=0 e) (x-2)2-(x-2).(x+2)=0
a, (\(x-2\))2 - (2\(x\) + 3)2 = 0
(\(x\) - 2 - 2\(x\) - 3)(\(x\) - 2 + 2\(x\) + 3) = 0
(-\(x\) - 5)(3\(x\) +1) = 0
\(\left[{}\begin{matrix}-x-5=0\\3x+1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\3x=-1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\) { -5;- \(\dfrac{1}{3}\)}
b, 9.(2\(x\) + 1)2 - 4.(\(x\) + 1)2 = 0
{3.(2\(x\) + 1) - 2.(\(x\) +1)}{ 3.(2\(x\) +1) + 2.(\(x\) +1)} = 0
(6\(x\) + 3 - 2\(x\) - 2)(6\(x\) + 3 + 2\(x\) + 2) = 0
(4\(x\) + 1)(8\(x\) + 5) =0
\(\left[{}\begin{matrix}4x+1=0\\8x+5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{5}{8}\end{matrix}\right.\)
S = { - \(\dfrac{5}{8}\); \(\dfrac{-1}{4}\)}
d, \(x^2\)(\(x\) + 1) - \(x\) (\(x+1\)) + \(x\)(\(x\) -1) = 0
\(x\left(x+1\right)\).(\(x\) - 1) + \(x\)(\(x\) -1) = 0
\(x\)(\(x\) -1)(\(x\) + 1 + 1) = 0
\(x\left(x-1\right)\left(x+2\right)\) = 0
\(\left[{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
S = { -2; 0; 1}
e, (\(x\) - 2)2- (\(x\) - 2)(\(x\) + 2) = 0
(\(x\) - 2)(\(x-2\) - \(x\) - 2) =0
-4 (\(x-2\)) = 0
\(x\) - 2 = 0
\(x\) = 2
S ={ 2}
Tìm x
(2x-3).(x+1)-2x^2+6x=0
(X^2-x+1).(x-3)-x^3+4x^2=0
(X^2-2).(x^2+2)-x^4-2x+5=0
(X-3).(x^2-3x+2)-(x^2-2x-7).(x-2)+2x^2-2x=0
( 2x - 3 )( x + 1 ) - 2x2 + 6x = 0
<=> 2x2 - x - 3 - 2x2 + 6x = 0
<=> 5x - 3 = 0
<=> 5x = 3
<=> x = 3/5
( x2 - x + 1 )( x - 3 ) - x3 + 4x2 = 0
<=> x3 - 4x2 + 4x - 3 - x3 + 4x2 = 0
<=> 4x - 3 = 0
<=> 4x = 3
<=> x = 3/4
( x2 - 2 )( x2 + 2 ) - x4 - 2x + 5 = 0
<=> ( x2 )2 - 4 - x4 - 2x + 5 = 0
<=> x4 + 1 - x4 - 2x = 0
<=> 1 - 2x = 0
<=> 2x = 1
<=> x = 1/2
( x - 3 )( x2 - 3x + 2 ) - ( x2 - 2x - 7 )( x - 2 ) + 2x2 - 2x = 0
<=> x3 - 6x2 + 11x - 6 - ( x3 - 4x2 - 3x + 14 ) + 2x2 - 2x = 0
<=> x3 - 6x2 + 11x - 6 - x3 + 4x2 + 3x - 14 + 2x2 - 2x = 0
<=> 12x - 20 = 0
<=> 12x = 20
<=> x = 20/12 = 5/3
a, \(\left(2x-3\right)\left(x+1\right)-2x^2+6x=0\)
\(\Leftrightarrow2x^2+2x-3x-3-2x^2+6x=0\Leftrightarrow5x-3=0\Leftrightarrow x=\frac{3}{5}\)
b, \(\left(x^2-x+1\right)\left(x-3\right)-x^3+4x^2=0\)
\(\Leftrightarrow x^3-3x^2-x^2+3x+x-3-x^3+4x^2=0\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)
c ; d tương tự nhé !
tìm x: part 1 : a,(x^3)^2-(x+1)(x-1)=1 b,(x-2)^2-3(x-2)=0 c,(x+2)(x^2-2x+4)-x(x^2+2)=15 d,(x+1)^2-(x+1)(x-2)=0 e,4x(x-2017)-x+2017=0 f,(x+4)^2-16=0 part 2: a,x^3+27+(x+3)(x-9)=0 b,(2x-1)^2-4x^2+1=0 c,2(x-3)+x^2-3x=0 d,x^2-2x+1=6x-6 e,x^3-9x=0
Tìm x
1.x^2-2x-1=0
2.x^2-x-1=0
3.x^2+x-3=0
4.4x^2-4x-1=0
5.4x^2-2x-1=0
6.4x^2-x-1=0
7.2x^2-2x-3=0
8.3x^2+3x-1=0
1)x^2-2x-1=0
<=> (x^2-2x+1)-2=0
<=>(x-1)2 =2
=>x-1 = \(\pm\sqrt{2}\)
=> x= \(\pm\sqrt{2}\) +1
2) x^2-x-1=0
<=> (x^2-x+1/4) -5/4=0
<=>(x+1/2)2= 5/4
=> x+1/2 = \(\pm\sqrt{\dfrac{5}{4}}\)
=>x=\(\pm\sqrt{\dfrac{5}{4}}\) - 1/2
3)x^2+x-3=0
<=> (x^2 + x + 1/4) -13/4=0
<=>(x+1/2)2 = 13/4
=> x+1/2 = \(\sqrt{\dfrac{13}{4}}\)
=> x= \(\sqrt{\dfrac{13}{4}}\) -1/2
4) 4x^2-4x-1=0
<=> (4x^2-4x+1)-2=0
<=>(2x-1)2 -2=0
<=> (2x-1)2 - \(\left(\sqrt{2}\right)^2\) =0
<=> (2x-1 - \(\sqrt{2}\) ) . (2x-1 +\(\sqrt{2}\) )=0
=> 2x-1-\(\sqrt{2}\) =0 hoặc 2x-1+\(\sqrt{2}\) =0
=> 2x= 1+\(\sqrt{2}\) hoặc 2x= 1 - \(\sqrt{2}\)
=> x=\(\dfrac{1+\sqrt{2}}{2}\) hoặc x=\(\dfrac{1-\sqrt{2}}{2}\)
Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0 d) x2.(x+1)-x.(x+1)+x.(x-1)=0
b) 9.(2x+1)2-4.(x+1)2=0 e) (x-2)2-(x-2).(x+2)=0
c) x3-6x2+9x=0 g) x4-2x2+1=0
h) 4x2+y2-20x-2y+26=0 i) x2-2x+5+y2-4y=0
1) Tìm x và y biết
a) (2x+1)^2 + y^2 = 0
b) x^2 +2x+1+(y-1)^2 = 0
c) x^2 - 2x+y^2 + 45y + 5 = 0
2) Tìm x biết
a) x(5-2x) - 2x(1-x) = 15
b) (x-3)^2 - 16+0
c) (2x-1)^2 + (x+3)^2- 5(x+7)(x-7) = 0
1) Tìm x và y biết
a) (2x+1)2 + y2 = 0
Ta có : \(\left(2x+1\right)^2\ge0;y^2\ge0\)
\(\Rightarrow\left(2x+1\right)^2+y^2\ge0\)
Để \(\left(2x+1\right)^2+y^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=0\end{matrix}\right.\)
b) x2 + 2x + 1 + (y-1)2 = 0
\(\Rightarrow\left(x+1\right)^2+\left(y-1\right)^2=0\)
Lập luận tương tự câu a ,ta có :
\(\left(x+1\right)^2+\left(y-1\right)^2\ge0\)
\(\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
c) x2 - 2x + y2 + 4y + 5 = 0
\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
Lập luận tương tự 2 câu trên
\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0
b) 9.(2x+1)2-4.(x+1)2=0
c) x3-6x2+9x=0
d) x2.(x+1)-x.(x+1)+x.(x-1)=0
a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)
\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)
Do \(\left(x+1\right)^2+1>0\)
\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
1) 2x+6=0
2)(x^2-2x+1)-4=0
3)x-2/x+2+3/x-2=x^2-11/x^2-4
4)x(x^2-1)=0
5)4x+20=0
6)x+3/x+1+x-2/x=2
7)1+2x-5/6=3-x/4
8x+2/x-2-1/x=2/x^2-2x
9)2(x+1)=5x-7
1)\(2x+6=0\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
Vậy : x=3 là nghiệm PT
2)\(\left(x^2-2x+1\right)-4=0\)
\(\Leftrightarrow\left(x-1\right)^2=4\)
\(\Leftrightarrow\hept{\begin{cases}x-1=2\\x-1=-2\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)
Vậy:....
3)\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)
\(\Rightarrow\left(x-2\right)^2+3\left(x+2\right)=x^2-11\)
\(\Leftrightarrow x^2-4x+4+3x+6-x^2+11=0\)
\(\Leftrightarrow-x+21=0\)
\(\Leftrightarrow-x=-21\)
\(\Leftrightarrow x=21\)
Vậy:......
4) \(x\left(x^2-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x^2-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x^2=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy:........
5)\(4x+20=0\)
\(\Leftrightarrow4x=-20\)
\(\Leftrightarrow x=-5\)
Vậy:...
6)\(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)
\(\Rightarrow x\left(x+3\right)+\left(x+1\right)\left(x-2\right)=2x\left(x+1\right)\)
\(\Leftrightarrow x^2+3x+x^2-2x+x-2-2x^2-2x=0\)
\(\Leftrightarrow-2=0\)(vô lí)
Vậy : PT vô nghiệm
7)\(\frac{1+2x-5}{6}=\frac{3-x}{4}\)
\(\Leftrightarrow\frac{-4+2x}{6}=\frac{3-x}{4}\)
\(\Rightarrow2\left(-4+2x\right)=3\left(3-x\right)\)
\(\Leftrightarrow-8+4x-9+3x=0\)
\(\Leftrightarrow-17+7x=0\)
\(\Leftrightarrow7x=17\)
\(\Leftrightarrow x=\frac{17}{7}\)
8) Làm tương tự
9) \(2\left(x+1\right)=5x-7\)
\(\Leftrightarrow2x+2-5x+7=0\)
\(\Leftrightarrow-3x+9=0\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
#H
1.\(2x+6=0\)
\(\Leftrightarrow2\left(x+3\right)=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của PT là \(S=\left\{3\right\}\)
2.\(\left(x^2-2x+1\right)-4=0\)
\(\Leftrightarrow\left(x-1\right)^2-4=0\)
\(\Leftrightarrow\left(x-1-2\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy tập nghiệm của PT là \(S=\left\{3;-1\right\}\)
3.\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)
ĐKXĐ :\(x\ne\pm2\)
Ta có ; \(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{x^2-4x+4+3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{x^2-x+10}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-x+10=x^2-11\)
\(\Leftrightarrow21-x=0\)
\(\Leftrightarrow x=21\)(Thỏa mãn ĐKXĐ)
Vậy tập nghiệm của PT là \(S=\left\{21\right\}\)
4.\(x\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=0\)
hoặc \(x-1=0\)
hoặc \(x+1=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
Vậy tập nghiệm của PT là \(S=\left\{0;\pm1\right\}\)
5.\(4x+20=0\)
\(\Leftrightarrow4\left(x+5\right)=0\)
\(\Leftrightarrow x+5=0\)
\(\Leftrightarrow x=-5\)
Vậy tập nghiệm của PT là \(S=\left\{-5\right\}\)
6.\(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)
ĐKXĐ : \(x\notin\left\{-1;0\right\}\)
Ta có : \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)
\(\Leftrightarrow\frac{x\left(x+3\right)}{x\left(x+1\right)}+\frac{\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=\frac{2x\left(x+1\right)}{x\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2+3x+x^2-x-2}{x\left(x+1\right)}=\frac{2x^2+2x}{x\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2+2x-2}{x\left(x+1\right)}=\frac{2x^2+2x}{x\left(x+1\right)}\)
\(\Rightarrow2x^2+2x-2=2x^2+2x\)
\(\Leftrightarrow0x=2\)(Vô lí)
Vậy PT vô nghiệm
7.\(1+\frac{2x-5}{6}=\frac{3-x}{4}\)
\(\Leftrightarrow\frac{12}{12}+\frac{2\left(2x-5\right)}{12}=\frac{3\left(3-x\right)}{12}\)
\(\Leftrightarrow\frac{12+4x-10}{12}=\frac{9-3x}{12}\)
\(\Leftrightarrow\frac{4x+2}{12}=\frac{9-3x}{12}\)
\(\Rightarrow4x+2=9-3x\)
\(\Leftrightarrow7x=7\)
\(\Leftrightarrow x=1\)
Vậy tập nghiệm của PT là \(S=\left\{1\right\}\)
8.\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
ĐKXĐ : \(x\notin\left\{0;2\right\}\)
Ta có : \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\frac{x^2+2x-x+2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\frac{x^2+x+2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+x+2=2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)(Không thỏa mãn ĐKXĐ)_(Thỏa mãn ĐKXĐ)
Vậy tập nghiệm của PT là \(S=\left\{-1\right\}\)
9.\(2\left(x+1\right)=5x-7\)
\(\Leftrightarrow2x+2=5x-7\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của PT là \(S=\left\{3\right\}\)
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^